Homework: Show that G has a normal subgroup of order nm

[gorcee]

EDIT: There is a new question posted below: viewtopic.php?f=17&t=80765#p2889951

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I am stuck on the following homework problem:

Determine whether K = \left\{ \rho_0, \left(1\ 2\right)\left(3\ 4\right), \left(1\ 3\right)\left(2\ 4\right), \left(1\ 4\right)\left(2\ 3\right)\right\} is normal in S_4.

I know that a subgroup is normal if one of the following properties holds:
- gK = Kg for all g in S_4
- gKg^{-1} \subset K for all g in S_4
- gKg^{-1} = K for all g in S_4

I also know that the index of K in S_4 is 6: [S_4 : K ] = |S_4|/|K| = 4!/4 = 6.

However, I don't know how to show this, short of actually listing out the left and right cosets of K. Is there a better strategy?

Edit edit: no that's not right.

[++$_]

You can use one of the other criteria.

Hint: What are all the conjugates of (1 2)(3 4)?

[gorcee]

You can use one of the other criteria.

Hint: What are all the conjugates of (1 2)(3 4)?

I don't know what you mean by conjugates. I've not seen that term used in this context.

[++$_]

Sorry; the conjugates (in G) of an element k in G are the elements gkg^-1, where g ranges over G.

[gorcee]

Not sure that I can use a conjugation argument, we're only really able to use what has been previously built upon, and we haven't done anything with group actions and their relationship to normal subgroups.

[++$_]

Well, you don't need any advanced stuff here.

A subgroup K of G is normal iff all the conjugates of elements in K are also in K. (This is one of your criteria.)

So all you have to do is find out what the conjugates of the elements of K are.

The shortcut is that the set of conjugates in S_n of any permutation is easy to write down, so it is easy to check whether that is contained in K.

[gorcee]

Well, you don't need any advanced stuff here.

A subgroup K of G is normal iff all the conjugates of elements in K are also in K. (This is one of your criteria.)

So all you have to do is find out what the conjugates of the elements of K are.

The shortcut is that the set of conjugates in S_n of any permutation is easy to write down, so it is easy to check whether that is contained in K.

So... the conjugates of (1 2)(3 4) would be what?

(1 2) (1 2)(3 4) (1 2)
(1 3) (1 2)(3 4) (1 3)
...
(1 2 3) (1 2)(3 4) (1 2 3)?

[jestingrabbit]

The shortcut is that the set of conjugates in S_n of any permutation is easy to write down, so it is easy to check whether that is contained in K.

Yeah, this is what you want to focus on. If you work out what all the conjugates of (1 2)(3 4) are you should be able to work out what is going on.

[++$_]

So... the conjugates of (1 2)(3 4) would be what?

(1 2) (1 2)(3 4) (1 2)
(1 3) (1 2)(3 4) (1 3)
...
(1 2 3) (1 2)(3 4) (1 2 3)?

Er, I don't agree with that last line. What is the inverse of (1 2 3)?

[gorcee]

So... the conjugates of (1 2)(3 4) would be what?

(1 2) (1 2)(3 4) (1 2)
(1 3) (1 2)(3 4) (1 3)
...
(1 2 3) (1 2)(3 4) (1 2 3)?

Er, I don't agree with that last line. What is the inverse of (1 2 3)?

Typo, meant to write (1 3 2).

[++$_]

OK, then I agree, as long as the "..." conceals 21 lines.

Now write them as the product of disjoint cycles so that you can find out who they really are.

[gorcee]

I think I've pretty much got it, but there really isn't any way that doesn't involve computing at least a handful of products, right?

[++$_]

Well, once you have discovered the general principle that all the conjugates of an element in S_n have the same ____________ as the original element, then you don't have to actually do any computation.

[gorcee]

Well, once you have discovered the general principle that all the conjugates of an element in S_n have the same ____________ as the original element, then you don't have to actually do any computation.

The conjugates would have the same structure. So conjugates of (1 2)(3 4) would be a product of two disjoint 2-cycles, all of which are in K.

However, I'm not sure that I can presently prove that and use it as a solution =(

[++$_]

Well, it shouldn't be so hard to prove. If you can prove it for a single cycle, you can prove it for any permutation because if k₁ and k₂ are two disjoint cycles, then gk₁k₂g^-1 = (gk₁g^-1)(gk₂g^-1) (and these are disjoint cycles).

So you just have to check it for a single cycle.

[gorcee]

Well, it shouldn't be so hard to prove. If you can prove it for a single cycle, you can prove it for any permutation because if k₁ and k₂ are two disjoint cycles, then gk₁k₂g^-1 = (gk₁g^-1)(gk₂g^-1) (and these are disjoint cycles).

So you just have to check it for a single cycle.

I still don't see that at all.

Edit: I still don't necessarily see it, but I see how to use it to get to where I need to be for this problem.

[++$_]

Sorry, maybe I wasn't clear.

Basically, the key insight is that if g is any permutation and k = (a b c ... z), then gkg^-1 is not just any cycle of the appropriate length, but specifically (g(a) g(b) g(c) ... g(z)). This is easy to prove -- just do it element by element.

Then suppose that (a b c ... z) and (A B C ... Z) are disjoint cycles. Using the previous result, the conjugate by g of (a b c ... z)(A B C ... Z) is (g(a) g(b) ... g(z))(g(A) g(B) ... g(Z)). These are still disjoint cycles, so the cycle structure of the permutation was again unchanged.

Hence, by induction on the number of cycles of the permutation, the cycle structure of any permutation is preserved by conjugation.

[gorcee]

Sorry, maybe I wasn't clear.

Basically, the key insight is that if g is any permutation and k = (a b c ... z), then gkg^-1 is not just any cycle of the appropriate length, but specifically (g(a) g(b) g(c) ... g(z)). This is easy to prove -- just do it element by element.

Then suppose that (a b c ... z) and (A B C ... Z) are disjoint cycles. Using the previous result, the conjugate by g of (a b c ... z)(A B C ... Z) is (g(a) g(b) ... g(z))(g(A) g(B) ... g(Z)). These are still disjoint cycles, so the cycle structure of the permutation was again unchanged.

Hence, by induction on the number of cycles of the permutation, the cycle structure of any permutation is preserved by conjugation.

Alright, I see that now. Thanks

I kind of hackneyed an argument in an ad hoc case. But then I took another look at the problem, and kind of went durrrr. I've got it now.

Thanks for all your help. Spent way too much time on this one problem.

[gorcee]

Ok, new question.

Let G be a group with a unique subgroup of order n and a unique subgroup of order m, where n and m are relatively prime. Show that G has a normal subgroup of order nm.

What I've done so far:

H n K (intersection) must be trivial, because we know that K and H are both subgroups of G, so H n K is also a subgroup of H and K both. By Lagrange's Theorem, |H n K| must divide |K|, and |H n K| must divide |H|, but since gcd(|H|,|K|) = 1, then |H n K| = 1, so H n K = {e}.

Therefore, |HK| = |H||K|/|H n K| = nm/1 = nm.

But, I have to show that HK is a subgroup. To do that, I must either show that one of H or K is normal, or that HK = KH. I am stuck.

Wow, sometimes just typing out the question makes me see it better.

If H is a subgroup of G, we know that gHg^{-1} is a subgroup of G with order |H|. But H is the unique subgroup of order n, so gHg^{-1} = H, therefore H is normal.

[Talith]

Yep, you pretty much hit the nail on the head. The only very small thing you now have to show is that your HK is normal in G, but I doubt you will have any problem with that.