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.

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Comments

  • 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.

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