Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer
@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. – TheBirdistheWord 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 – TheBirdistheWord Dec 26 '11 at 20:44
@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. – TheBirdistheWord Dec 26 '11 at 23:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.