Subgroups and Normal Subgroups

**leshields** · April 18th 2012, 08:37 AM

Suppose that $A$ and $B$ are subgroups of a group $G$ with $A\subseteq B$ . Prove that $[G:B]$ divides $[G:A]$

Let N be a normal subgroup of a group G and let H be a subgroup of G that contains N
i) Show that $H/N$ is a subgroup of $G/N$
ii) If $H/N \triangleleft G/N$ prove that $H \triangleleft G$

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 $|G|=|G:B||B|$ and $|G| = |G:A||A|$ then as $A\subseteq B$ are both subgroups $|B| = |B:A||A|$ so $|G:B||B:A||A| = |G:A||A| \Rightarrow |G:B||B:A| = |G:A|$ so that $[G:B]$ divides $[G:A]$

For i) as all $h \in H$ also $\in G$ clearly $H/N \subseteq G/N$

to show it has an inverse say $h, h' \in H$ such that $hh'=e$ then $hNh'N = hh'NN = e$ (can do this as $N$ is normal)
then following same argument for $h, k \in H, hk \in H$ also so $hNkN=hkN \in H/N$

For ii) as $H/N \triangleleft G/N$ we have $gNhN(gN)^{-1}=gNhNg^{-1}N=(ghg^{-1})N \in H/N$ thus $ghg^{-1} \in H$ so $H \triangleleft G$

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

**Deveno** · April 18th 2012, 12:34 PM

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^-1N) = (hk^-1N).

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

ii) this is good.

**leshields** · April 19th 2012, 06:03 AM

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

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