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"Let G be a finite group. For a subgroup $H \subset G$ defing the normalizer $N_G(H) \subset G$. Show that the normalizer is a subgroup, that $H \unlhd N_G(H)$ and that the number of subgroups $H'$ conjugate to $H$ in $G$ is equal to the index of $|G:N_G(H)|$ of the normalizer".

For the normalizer, I have the definition as the biggest subgroup $\supset H$, such that$H \unlhd$ in it: $H \unlhd N_G(H)$. What does this exactly mean though? Is it basically the biggest normal subgroup in G?

Also, I don't understand how I would show the other stuff.

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    Maximal wrt set inclusing: for any $\,H\leq N\leq G\,$ s.t. $\,H\triangleleft N\,$ , then $\,N\leq N_G(H)\,$2012-12-08

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Make the group $\,G\,$ act on the set $\,X:=\{K\;\;;\;\ K\leq G\}\,$ by conjugation. Thus, by the orbit-stabilizer theorem:

$|\mathcal Orb(H)|=[G:Stab(H)]$

but $\,\mathcal Orb(H)\,$ is just the set of all subgroups of $\,G\,$ conjugate to $\,H\,$ , and $\,Stab(H)\,$ is just $\,N_G(H)\,$, so...

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Given a subgroup $H$, there are possibly a bunch of intermediate subgroups $K$ lying in between $H$ and $G$. $H$ has to be normal in at least one of these, since it's normal in itself. So we can keep going up the chain of subgroups until we arrive at a "largest" subgroup that $H$ is normal inside. Now, that subgroup need not be normal in $G$. It's also not always the largest normal subgroup of $G$ because the largest normal subgroup in $G$ is just $G$!.

Now, in order to show the number of conjugates is equal to the index of the normalizer, I would start by writing out the conjugates of $H$: $\{H, g_1Hg_1^{-1}, \ldots, g_nHg_n^{-1}\}$. Now try and show that $\{N(H), g_1N(H), \ldots, g_nN(H)\}$ are precisely the left cosets of $N(H)$. That is, you need to show no two cosets in that list are equal, and that every coset appears on that list. Now you just use the fact that $|G : N(H)|$ is by definition the number of cosets.

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    I wasn't trying to give a precise explanation, but just the idea. Since the group is finite, we can certainly pass up the chain of subgroups in this way. Perhaps the tricky part with this approach is showing that a "maximal" subgroup in the chain actually exists. There's no reason a priori that I must get a unique normal subgroup of largest order using this method.2012-12-08