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

Suppose is a finite simple group in which every proper subgroup is abelian. If and are distinct maximal subgroups of , prove that .

Let be a simple group in which every proper subgroup is abelian, and suppose and are distinct maximal subgroups of . Consider the normalizer in of the subgroup . Since and is abelian, . Similarly, . Thus . Note that is a subgroup of which properly contains, say, . Thus , and we have that is normal in . Since is simple and is a proper normal subgroup, .

(Old proof)

Let and be distinct maximal subgroups. Since and and is simple, we have and . Moreover, because and are abelian, we have and . Finally, note that .

Now let and . We can write for some elements and . Since and both of and are abelian, , so that . In fact, is normal in . Since is a proper normal subgroup and is simple, .

### Like this:

Like Loading...

*Related*

or leave a trackback:

Trackback URL.

## Comments

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…

You’re right- 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 . 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?

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?

It’s because is a subgroup of which properly contains . (Properly because .) Since is maximal, it has to be all of .

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

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|.

Interesting! It seems that in fact , since and are isomorphic.

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