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welcome to my abstract algebra exam 2 review problems and solutions video in this video we're going to go over definitions theorems calculations and proofs related to intermediate group theory that means we'll be looking at alternating groups symmetric groups cosets lagranges theorem homomorphisms isomorphisms we'll also prove something called the first isomorphism theorem at the very end of the video and we'll also do calculations related
0:27
to external direct products for example and look at some basic group classification problems the first problem is about normal subgroups let g be a group and let h be a subgroup of g it's going to be a special kind of subgroup called a normal subgroup first we want to state the definition the original definition of what it means for h to be a normal subgroup of g and then secondly state the normal subgroup test which is really a theorem
0:55
to verify that the given subgroup is in fact a normal subgroup so the definition starts this way
1:03
the same kind of thing that we have up here let g be a group and let h be a subgroup of g we say
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that h is normal in g
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if something extra happens if something extra special happens
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if a h the left coset of h in g containing the element a
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which i'll explain here in a second equals h a the right coset of h and g containing the element of a a if that's true for all little a in g okay what is this thing right there that's again the left coset of h in g containing a it is the set of all possible products of a times elements in h as h ranges over the subgroup capital h and this thing h a is called the right coset of h in g containing a it's the set of all products h times a where a is now on the right as little h varies
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over capital h when those are equal that's a special condition it certainly happens if g happens to be an abelian group but in general this special condition doesn't always happen when it happens that's a special kind of subgroup we say h is normal in g and notationally we also write h is not just a subgroup of g but normal we make this little triangle here to indicate that it's normal in g that's the definition and again this stuff right here goes at the front of the definition
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what is the normal subgroup test that really is a theorem
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a normal subgroup test so this was part one and this is part two of this problem
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we're once again going to let h be let g be a group and let h be a subgroup of g so that gets put down here as well then h is normal in g i'll go ahead and use the symbolism again
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if and only if i f stands for if and only if x h x inverse is a subset of h for all x in g
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okay which actually ends up meaning it actually equals that but it's sufficient to prove that it's just a subset what is this expression sometimes called the conjugate of h in g by x in general it's not necessarily well okay it's not necessarily going to be a sub
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set of h but when that special case happens then we say h is normal and g this is a theorem the normal subgroup test that can prove h is normal in g what is this thing it's the set of all products of the form x little h x inverse as little h varies over capital h okay this is usually easier to verify than the definition so it's worthwhile having a theorem problem two says fully state lagrange's theorem and at least in the textbook that i use by
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gallian it is in two sentences lagrange's theorem is sometimes called the fundamental theorem of finite group theory it's really really important it states that if g is a finite group
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and h is a subgroup not necessarily a normal subgroup but any subgroup
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h is a subgroup of g that's notationally written like this without a full triangle make it look like a less than sign h is a subgroup of g
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then the order of h the number of elements in h divides the order of g the number of elements in g
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in the book by galleon that's written the order of h like this divides
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the order of g written like that you could also write this with divides notation but then there's lots of vertical lines like this in some other books instead of using these absolute value kinds of things for the order of the group or the subgroup there's a little o used for the order you could write o of g o o of h excuse me divides o of g that's the first sentence of lagrange's theorem it gives you the possibilities for orders of
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subgroups the second sentence of the grandeur's theorem says how many left cosets of h and g there are which is also related to the concept of a factor group also called the quotient group
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uh let's say it like this moreover
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the number all right number like that number of distinct
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left cosets
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and also the number of distinct right cosets because they'll be the same number though they're not necessarily equal unless h happens to be normal
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of h in g is equal to the order of g divided by the order of h and again if your book uses o notation you could write that as o of g divided by o of h and if you want to say a little extra here galleon does not do this you could also say this is the same as the order of the factor group of g by h also called the quotient group of g by h whose elements in fact are left cosets okay my students in galleon's book chapter seven you don't have to write this last thing here you
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can stop right there okay that's the number of distinct left codes and also right cosets in problem three we are essentially using lagrange's theorem for a fairly trivial application
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let g be a group whose order is 42 lists the possible orders possible orders not guaranteed orders of subgroups of g the converse of lagrange's theorem is false just because the order of some number divides the order of g doesn't mean there's some subgroup of that order and we're going to actually look at a an example that illustrates that that's false later on so it only gives you possible orders we need to know the factors
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of 42 42 whose prime factorization is 2 times 3 times 7.
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what are those factors of 42 well don't forget the trivial ones like 1 and 42 in fact i think i'll group these vertically 1 and 42 2 and 21
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3 and 14.
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what comes next how about 6 and 7 and that will be it there are no other factors of 42 and so these are this is the answer these are the possible orders
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of subgroups of g because those are all the divisors positive divisors
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and by lagrange's theorem
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again it doesn't guarantee that for any one of these except the trivial ones in one one and 42 that there will be a subgroup of that order there is definitely always a subgroup of order one the trivial subgroup just containing the identity element often called e and there is always a subgroup whose order equals the order of the entire group the group itself the entire group itself number four about isomorphism decide whether u10 and u12 are
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isomorphic or not that's supposed to be an or there or not briefly explain your answer okay
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so what is an isomorphism an isomorphism between two groups from one group to another is a one-to-one and onto function there's also what's called operation preserving and in a nutshell what that means is these two groups if they're going to be isomorphic will be indistinguishable from each other as far as their group properties go their cayley tables can be put in correspondence with each other in particular that means they have the same number of elements
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it also implies more for example there are the same number of elements of order two if there are any elements of order two in these of order four for example if there are any elements order four etcetera u10 is the group of units modulo 10 meaning the positive integers less than 10 that are relatively prime to 10 1 3 7 and 9 under multiplication mod 10. u12 is the group of units um whose elements are relatively prime to 12. positive integers are relatively prime to 12
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under multiplication mod 12. its elements would be 1 5 7 and 11.
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they have the same number of elements they both have four elements it doesn't matter that the operations are different and it doesn't matter that the elements are different they still could be isomorphic however if you test out the orders of these elements and maybe test whether one of these groups is cyclic or not you will end up seeing they are not the same they are not isomorphic that is going to be the answer let's go ahead and figure out the orders of these things
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one of course in both groups is the identity and has order one three in u10 what is its order well it's not two because three squared is nine which is not the identity in u10 by lagrange's theorem that must mean 3 has order 4. this is going to be a cyclic group right here 3 cubed is 27 mod 10 is 7 also not equal to one three to the fourth must be back to the identity three to the fourth
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in ordinary arithmetic is 81 but then mod by 10 you get one the identity yes three has order four in u10 you get funny looking equations like this that's not the absolute value of 3 that's the order of 3 as an element in u10 and that means u10 is cyclic and 3 is a generator of u10
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i could write this another generator of u10 would be seven for example seven would end up having order four as well um yeah and nine one has order one nine would have order two nine squares 81 81 mod 10 is 1. the order of 9 is order 2. but this is enough to say u10 is cyclic
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evidently u12 is not cyclic one certainly has order one what about five five squared is 25 mod 12 is one five has order two what about seven seven squared is 49 mod 12 is also one right away you can now say these groups are not isomorphic both 5 and 7 have order 2 but only
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9 has order 2 in u10 just for extra emphasis 11 squared's 121 121 mod 12 is 1.
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all these elements non-identity elements have order 2 in u12 and that means u12 is not cyclic
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so there's various reasons you could give here you could say no
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for example
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u10 is cyclic
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but u12 is not
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you could also say u10 has just one element of order two again that's the element nine whereas u12 has three elements of order two that's also enough to say they're not isomorphic okay various kinds of reasons you could give that's the simplest thing to do in this case
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onto the second page
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how many elements of order four does this external direct product z4 z2 o plus z4 have by the way with external direct products not every book and or professor uses this plus sign with a an o around it called o plus uh sometimes they'll use a cross a time symbol for an external direct product but in galleon's book this is what they use as a set this is the set of all ordered pairs a comma b where the first component or coordinate
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if you prefer of the ordered pair is from the first factor there in the external direct product
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and the second component or coordinate is from the second factor in the external direct product
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and since there are two choices for a either 0 or 1 and 4 choices for b either 0 1 2 or 3
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the group operation for each factor here is addition mod 2 in the first case in mod 4 in the second case there are altogether eight elements in this group the order of the group z2 external direct product with z4 is 2 times 4 is eight so um
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there are most eight elements of order four and in fact any element of course generates a cyclic subgroup generated by that element and by lagrange's theorem that's going to mean the only possible orders of elements in here are the divisors of eight one two four and eight turns out because two and four are not relatively prime that this external direct product is not isomorphic to z z8 the cyclic group of order eight so there will be no elements of order eight
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this is all extra stuff i'm telling you come back here to the question at hand we wanna have we're after elements of order four the other fact that you need is that the order of an element in an external direct product this ordered pair is the least common multiple when you're dealing with finite groups of the order of the first component and the order of the second component we want that for this problem to equal four we want to choose a and b so that that happens
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a is either zero one and has order either one or two
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and b is either zero one two or three and has order either one two or four so we want this lcm to be four there's only two possible ways that can happen that implies either the order of a is one and the order of b is four or the order of a is two and the order of b is four again
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there's only one way to choose an element from z2 of order 1 make it be the identity 0 in this case again the operation is addition mod 2 for the first factor
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and there's only there's two ways to make elements of order 4 from z4 the elements 1 and 3 have order 4 under addition mod 4. in the second case here the order of a is 2 that means a is the element um 1. again the operation is addition hearing out multiplication 1 plus 1 one is not zero and z two but one plus one is zero one has order two in z two
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once again b would have to be one or three in each case there are two possibilities two possibilities
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two possibilities in total giving four elements two plus two equals four elements of order four
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okay all together and you can even list them out zero zero um excuse me 0 1 0 3
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1 1 and 1 3 would be the four elements of order four in this external direct product
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number six let h and k be subgroups of g
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g as a group order of h is eight order of k is twelve and the order of their intersection is two is a general fact as an extra common here that when h and k are subgroups of g that their intersection will also be a subgroup of g
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question is how many elements are in this set h k which equals the set of all products little h times little k as little h ranges over the subgroup capital h and little k ranges over the subgroup capital k now that notice i call this a set in general h k this set is not a subgroup of g it will be if h and k happen to be normal subgroups for example but in general it's not
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show the necessary calculation to get the final answer for how many elements are in this set you could call it the order of the set but that's a could be confusing because again it's not necessarily a sub group for my students when i give this kind of problem i'm expecting they know this from a fact in galleon's book that we do talk about the order of this set the number of elements in the set turns out to equal through a proof that's about a paragraph long the order of
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the subgroup h times the order of the subgroup k i could put a data in there for times divided by the order of their intersection h intersect k okay that's a fact that can be proved it's true even if h k is not a subgroup of g it's true even if h and k themselves are not normal subgroups of g we are assuming finite answers here and we're given that the order of eight h is eight and the order of k is 12 and the order of h intersect k is two so this ends up being ultimately
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8 divided by 2 is 4 4 times 12 is 48 and that is the answer for the number of elements in this set which again in general is not a subgroup but could be if for example h and k are normal subgroups of g it will it will be in that case
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number seven there's a long description of something going on here but it's actually a fairly short problem as far as what is needed to be done but it's worth reading through and thinking about based on things you should have learned if you're at this point in an abstract algebra class given a group g and a normal subgroup h of g like i said before written that way we have defined the factor group also called quotient group
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in some books galleon's book uses factor group of g by h sometimes said g mod h but it's not modular arithmetic as follows as a set the elements
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are the left cosets of h and g these expressions of the form a h which by definition is the set of all products of little a times little h where little h ranges over capital h
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those are the elements of the factory group you could write this the group operation is coset multiplication defined by this product the goal of this problem is to see why this is true or why this is what excuse me why this is called well defined well defined right there in the proof that this factor group is a group this the hardest part is to show that the this cult closet multiplication ah times b h the left coset of h and g containing a times the left coset of h
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and g containing b is this coset a b times h the left cosine of h and g containing a b
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why is that an issue what do you mean by well-defined i mean if you used different coset representatives would you get the same answer ah could equal for example
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a prime times h where a prime is different than a okay that could happen or it won't necessarily
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a and a prime in that case would be representatives for the same coset of h and g
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do we get the same answer in the end if i assume this is true and this is true is it does it follow that this is true if it is then this operation is well defined if i've got different coset representatives here and here do i get the same coset in the end that's a big issue
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below is a string of equalities that proves the operation is well defined
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just as equalities without explanations also based on the fact that if a h equals a prime h and bh equals b prime h then you can say there's elements little h1 and little h2 of capital h so that a prime equals a times h1 and b prime equals b times h2 that's something to know about when cosets are equal why does it work well for example with the first one here
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a certainly well okay let's write it think of it this way we're looking at this equation here a prime is certainly an element of a prime h because e is an h h is a subgroup of g the identity element e is an h a prime is certainly in a prime h since a prime equals a prime times e
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but we're saying
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a prime h that left coset equals ah this left coset
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and that would mean a prime also equals little a times something else in h
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in other words a prime equals a prime e also equals a times some other element of h
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and you could say something similar between b prime and and b your task is to identify circle the two places where the string of inequalities here where the normality of h and g is used h does need to be a normal subgroup of g okay that's necessary for this to work so let's look at these equalities
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cosets a prime b prime h how do i go from here to here i'm just using that property and this property if you i'm replacing a prime by a h1 and b prime by bh2 so that's not using the normality
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it's using the equality the assumed equality of these cosets next step ah1 bh2 capital h equals ah1bh what am i doing there i am really letting the h2 go away how because it gets sucked into h
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h2 h would equal h once again for a similar kind of reason is up here h2 does equal h2 times e
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that allows it to get sucked in so to speak now i've got a h1b equal times capital h equals a h1b capital h b i switch the left coset to a right coset well kind of
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like i got a combination of left and right cosets here this is justified it turns out the bh got switched to an hb think back to the previous page in the definition of normality that was the definition of the normality there is one spot where it's used okay put a big star there
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move from here to here what happened the h1 goes away it got sucked into h
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and here to here i switch around the a uh the the hb and the bh excuse me once again there is where normality is being used again bh equals hb and hb right here equals bh those are the two spots gal i'm mimicking galleons proof here i'm doing it on a and a coset level this proof of
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the um multiplication being well defined you can also do this at a more element level i'll leave that as an exercise for you a bunch of true false questions before we move on to some calculations
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this is based on some knowledge
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true or false a finite abelian group g with the whose order is 147 has an element of order seven
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first of all we should probably double check and see if lagrange's theorem might be violated here an element of order seven is going to generate a subgroup of order seven how do you factor 147 is seven divisor seven is a divisor it's seven times 21 in fact ultimately it's seven times that's three times seven squared seven is a divisor and here's the key seven is also prime
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there's a theorem called koshi's theorem
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since seven divides 147 and seven is prime and seven is prime that would say yes it does have an element of order seven and it would also have a subgroup a cyclic subgroup of order seven okay koshi's theorem to this point in intermediate group theory requires the assumption that the group is a billion
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in order to prove it early in the book koshi's theorem actually can be generalized to non-abelian finite groups it still is true but the proof is harder okay i'll just put not necessary in general
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that word's supposed to be necessary but earlier in the book in gallian's book you need that assumption if you're going to prove it more easily but koshi's theorem does apply to non-abelian finite groups it's true in either case you got a prime that divides the order of the the order of the group it's going to there's going to be an element of that order and also a cyclic subgroup of that order
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more factual knowledge if g is a group whose order is 26 then g is isomorphic z26 26 is 2 times 13.
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there's a theorem about classifying groups of order 2p where p is a prime bigger than 2.
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and the answer is false it's not going to be necessarily cyclic but there are only two possibilities up to isomorphism either that the group is cyclic like z26 in this case or that the group is isomorphic to dp in this case d13 the dihedral group of order 26 which corresponds to what would be symmetries of a a regular 13 gone which are rotations and reflections i'm not going to try drawing anything there
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the theorem says it's going to be isomorphic to one of those when your order is two times a prime where p here the prime is bigger than two so that's false
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ten if p is a prime and the order of g is p then g is isomorphic dp the dihedral group of order 2p that's definitely false if i had zp there instead of dp then it would actually be true up to isomorphism if the order of g is prime it's got to be cyclic it's got to be isomorphic to zp the integers from 0 to p minus 1 under addition mod p though the operation of g itself might not be addition 11. so this would be true if i had a zp there instead of a dp
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11. if g is a non-abelian group then this factor group of g by its center is not cyclic this is something just to factually no this is true it turns out there's something called the gz theorem
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which is a theorem so it's also true which is essentially this exact same thing except as it's contrapositive
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it says if g mod zg the factor group of g by its center which is the set of all elements in g that commute with everything in g is cyclic
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then g is a billion
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a billion that's what i'm trying to write there and the statement in number 11 is the contra positive that if g is not a billion then this factor group is not cyclic and so if this is true then this is true as well it's kind of interesting a little theorem if this factor group is cyclic then g is a billion but that would actually imply more that would imply that the center is the entire group and it would also imply that the factor group must be trivial
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factor groups like this are never non-trivial cyclic groups trivial meaning just one element and that one element is the generator so kind of a weird thing that way but this is still true
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12 is also related to the idea of a contrapositive if g and g bar are groups and the group of automorphisms of g is not isomorphic to the group of automorphisms of g bar then g is not isomorphic to g bar that's the contrapositive of the statement the implication that if g is isomorphic to g bar then otg is isomorphic to ought g-bar that is an easier implication to prove since this one's the contrapositive of this one
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i've proved that one this one is true as well because a contrapositive means the same thing what this ultimately means is ought sometimes called a functor is um an invariant for group isomorphisms if you've got two isomorphic groups their groups of automorphisms are isomorphic so more usefully if the groups of automorphisms are not isomorphic then the original groups are also not isomorphic
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more knowledge here 13 ought of z8 the group of automorphisms of z8 by the way automorphisms are isomorphisms from the group to itself and the so the elements of this are isomorphisms of z8 to z8 and the func of the group operation is function composition this is isomorphic to u8 and so now the question is is u8 cyclic okay that's just a fact to know well u8
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has four elements one three five and seven um positive integers less than eight that are relatively prime to eight is the cyclic you could just do brief computations three squared is nine mod eight is one so three has ordered two five squared's 25 mod 8 is 1 yeah this can't be cyclic this is going to be false 7 squared is 49 mod 8 is also 1. 3 5 and 7 all have order 2.
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no no element has order 4. this is not cyclic therefore this is not cyclic because it's isomorphic to this the answer is false 14 the external direct product of z2 and z5 is cyclic that's true 2 and 5 are relatively prime i talked about that briefly back up here
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that i mentioned that this is not going to be cyclic because 2 and 4 are not relatively prime
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not going to bother coming up with the generator this would have 10 elements in it and that would also mean it turns out four generators but i'll let you figure that out if you want 8 and 14 are not relatively prime so this one's false
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onto page three
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recall that z of d6 the center of d6 the dihedral group of order 12 which would be symmetries
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of a regular hexagon
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consists of r0 which is the identity rotate by 0 degrees and r180
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rotation by 180 degrees is going to be a symmetry of a regular hexagon
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d6 itself has 12 elements 6 rotations if you think of our zeros being a rotation and six reflections reflecting about lines of symmetry the question is what is the order of the elements r 60 z d6 this is a left coset a left coset our 60 is an element in d6 zd6 that's the center is a subgroup of d6 what's the order of that element as an element of the factor group of d6 by its center
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it's not one this is not the identity because r60 is not in the center hopefully pretty quickly you can see the answer has got to be 3. r 60 z d 6 squared by the definition of coset multiplication from the previous page talking about the group operation and the factory group being well defined you've got to multiply the representative times itself or in this case compose it with itself
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and r60 composed with itself is r120 r120 is also not in zd6 this is not the identity element
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of this factor group
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but if you cube the original element here compose it with itself with three factors so to speak you are going to get back to the identity
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because r60 cubed would be r180
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and r180 is an element of the center this does equal z d6 which is the identity element of the factor group therefore the order that we are after is 3. 3 is the smallest positive integer such that when you raise this element to that power you get back to the identity element of the group
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17 list all the possible abelian groups up to isomorphism
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of order 27. in addition for each of these groups determine the number of elements order three
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okay this is a classification theorem related to finite ability in groups
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z27 cyclic group of order 27 certainly is one answer elements are 0 through 26 group operation is edition mod 27 an example of a generator is the number one the group operation is addition mod 27. that's cyclic that's one answer turns out this external direct product is another answer up to isomorphism and since 3 and 9 are not relatively prime it's not isomorphic to z27 this is not cyclic but it does have 27 elements and final answer if you think about the factorizations of 27
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in terms of products powers of 3 27 is 3 times 9 it's also 3 times 3 times 3. the last possibility is the external direct product of z3 with itself with 3 factors
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number of elements
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of order three
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in z27 what's that going to be
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well it is a cyclic group you could refer to the fundamental theorem of cyclic groups of finite cycle groups it's going to going to have one subgroup of each order that divides the order of the entire group 27 here and that means there's going to be one subgroup of order three
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the converse lagrange's theorem does apply to cyclic groups finite cyclic groups
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one subgroup order three and the subgroups of order three have two generators so there's going to be two elements worth of three three what are they nine and eighteen if you think about it nine plus nine is eighteen not equal to zero mod twenty-seven 9 plus 9 plus 9 is 27 which does equal 0 mod 27 18 plus 18 is 36 mod 27 is 9. 18 plus 18 plus 18 is 57 no excuse me 54. mod 27 is zero the identity element is zero so there's two elements of order
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three in z27 how about in here it's good to think about the fact that the order of an ordered pair is once again the least common multiple the orders of the components and we want that to equal three
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how can that happen it can happen if the order of a is one and the order of b is three
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it could happen if the order of a is three and the order of b is one and it could happen if the order of a and b are both three now b can also have order nine but we're not thinking about that
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in each case there are two elements that fit these equations two two two giving us a total of six elements of order three what's one example
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if i think think of this case one example
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would be for example a's and z3 it could be one or two b is in z nine has order three it could be three or six so one example would be for example uh one six one three two six those are all examples that fit this but overall there are six elements of order three how about this one
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now we are talking about ordered triples for elements the order of that ordered triple or the group operation is defined component wise i forgot to emphasize that before is the least common multiple of the orders of their groups of the elements excuse me in the components
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and i want that to equal 3 and actually that's always going to be the case except when a b and c have order 1. a b and c all have to either have order 1 or 3 and the only way this will
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not equal 3 is when these all have order one when they're all zeros
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in other words all of the non identity elements in here have order three
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the answer is the number of elements of order 3 is 27 minus 1 is 26.
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okay but you also can think about this and think about cases like we did up here as well
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number 18.
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let g be a group with order 21. if g has only one subgroup of order three and only one subgroup of order seven prove our first proof prove that g is cyclic
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okay so let's go ahead and write a careful proof of this in sentences let's think intuitively first
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you could say let hbs out the unique subgroup over 3 and k be the unique subgroup of order of 7.
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by lagrange's theorem the only possible orders of subgroups
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are 1 3 7 and 21.
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are there any elements of order 21 i hope it's pretty clear that there there should be just take anything that's not in h and not in k and it would have to have order 21 because it would generate a subgroup of order 3 or 7 or 21 and but h and k use up all the elements of orders 3 and 7.
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that's the intuition so let's see here how should we write this as a proof
43:43
let so here's our proof let h be the unique subgroup
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of order 3 and k be the unique subgroup of order seven
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since uh
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let's see the order of g is 21 the order of h is three and the order of k is seven there would exists an element of g call it a
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such that a is not in h or k it's not in their union now be careful
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h union k is not necessarily a subgroup of g but as a as an element in a set it's not necessarily in it's it's not going to be in their union we can find some element that's not in their union
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there are what would it be here 21 minus elements of order three and h are two of them
45:13
two generators cyclic subgroup order three three is prime there would be 6 elements of order 7 in k
45:23
and we also want to subtract off the identity element 20 21 minus 9 or 12 elements
45:35
in g but not in this union you could write in g set minus h union k
45:44
12 12 elements to choose from but i only have to choose one of them
45:50
by lagrange's theorem
45:58
the order of a which is the same as the order of the cyclic subgroup generated by a
46:04
divides 21 so the order of a is 1 3 7 or 21.
46:16
but by uniqueness of h and k uniqueness
46:23
a could not have order 3 or 7 because then it would generate a cyclic subgroup of order three or seven and that would have to either be h or k but a is not in h or k
46:37
a is also not the identity element
46:41
by elimination you could say
46:45
what we've always already said the order of a must then be 21 and done qed
46:54
this would also prove there's a subgroup of order 21 and that means geocyclic i guess i forgot the final conclusion i should say therefore
47:03
g is generated by a so g is cyclic i forgot what the final conclusion was there
47:12
excuse me okay let's move on
47:29
here we come to our promised verification that the converse of lagrange's theorem is false
47:35
a long-looking problem but it's actually fairly easy when you go through it let's just fill in the blank
47:43
the group a4 alternating group we haven't talked about the alternating group or symmetric group yet
47:50
s4 would be the called the symmetric group of order for which has four factorial or 24 elements it's excuse me all the permutations on the set 1 2 3 4. a4 is a subgroup of s4 it's got 4 factorial divided by 2 or 12 elements it's all the even permutations in s4
48:16
not going into a ton of detail about what permutations are i'm assuming you've dealt with them and look at the proof we are dealing with them in cycle notation
48:27
it can be used to prove that the converse of lagrange's theorem is false in particular a4 which has order 12 has no subgroup of order six despite the fact that six divides twelve below our steps to prove this fact your task is to fill in the blanks
48:46
okay all right let's go through it here prove that a4 has no subgroup order six first note that any three cycle a b c where a b and c are all numbers one two three or four i'm not thinking of this as being a group just thinking of it as being a set and elements like three cycles like this permute
49:09
these numbers they're one to one and onto functions from this set to itself
49:14
a would this notation means a maps to b b maps to c and c maps back to a i'm assuming you know that where a b and c are all distinct numbers from the set should emphasize they are all distinct this is in a4 because we can write it like that it's in a4 as a product of two
49:36
two cycles and because it's an even number of two cycles that's why it's an even permutation despite the fact that it's as a cycle it's an odd cycle it's an even permutation because you can write it as an even number of two cycles what does this mean these are functions remember and you work from right to left this one says a maps to b but then this one says b maps to itself so ultimately in the composition right to left a maps to b which this notation means how about b this
50:05
first permutation maps b to a then this one maps a to c so ultimately b gets mapped to c with c this one keeps it fixed maps c to itself this one maps c to a so ultimately c gets mapped back to a
50:21
that's a factorization of this three cycle into the product of an even number of two cycles and that's why three cycles are even permutations and that's what the answer to this is
50:35
when you write permutations in cycle notation there are various ways to write the same permutation all three of these things mean the same thing a maps to b whether you think of it here or here or here b maps to c like that or like this or like that and c maps to a
50:56
c maps a over here by going backwards maps to a here like that and maps to a here like that
51:02
these each three cycles has three ways of representing it representing them and because of that how many three cycles are there the number of three cycles is four times three times two because there's four choices for a three choices for b and two choices for c from this set but you got to divide that by three because for each three cycles there's three ways of doing it and that does equal eight and so a four has eight elements of order
51:29
three three cycles have order three they're even permutations but they have order three
51:37
each of these has order three the order of such a permutation
51:44
is three you you gotta cube this thing to get back to the identity
51:49
here's our argument though suppose the contrary it's an argument by contradiction that there exists a subgroup h of a4 with order six we've got to get a contradiction out of this we claim that h contains all the elements of order three and since there's 8 of them
52:06
8 elements of order 3 that'll contradict the assumption that h has order 6.
52:14
and we'll be done to prove this claim let a be an arbitrary element of order three don't get confused with the a here maybe i should have used a different letter like alpha it's not the same as an a from the set a here is a permutation for example a three cycle
52:33
arbitrary element of order three what i'm trying to show is that it's an h
52:37
if i assume for sake of contradiction that it's not an h i must have to get a contradiction
52:43
then h in the left coset of h and a and g containing a is are distinct it's property of cosets if a is not in h these must be distinct
52:56
and their intersection is moreover their intersection is empty property of cosets
53:03
and the index of a4 in h i forgot to use index notation when i did lagrange's theorem i think i wanted to but i forgot to
53:15
in other words the number of left cosets of h in a4 by lagrange's theorem is 2 12 divided by 6.
53:25
and um that'll mean a4 can be written as h union ah
53:40
i guess i'm not really using the fact that a is not an h to say that but i am using this fact here
53:46
trying to get a contradiction consider a squared by the conclusion of the preceding paragraph we know either a squared is in h or ah h and a h form a um
54:06
partition of g of a4
54:11
their intersection is empty and their union is the entire thing so any element in a4 has to be in one or the other and actually you could even say not both if a squared is in the first one then by closure of h if you square that you'll also stay in h by closure because h is a sub group of a4
54:32
and since a has order three a to the fourth will equal a
54:37
that's where we use the fact that a has ordered 3.
54:41
in other words a is an h but wait a minute i assumed a was not an h
54:47
up here
54:49
so we get a contradiction
54:53
if on the other hand a squared is not an h but in a h then a squared will equal a times little h for some little h in capital h and you cancel the a's and say ultimately a equals h
55:10
but that contradicts the fact that i'm assuming a is not an h
55:14
so it's kind of convoluted it seems but it really is accurate a valid argument
55:23
therefore the since the claim is proven and h the sub order six contains all elements order three that's absurd no such subgroup can exist 20 symmetric group
55:37
let gbs6 symmetric group of order 6 which really has 6 factorial or 720 elements
55:45
per day find the number of elements of order g of ng of order six that are six cycles
55:54
six cycles would have the form a b c d e f where these numbers are all distinct and they're from the set one two three four five six a gets mapped to b b gets mapped to c c gets mapped to d d gets mapped to e e gets mapped to f and f gets mapped back to a
56:10
one to one and on two functions from the set of numbers one two three four five six to itself how do you figure out how many elements of this form there are
56:20
well there's six choices for a five choices for b they have to be all distinct four choices for c three choices for d two choices for e and one choice for f but that's not the answer that's the number of elements in s6 since each one of these six cycles can be represented in six different ways you gotta divide this by six instead of the a being first i could have the b being first or i could have the c first or the d or the e or the
56:49
f you've got six ways of doing each of those so altogether this is giving you 120 elements that are six cycles and those six cycles would have order six as an element of the group
57:03
b find the number of elements of g of order six or order six that are products of two disjoint cycles
57:14
and this is kind of tricky we also want to remember that disjoint cycles commute with each other i can put in either direction compose them either direction
57:23
this joint cycles means in cycle notation they don't have any numbers in column in common
57:29
how are you going to get it to have order six
57:34
when you multiply or compose two disjoint cycles they're gonna have order equal to the least common multiple of the orders of each cycle individually it's not the same thing as the external direct product fact we're talking about s6 here no external direct product adult
57:55
so if i'm going to get order six one of them is going to have to have order two and one's gonna have to have order three say it looks like this
58:06
there's a two cycle it's got order two is an element of s6 there's a three cycle it's got order three is an element of s6 i'm composing these the order of this is going to be two times three or the least common multiple of two and three which is two times three or six
58:25
you've got six choices for a five choices for b these numbers have to be different you've got four for c three for d and two for e
58:34
we gotta divide there's two ways to write this two cycle here either a first then b or b first then a so i gotta divide by two
58:43
there's also three ways to write a three cycle either c d e
58:48
d e c or e c d those all mean the same thing so i've got to divide by 3 as well
58:56
do i need to divide by anything else that may not be clear here i need to think about it
59:05
it's a little tricky if these cycles have the same length if they were
59:10
uh two three cycles the order wouldn't be six anymore but switching them around um would not change the form of these things as a two cycle first then a three cycle as i have here
59:25
it would be indistinguishable is what i'm trying to say if i switch these around put the cde over here and the a b over here i have not created a new permutation because these are disjoint cycles they commute it would be the same permutation so the upshot of what i'm trying to say is you do not need to for example divide by two because switching these around would not change anything
59:49
this is the answer and it ultimately equals 6 times 5 times 4
59:55
30 times 4 is 120 120 elements of order six that are product of two disjoined cycles
1:00:05
part c using your answers from parts a and b
1:00:11
find the number of distinct cyclic subgroups of s6 of order six
1:00:17
cyclic subgroups of order six are generated by elements of order six
1:00:23
and there's always two such element elements for every any given cyclic subgroup of order six that's something you should know with working with cyclic groups
1:00:37
first of all the number of elements of order six
1:00:44
it's got to be the sum of these two things 120 plus 120 is 240.
1:00:51
but we're after the number of distinct cyclic subgroups of s6 of order six
1:00:58
for each element of oro6 it generates a cyclic subgroup of order six but
1:01:04
any such cyclic subgroup has two generators
1:01:08
each of which has order 6. and so therefore the answer to the number of cyclic subgroups
1:01:19
of order 6 is 240 divide by two number of generators for any cyclic sub group order six 120 cyclic subgroups of order six so the answer to each product part in this problem is the same 120 in each case the way uh s6 could have non-cyclic subgroups of order six there are not every group of order six is cyclic
1:01:49
there are plenty of non-cyclic subgroups or six and they're all isomorphic to s3 they're not equal to s3 but they're isomorphic to s3 i'll let you think about that also isomorphic to d3
1:02:04
i'm getting close here two more pages
1:02:11
the very last page is a proof of the first isomorphism theorem
1:02:15
this page has got calculations with you groups
1:02:19
here's a fact from chapter 8 i probably shouldn't have said that in this video this is from galleon
1:02:28
galleon's book g is this u-group group of units in this case positive integers less than 64 that are relatively prime to 64.
1:02:40
under multiplication mod 64 multiplication not addition use the fact from gallium's chapter 8 to find positive integers m n such that u 64 is isomorphic to z m external direct product with zn
1:02:57
that's a memorization thing okay in general um
1:03:06
u of 2 to the n
1:03:09
is isomorphic to the external direct product of uh it's z sub 2 to the n minus 2 with z2 it turns out that's a memorized fact
1:03:25
uh and this is true for all n greater than or equal to 2 yeah
1:03:33
all integers are n greater than or equal to 2. therefore in our case n is 6. therefore u 64
1:03:41
which is u 2 to the 6th is isomorphic to the external direct product of z sub 2 to the 4th with z2 which is not cyclic because 16 and 2 are not relatively prime that's just a factor remember
1:04:02
part b says use your answer from part a to state how many elements g has it's not so obvious thinking about u64 but once you think of it as isomorphic to this you can say u64 has order
1:04:20
equal to 16 times 2 32.
1:04:27
it's a non-cyclic group of order 32.
1:04:32
c is g g cyclic no just set it
1:04:39
since z16 external direct product with z2 is not cyclic because 16 and 2 are not relatively prime the gcd of 16 and 2 is 2 which is not 1 they're not relatively prime
1:04:59
d use your previous work to determine how many elements of order 16 that g has
1:05:09
order 16
1:05:13
element i can think of in terms of z six this external direct product and i can think in terms of this fact
1:05:26
i want this to equal 16. that's the want how do i choose order of a in order of b if i'm thinking of b as being from z2 it's either 0 1 and a as being from z16
1:05:43
i could either have a has order 16 and b has order 1 or a has order 16 and b has order two
1:05:54
for b in each case there's only one element in this case b is zero in this case b is one under addition mod two in z two so the question is how many choices are there for a how many positive integers less than 16 in fact are relatively prime to 16.
1:06:14
not negative integers starting with zero i should i want to think about the generators of of z16
1:06:21
16 is 2 to the 4th generators of z16
1:06:28
are going to be
1:06:31
numbers less than 16 relatively prime to 16 i that's going to be all the odd numbers
1:06:40
one two three four five six seven eight of them so in each case you get eight elements
1:06:49
altogether giving you eight plus eight or sixteen elements of order sixteen
1:07:02
in z16 an external direct product with z2 so also in the u64 because it's isomorphic to that
1:07:13
e let h be this subgroup the cyclic subgroup of g which is you u64 sorry
1:07:22
you're trusting me that that is a subgroup and in fact a normal subgroup which it would have to be because g is a billion do calculations to find the order of 3h in the factory group
1:07:34
by lagrange's theorem
1:07:38
um
1:07:42
let's see here we can say the order of this factor group g by h is the order of g divided by the order of h which is going to be 32 divided by what's the order of h here 1 2 3 4 5 6 7 8. 32 divided by 8 is 4 and therefore the answer has got to be 1 2 or 4.
1:08:05
now 3 h is not equal to h because 3 is not an h 3h does not equal h since 3 is not in h and what that means is 3h is not the identity element of the factor group 3h does not have order 1 not equal to 1.
1:08:32
does it have order 2 square it you get 9h is that h go back up here 9 is not in h
1:08:48
and therefore 3h does not have order 2. and you could then right away say by lagrange's theorem has got to have order 4.
1:09:01
since the factor group has order 4 and 1 2 and 4 are the only divisors of four
1:09:06
you could also very verify it computationally
1:09:15
3h cubed there's 27 h is not equal to h because 27 is not in here
1:09:27
but 3 to the 3 h to the 4th 81 h do need to mod 81 by 64. that's the same as 17 h
1:09:40
that does equal h because 17 is in h
1:09:45
we are back to the identity element of the factor group that's another way to see ultimately that the order of 3h in the factor group is 4.
1:09:55
22 let phi be a homomorphism from u15 this group of order eight
1:10:03
to itself whose kernel consists of the cyclic subgroup generated by four
1:10:09
under multiplication mod 15 4 squared 16 mod 15 is one that's a cyclic subgroup of order two in this group of order eight if phi of seven equals seven find all elements of u15 that map to seven
1:10:27
based on a fundamental fact about homomorphisms the inverse image or pre-image of any element
1:10:36
is in the domain um the left coset of the subgroup of the kernel
1:10:55
now in general this is a little funny since 7 is getting mapped to itself in general what you've got here is if p is mapping g to g bar
1:11:07
and you're looking at the inverse image or pre-image of the element little g bar
1:11:15
the answer would be that it's the left coset of the kernel containing g
1:11:23
where phi of g maps to g bar equals g bar
1:11:31
the g and the g bar are the same here they're both seven so it's a little weird that way this is like the g bar here and this is like the g
1:11:40
but phi of g equals g bar but that is the answer so what i need to do is figure out what that coset is the kernel is consist of one and four so this coset consists of seven and seven times 4 mod 15
1:11:57
7 and 28 mod 15 is 13. and that would be the answer those are the elements
1:12:05
in the domain that get mapped to 7. it's not a subgroup of u15 it's a left coset
1:12:12
that's the way homomorphisms and kernels relate in this context
1:12:17
last problem is a proof of the first isomorphism theorem
1:12:24
ideally you'd write this in sentences but for sake of time i'm just going to give you the idea of the proof what it says let fee from g to g bar be a group homomorphism i haven't said what homomorphism is it's a mapping from one group to another that's operation preserving
1:12:53
phi of g1 times g2 is phi of g1 times phi of g2
1:13:01
isomorphisms are homomorphisms that are also one to one and onto
1:13:07
the goal is to prove that the mapping that i call fibar galleon doesn't call it that but i call a fee bar from this factor group to the image of g under phi this is uh the set of all outputs phi of little g as little g ranges over capital g and it does not necessarily equal g bar this is not necessarily onto defined by this formula is an isomorphism okay there's four steps to do
1:13:40
step one show this mapping is well-defined the well-defined issue came up earlier with factor group multiplication it's coming up now in defining this proposed homomorphism isomorphism in fact iso so it's going to be not just operation preserving but 101 and onto step two is to prove it's one to one step three is to show it's on to not g bar itself but the image fee of g
1:14:06
if you assumed fee was on two to begin with then this fee of g would equal g bar and sometimes the first isomorphism theorem is stated assuming fee is on to and for show it's operation preserving so again i'm just going to give you the ideas of these things not full details
1:14:27
step one
1:14:31
so it's well defined what a well defined mean with
1:14:34
factor groups that meant if a h equal to a prime h and bh equal b prime h
1:14:41
that you get the same answer when you multiply cosets abh equals a prime b prime h
1:14:49
to show this fee bar function is well defined you're trying to show it's a function
1:14:55
if you had two representatives of the same coset you get the same output
1:15:01
if you assume phi of let me call it g1 curve fee
1:15:08
equals phi of g2 curfee what am i trying to show um excuse me that's that's one that's the one to one if i assume g1 curve fee equals g2 curve fee
1:15:27
does it prove that i get the same answer in the end v bar of those things is the same thing it should that's what it means to be well defined
1:15:40
these are these are left cosets to say they're equal is equivalent to implying for example
1:15:47
for example that if i pretend i'm multiplying both sides of this on the left by g1 inverse
1:15:56
that this is true
1:16:00
these are properties of cosets and that's equivalent
1:16:04
in fact i'm going to go ahead and do steps one and two at the same time here and i'll just explain them the left to right direction is going to be step one the right to left direction is going to be step two that's one to one these are equivalent they mean the same thing so the implication goes both ways and that's equivalent to g1 inverse g2 being an element of the curve granola fee
1:16:33
now you could just jump from here to here that's allowed okay but i did this intermediate step and what is that equivalent to it it's equivalent to fee mapping those things that product to the identity element of g bar call it e bar
1:16:52
by the fact that phi is operation preserving phi is op that's really the key step that's equivalent to phi of g1 inverse phi of g2 equals e bar
1:17:06
it's like the fee factors through not literally it's not a multiplication but it's kind of like that you can go from here to here and you also can say
1:17:18
by the fact that phi is operation preserving that fee of g1 inverse is phi of g1 inverse in g bar
1:17:29
what's my goal in all this my goal is to get 2 showing this was well defined and therefore fee of g1 equals phi of g2 and i'm almost there because i just need to now multiply both sides of this equation on the left by phi of g1 and i'll get phi of g2 equals phi of g1 which is the same thing as phi of g1 equals phi of g2 so the left to right implications here starting by assuming you've got two different representatives of the same coset
1:17:57
showing that you get the same answer in the end is enough to say this is well defined that it makes sense the right to left implications are enough to show fee bars one to one i should maybe add one more step here this is means fee bar of g2 curfee equals fee bar g1 curfee
1:18:21
go right to left for one to one assume fee bar of these inputs are the same and try to show the
1:18:29
inputs themselves are the same and all these implications all these things are equivalencies two-way implications we can go from right to left in each step
1:18:42
and that would prove fee is one to one
1:18:47
step three
1:18:53
show fee is on to its image by the way after i finish this proof i'm going to draw kind of a fun picture that people like to draw on the situation so hang on for that show feedball is on to it's image well fee is on to its image by definition and that's going to apply fee bars onto
1:19:13
let little g bar be in
1:19:18
this direct image of g under phi which is a subgroup of g bar
1:19:28
we're trying to show there's something in the factor group
1:19:33
that gets mapped to the given g bar such that phi of g equals g bar here
1:19:40
well you just got to use the fact that phi is onto its image therefore there exists a little g in capital g such that g bar equals phi of little g that's what it means for that to be true
1:19:57
and that's the same g that you can use for this left coset here therefore phi bar of g curve e
1:20:07
by the formula for v bar right there
1:20:12
is phi of g but that is g bar done that's enough to say feet bars onto
1:20:20
not all of g bar but onto the image of g under phi
1:20:27
if v were itself or onto that would equal g bar step four
1:20:37
chauffe is operation preserving isomorphisms are homomorphisms so give me two arbitrary left cosets
1:20:53
like this multiply them and plug them into phi bar do i get the bar of the first one times february the second one that's analogous to this thing up here
1:21:08
well you just gotta plug away how do you multiply left cosets you multiply the representatives
1:21:17
that's just coset multiplication
1:21:24
now the only thing you can do is just use the formula for b fee bar that's a lot of what you do in this class is just do the only thing you can do but you got to know what you you should do before you do it use the formula for feedback that's all you can do here this formula here using fee applied to the coset representative this equals phi of g1 g2
1:21:48
that's the definition of fee bar
1:21:52
but now hey i must have to use the fact that phi is assumed to be a homomorphism phi itself is assumed to be operation preserving this is fee of g1 fee of g2 but hey
1:22:05
that's the same thing i'd get if i applied fee bar to the corresponding left cosets
1:22:13
and that means i'm done
1:22:19
that's the definition of p bar once again okay the diagram i want to make to close out the video
1:22:30
is in this context if you've got g mapping to its image which is a subgroup of some other group g bar uh the way people think about defining the map fee bar here with this formula they think visually in turn like this they imagine mapping g bar onto this factor group
1:22:56
which is what the domain of p bar is
1:23:01
how by taking an arbitrary element little g in capital g and mapping it to
1:23:09
its left coset the left coset of the kernel containing little g
1:23:17
sometimes called a projection mapping sometimes labeled with the greek letter pi but that's not the number pi it's a mapping symbol in this case and they call v bar the induced map
1:23:33
taking this element and mapping it to phi of g and this diagram is said to commute
1:23:43
you draw three lines there to say the diagram commutes
1:23:50
and what that means is phi itself can be factored so to speak as the composition of fee bar with pi um it doesn't really prove much in and of itself but it helps you understand how fee and fee bar are related i think i got everything right here okay and again here was the statement of the first isomorphism theorem if you want to look at that again thanks for watching
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