# Quotient groups related problem

1. Jun 1, 2010

### zcahana

1. The problem statement, all variables and given/known data

Let G be a finite group and N$$\triangleleft$$G such that |N| = n, and gcd(n,[G:N]) = 1.
Proof that if x$$^{n} = e$$ then x$$\in$$N.

2. Relevant equations

none.

3. The attempt at a solution

I defined |G| = m and and tried to find an integer which divides both n and m/n.
I went for some X which is not in N, for which X^n = e.
I defined o(X) = q and then q | n.
No luck showing that q | m/n.

Any other ideas?

Zvi

Last edited: Jun 1, 2010
2. Jun 1, 2010

### mrbohn1

This is not true! Take some subgroup H of the Klein group K of order 2. It is normal as K is abelian. K/H also has order 2, and so gcd(H,[K]) = 2. But there is certainly an element of order 2 in K that is not in H.

Did you mean to write that the gcd is 1?

3. Jun 1, 2010

### zcahana

Yes I did, sorry for the mistake...
*Corrected*

4. Jun 1, 2010

### mrbohn1

Here's a rough sketch:

Suppose x is not in N. Let M=<x>. If xn=e then M has order dividing n. |MN|=mn/|M$$\cap$$N|, where m is the order of M. MN is a subgroup of G by the second (or third, depending on your numbering!) isomorphism theorem. So mn/|M$$\cap$$N| divides |G|.

Now, |G| = tn, where t=[G:N] is coprime to n. So m/|M$$\cap$$N| must divide t. But this is impossible, as m divides n, and t is coprime to n. So x is in N.

As a disclaimer, it's a while since I did any group theory! So you should check all this carefully.

Last edited: Jun 1, 2010
5. Jun 1, 2010

### zcahana

Thanks for the response.

By stating that M$$\cap$$N = <1> did you mean M$$\cap$$N = e?
If so, it seems to me that it is not necessarily true.
If you meant something else, please elaborate.

Zvi.

6. Jun 1, 2010

### mrbohn1

hmmm..yes, sorry about that. I've modified the argument above. Still no guarantee it is right!

Yes, when I say <1> I mean e....itis just to distinguish the identity element from the trivial group.

7. Jun 1, 2010

### zcahana

Sorry... I still see a problem with the conclusion that mn divides |G|...

8. Jun 1, 2010

### mrbohn1

That was just an error with my previous editing. Hopefully it is fixed now.

If you spot another problem, try filling in the details yourself...it's the only way to learn! I have done this very much off the cuff, and intended it to be a "suggested method" rather than a full solution to be copied down.

9. Jun 1, 2010

### zcahana

Thanks for you help!...