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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.2012-12-08
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    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?2012-12-08
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    *If* it is the maximal such one, yes.2012-12-08
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    What do you mean by "maximal" one? The one with the most elements?2012-12-08
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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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