## In a finite simple group whose every proper subgroup is abelian, distinct maximal subgroups intersect trivially

Suppose $G$ is a finite simple group in which every proper subgroup is abelian. If $M$ and $N$ are distinct maximal subgroups of $G$, prove that $M \cap N = 1$.

Let $G$ be a simple group in which every proper subgroup is abelian, and suppose $M$ and $N$ are distinct maximal subgroups of $G$. Consider the normalizer in $G$ of the subgroup $M \cap N$. Since $M \cap N \leq M$ and $M$ is abelian, $M \leq N_G(M \cap N)$. Similarly, $N \leq N_G(M \cap N)$. Thus $\langle M, N \rangle \leq N_G(M \cap N)$. Note that $\langle M, N \rangle$ is a subgroup of $G$ which properly contains, say, $M$. Thus $\langle M, N \rangle = G$, and we have that $M \cap N$ is normal in $G$. Since $G$ is simple and $M \cap N$ is a proper normal subgroup, $M \cap N = 1$.

(Old proof)

Let $M$ and $N$ be distinct maximal subgroups. Since $M \leq N_G(M)$ and $N \leq N_G(N)$ and $G$ is simple, we have $N_G(M) = M$ and $N_G(N) = N$. Moreover, because $M$ and $N$ are abelian, we have $C_G(M) = M$ and $C_G(N) = N$. Finally, note that $G = \langle M,N \rangle$.

Now let $g \in G$ and $x \in M \cap N$. We can write $g = m_1n_1m_2n_2 \ldots m_kn_k$ for some elements $m_i \in M$ and $n_i \in N$. Since $x \in M \cap N$ and both of $M$ and $N$ are abelian, $gx = xg$, so that $gxg^{-1} = x$. In fact, $M \cap N$ is normal in $G$. Since $M \cap N$ is a proper normal subgroup and $G$ is simple, $M \cap N = 1$.

• J  On November 4, 2010 at 8:02 pm

Why would MN be G because you can’t even be sure it’s a group I think. But you can do the same with the group generated by M and N, this certainly equals G, and since G is finite (your proof doesn’t use this!) this means that every g in G has a representation g=m_1n_1m_2n_2 … m_kn_k, and then the proof works just the same…

• nbloomf  On November 5, 2010 at 9:18 am

You’re right- $MN$ is not necessarily a subgroup. Thank you also for suggesting a fix.

I am more disturbed by the fact that the proof doesn’t use the finiteness of $G$. I am having trouble seeing where this is needed. I also found a more order-theoretic proof that doesn’t seem to need finiteness. I will leave the old proof and the new proof up- does anyone see why finiteness is needed?

• Nish  On February 21, 2011 at 8:41 pm

I did not quite understand your step “Finally, note that G = \langle M,N \rangle” in old proof.Why product of any two maximal sub groups will necessarily span G?

• nbloomf  On February 21, 2011 at 8:44 pm

It’s because $\langle M, N \rangle$ is a subgroup of $G$ which properly contains $M$. (Properly because $M \neq N$.) Since $M$ is maximal, it has to be all of $G$.

• Nish  On February 21, 2011 at 8:58 pm

Please ignore the last comment.G indeed is finitely generated by M,N as M,N are maximal.

• nhanttruong  On July 6, 2011 at 7:17 pm

Corollary. Let M be a nonnormal maximal subgroup of the finite simple group G with all subgroups abelian. Then |G| has the same prime factors as |M|.

Proof. Since M not normal, it has a distinct conjugate N, which is also maximal and has the same prime factors as M. Then |G| = |MN| = |M||N| has the same prime factors as |M|.

• nbloomf  On July 7, 2011 at 11:45 am

Interesting! It seems that in fact $|G| = |M|^2$, since $M$ and $N$ are isomorphic.

• nhanttruong  On July 8, 2011 at 12:33 am

Ah, I’m wrong. MN doesn’t have to be a subgroup.