# Normalizer/Centralizer theorem

This is exercise 3.2.24 from Scott, Group Theory.

If $H$ is a finite maximal abelian normal subgroup of $G$ and $K$ is a normal abelian subgroup of $G$, then $K$ is finite.

The hint is to use Normalizer/Centralizer theorem.

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Hint: The fact that $H$ is MAXIMAL Abelian is crucial. –  Geoff Robinson Dec 23 '11 at 11:59
Note that $Aut(H)$ is a subgroup of $S_{|H|}$. –  Babak S. Dec 23 '11 at 13:46
I like giving hints too! What can you say about abelian normal subgroups of $C_G(H)$? –  user641 Dec 23 '11 at 17:44
@SteveD: $C_{G}(H)=G$? –  Babak S. Dec 23 '11 at 18:51
@BabakSorouh: Where are you getting that from? –  user641 Dec 23 '11 at 19:15

Since $H$ is finite, $Aut(H)$ is finite. By the Normalizer/Centralizer theorem, $\frac{N_{G}(H)}{C_{G}(H) }= \frac{G}{C_{G}(H)} \$ is isomorphic to a subgroup of $Aut(H)$ and so is finite. Now we note that if $M \ \trianglelefteq \ G \$, $M$ is abelian and $M \leq C_{G}(H) \$, then $HM$ is abelian and normal in $G$, but $H$ is maximal abelian normal so $HM\leq H \$ and then $M\leq H \$. Note that $K \cap C_{G}(H) \$ is abelian because $K$ is abelian and $K \cap C_{G}(H) \trianglelefteq G \$ because $K \trianglelefteq G \$ and $C_{G}(H) \trianglelefteq G$. Then $K \cap C_{G}(H) \subseteq H \$ and so $K \cap C_{G}(H) \$ is finite because $H$ is finite. $\frac{KC_{G}(H)}{C_{G}(H)} \simeq \frac{K}{K \cap C_{G}(H)} \$ is a subgroup of $\frac{G}{C_{G}(H)} \$ and so is finite. Then $|K| = |\frac{K}{K \cap C_{G}(H)}| \cdot |K \cap C_{G}(H)| \$ is finite.
@Andrew you can conclude that K is H or G only if $H \leq K \$ –  WLOG Dec 26 '11 at 18:52