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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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There are many groups which have subgroups that are not normal. The normalizer of a sbgp. $\,H\,$ is a subgroup (i) containing $\,H\,$ and (ii) in which $\,H\,$ is normal, and this normalizer sbgp. is the maximal one wrt these two properties. Note that the normalizer itself is NOT, in general, normal in the big group. –  DonAntonio Dec 8 '12 at 16:39
    
So G has a subgroup $G_1$. Within this subgroup, there is another subgroup H which is normal to $G_1$. Therefore $G_1$ is the normaliser of G? –  Kaish Dec 8 '12 at 16:57
    
If it is the maximal such one, yes. –  DonAntonio Dec 8 '12 at 17:03
    
What do you mean by "maximal" one? The one with the most elements? –  Kaish Dec 8 '12 at 17:10
    
Maximal wrt set inclusing: for any $\,H\leq N\leq G\,$ s.t. $\,H\triangleleft N\,$ , then $\,N\leq N_G(H)\,$ –  DonAntonio Dec 8 '12 at 17:15
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2 Answers

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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The "Going up the chain of subgroups" can prove to be a rather difficult task if the cardinality of all subgroups "between" $\,H\,$ and $\,G\,$ is, say more than $\,\aleph_0\,$ . Kurosh, in his very interesting and important book, talks of this stuff, but it is far from being obvious or trivial. –  DonAntonio Dec 8 '12 at 16:54
    
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. –  Zach L. Dec 8 '12 at 17:04
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