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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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What have you tried? –  lhf Dec 23 '11 at 11:47
    
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
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@BabakSorouh: Where are you getting that from? –  user641 Dec 23 '11 at 19:15

1 Answer 1

up vote 3 down vote accepted

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.

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@stefHi-I've been following this with interest. I am a beginner, so maybe I could pose a question. If H is maximal normal and abelian, then K, being normal and abelian is either in H or is equal to G. In the first case you're done. So is there any way to say that G is finite - along the lines that G can't be isomorphic to the integers since all subgroups are of the form nZ. Or if G were to be infinite, K could not equal G, because a normal abelian group (i.e. the integers) wouldn't have a finite normal abelian subgroup. Or maybe some other way to force either G to be finite or K to be in H. –  Andrew Dec 26 '11 at 18:11
    
@Andrew you can conclude that K is H or G only if $H \leq K \ $ –  WLOG Dec 26 '11 at 18:52
    
@StefThanks. Maybe you would please help me with what "maximal" means. I was thinking no normal abelian subgroups between H and G. Which would force K to be in H or equal to G. Regards, Andrew –  Andrew Dec 26 '11 at 20:44
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@Andrew If K is normal and abelian you can say that HK is normal but not that HK is abelian –  WLOG Dec 26 '11 at 23:17
    
@StefThanks again. –  Andrew Dec 26 '11 at 23:45

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