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Thread: Subgroups and Normal Subgroups

  1. #1
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    Subgroups and Normal Subgroups

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

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

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

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

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

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

    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
    Last edited by leshields; Apr 18th 2012 at 09:11 AM.
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  2. #2
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    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-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.
    Thanks from leshields
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  3. #3
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    Re: Subgroups and Normal Subgroups

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