Proving g^m is an Element of K in G: Factor Group Question Homework

[PsychonautQQ]

Homework Statement

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?

 

[HallsofIvy]

Homework Statement

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

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

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

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

is this work legit?

 

[PsychonautQQ]

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

 

[jbunniii]

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

 

[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

 

[jbunniii]

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

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 $G$.

 

[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

 

[jbunniii]

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

Yes, that's all there is to it.