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Let $G$ be a non-abelian finite group such that every proper subgroup of $G$ is abelian. Suppose $M$ is a maximal subgroup of $G$ which is not normal in $G$. I was asked to show that

$\bigcup_\limits{g \in G} gMg^{-1}$

has at least $ 1 + |G|/2 $ many elements.

We can show that $N_G(M) = M$ and $|M| \geq 2$. I was considering the intersection between $M$ and $gMg^{-1}$, where $g \notin M$. If the intersection is trivial we are done. However, I could not prove whether it is true.

Let's say if $g$ is a nontrivial element in the intersection, then we can find $m$ and $m'$ in $M$ such that $m = g m' g^{-1}$. What can we say about it?

Is it the right way to tackle this question? Thank you very much.

E.E.
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    All proper subgroups are abelian, so $M$ is abelian, so $gMg^{-1} \cap M$ is centralized by both $M$ and $gMg^{-1}$ and hence by all of $G$. So $M \cap gMg^{-1} \le Z(G)$. But we also must have $Z(G) \le M$. So intersecions of conjuagtes of $M/Z(G)$ are trivial in $G/Z(G)$. – Derek Holt Sep 23 '17 at 07:13
  • What does 'centralized by M' mean? – E.E. Sep 24 '17 at 01:43
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    @E.E. it means every element of $M$ commutes with every element of $gMg^{-1}\cap M$. – pisco Sep 24 '17 at 06:00

1 Answers1

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The hint by Derek almost says it all.

Denote $M' = M/Z(G)$. As a consequence of $N_G(M)=M$, $g_1M'g_1^{-1}$ and $g_2M'g_2^{-1}$ are distinct subgroups in $G/Z(G)$ iff their intersection is trivial, this happens iff $g_1,g_2$ are in different cosets of $M$. Hence $$\left| {\bigcup\limits_{g \in G} {gM'{g^{ - 1}}} } \right| = \frac{{\left| G \right|}}{{\left| M \right|}}\left( {\left| {M'} \right| - 1} \right) + 1$$ From which you can easily derive $$\left|\bigcup\limits_{g \in G} {gM{g^{ - 1}}} \right| = \left| G \right|\left( {1 - \frac{{\left| Z \right|}}{{\left| M \right|}}} \right) + \left| Z \right| \ge \frac{{\left| G \right|}}{2} + 1$$

pisco
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