4.) Describe all subgroups of Z under ordinary addition. Let...

4.) Describe all subgroups of Z under ordinary addition. Let b be a group element of infinite order. Describe all the subgroups of <b>. 5.) For any element a in any group (G, ?), prove that <a> is a subgroup of C(a) (the centralizer of a.) 6.) Let (G, ?) be a group of permutations on a set X. Let a ? X and define stab(a) = {? ? G | ?(a) = a}. The set stab(a) is called the stabilizer of a in G (since it consists of all permutations of G that leave a fixed.) Prove that stab(a) is a subgroup of G. 7.) Let C be a cyclic group with generator c. What other elements of C generates C? (Include the case in which C is infinite.)

Transcript

00:01 So we would like to describe subgroups of z.

00:04 A couple things.

00:04 First, subgroups of the integers under addition.

00:09 Let's see, let's let h be a subgroup of z.

00:14 We're going to notice here z is cyclic, so it is generated by the element 1.

00:26 And i will claim in general that h also has to be cyclic with some generator.

00:33 Those are of course all subgroups, right? i mean, these are cyclic subgroups, you just pick a generator for h, you're going to get a subgroup.

00:47 But to show that these are the only subgroups, let's do that.

00:52 Let's say that h is a subgroup of g, right? a subgroup of z.

00:59 In particular, it is a set, so that's going to be, say, 1 times n, 1 times m, 1 times k, four integers, n, m, and k.

01:13 I want to let you notice, by bazoo's theorem, the greatest common divisor of n, m, k, etc.

01:28 Is going to be an element of h.

01:32 You're going to be able to make some integer linear combination of each of these numbers in h that makes their greatest common divisor, by bazoo's theorem.

01:52 Which in particular, we'll also notice that what this means is that the group generated by the gcd is a subgroup of h.

02:07 Furthermore, by definition of the greatest common divisor, every element of h is some multiple of the gcd, which means that h is in fact equal to the subgroup generated by the gcd.

02:25 And as such, h must be cyclic.

02:28 The only subgroups of z are cyclic subgroups generated by some value a.

02:34 So like 2 or 2z is a subgroup, 3z is a subgroup, 19z is a subgroup.

02:42 They're all subgroups, they're all in fact isomorphic to the integers.

02:46 In fact, in general, for a finite cyclic group, it's not that hard to check that just all subgroups are cyclic in the same way.

02:58 And for an infinite cyclic subgroup, that's what we've shown.

03:05 Great, so for number five here, any element of any group, the cyclic group generated by a is a subgroup of c of a, the centralizer of a.

03:18 So real quickly, pull the thing up in here.

03:25 So we have a group g, and we have some a in g.

03:31 We'd like to show that the cyclic subgroup generated by a is a subgroup of the centralizer of a.

03:44 So we'll remind you here that the centralizer of a is all of those elements g, such that g a is equal to a g.

03:59 So all those that commute with a.

04:02 And the cyclic subgroup a, generated by a, is just a to the n for any natural n.

04:11 We'll note that a times a to the n is equal to a to the n plus one, which is equal to a to the n times a, by associativity, really.

04:21 And thus, any power of a commutes with a.

04:28 So any element of the cyclic group generated by a is in the centralizer qed.

04:35 Now here, g is now a group of permutations on a set x, so it's a finite group.

04:40 Let a be an element of x, and the stabilizer of a is all those permutations that fix a here.

04:50 That's the stabilizer.

04:52 Prove that stabilizer a is a subgroup of g.

04:55 Well, it's enough to show that it is closed under inversion and has zero in it...

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