# Factor Group Question

1. Aug 7, 2014

### PsychonautQQ

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

if K is normal in G and has index m, show g^m is an element of K for all g in G

Work (I haven't done much with proofs so bear with me):
|G/K| = |G| / |K| = m
|G| = x
|K| = y

g^m must be an element of G since m|x

if g^m is an element of G and K is normal to G then
(g^m)K = K(g^m) --> (g^m)K(g^m)^-1 = K for all g^m in G

is this work legit?

Last edited: Aug 7, 2014
2. Aug 7, 2014

### HallsofIvy

Is this what you meant to say? g to any integer power is an element of G because G is closed under multiplication.

This is pretty much the definition of "normal subgroup". But you haven't even said, much less proved, that g is in K.

3. Aug 7, 2014

### PsychonautQQ

Ahh okay yeah I don't know what I thought I was doing there lol. Help a noob out? Hint hint possibly?

4. Aug 7, 2014

### jbunniii

Hint: $g^m \in K$ if and only if $g^mK = K$.

5. Aug 9, 2014

### PsychonautQQ

I just saw your clue and will think about it and try to use it to help me figure the problem out, but meanwhile here is what I've done so far on my own.
|G| / |K| = m
|G| = x
|K| = y
x = my

G = {1,g,g^2.....,g^(x-1)} are all elements of G so
{1,g,g^2.....,g^(x-1)}^x are all elements of G so
{1,g,g^2.....,g^(x-1)}^my are all elements of G

is this going in the right direction? I'll start working on this problem again now with the hint you've given

Last edited: Aug 9, 2014
6. Aug 9, 2014

### jbunniii

I don't see where this is taking you. Try using my hint to reword the problem in terms of the group $G/K$, i.e. work with cosets of $K$ instead of individual elements of $G$.

7. Aug 9, 2014

### PsychonautQQ

Solution: There are m cosets of K in G. This means that the group G/K has m elements in it, thus any element of that group raised to the m power would equal the identity element, K. (Kg)^m = Kg^m = K

Is this correct? If it is then wow,, I overthinking this

8. Aug 9, 2014

### jbunniii

Yes, that's all there is to it.