Quotient groups related problem

[zcahana]

Homework Statement

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

Homework Equations

none.

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?Thanks ahead,
Zvi

 

[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?

 

[zcahana]

Did you mean to write that the gcd is 1?

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

 

[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\capN|, 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\capN| divides |G|.

Now, |G| = tn, where t=[G:N] is coprime to n. So m/|M\capN| 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.

 

[zcahana]

Thanks for the response.

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

 

[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.

 

[zcahana]

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

 

[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.

 

[zcahana]

Thanks for you help!...