Subgroups and Normal Subgroups

Suppose that and are subgroups of a group with . Prove that divides

Let N be a normal subgroup of a group G and let H be a subgroup of G that contains N

i) Show that is a subgroup of

ii) If prove that

I know these should be relatively straightforward and I have some answers but think they're really messy and not sure if what I've done is viable!

For first part using LaGrange I have and then as are both subgroups so so that divides

For i) as all also clearly

to show it has an inverse say such that then (can do this as is normal)

then following same argument for also so

For ii) as we have thus so

If anyone can let me know if what I've put is correct and or whether I've missed crucial steps I'd really appreciate it!

Thanks

Re: Subgroups and Normal Subgroups

ok, for your first part, there is a bit of a wrinkle: we have no idea whether or not G is, in fact a finite group. Since the statement of the result you are asked to prove doesn't even make sense unless the index of A is finite, i will assume that [G:A] is finite, but i will make no assumptions as to whether [G:B] is finite, or whether |G| is finite.

So we have a finite number of cosets gA. note that every coset gA lies entirely within a coset of B, gB. since a finite number of such gA account for every element of G, there can only be at most [G:A] cosets of B (there could be fewer, because there's room in a coset of B for more than one coset of A, perhaps).

This shows that [G:B] is likewise finite (we proved this, see?).

now, let's consider cosets bA, where b is in B. The number of these is also bounded by [G:A], so [B:A] ≤ [G:A]...in particular, it is finite as well.

now [G:A] = [G:B][B:A] (THIS is actually what Lagrange's theorem says), therefore [G:B] divides [G:A].

i) this is OK. you could also just use the one-step test, let hN, kN be in H/N (so that h,k in H). then (hN)(kN)^{-1} = (hN)(k^{-1}N) = (hk^{-1}N).

since H is a subgroup hk^{-1} is in H, thus (hk^{-1})N is in H/N.

ii) this is good.

Re: Subgroups and Normal Subgroups

Thanks, yeah that makes sense. As usual, I'm on the right track but not rigorous enough!