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Let $G$ be a group. Let $M$ be a maximal subgroup of $G$. I have to solve the following questions:

(a) Prove that $M$ is normal if the center $Z$ of $G$ satisfies $Z\not\subseteq M$.

(b) Prove $G/M$ is a cyclic group of prime order if $M$ is a maximal normal subgroup of $G$.

I found the proof for (b) in here. My goal now is to look for a proof for (a) that does not use (b). I guess the correspondence theorem and the fact that center is not a maximal subgroup of $G$ may help to deal with (a). However I haven't gotten the answer.

Any help would be much appreciated.

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user
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1 Answers1

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a) Hint : The normalizer of $M$ contains $M$ and contains $Z$.

Clément Guérin
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  • As for your hint, we write $M \le N_G(M) \le G$. Since $M$ is maximal, we have $M=N_G(M)$, for otherwise we're done. How can I go further? – user Sep 14 '15 at 12:17
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    @user From $M\subseteq N_G(M)$ and $M$ maximal we get that $N_G(M)=M$ or $N_G(M)=G$ (i.e. $M$ normal). Furthermore it is clear that $Z\subseteq N_G(M)$ so if $Z$ is not included in $M$ then $N_G(M)\neq M$. – Clément Guérin Sep 14 '15 at 12:35
  • It's correct. Thanks a lot @ClémentGuérin. – user Sep 14 '15 at 12:53