# Defining the normalizer, showing its a subgroup and $|H| = |G:N_G(H)|$

"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.

• 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

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...
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.
• 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