A nonabelian group whose every proper subgroup is abelian is not simple

Prove that if G is any nonabelian group in which every proper subgroup is abelian then G is not simple.


We begin with a lemma.

Lemma: If M \leq G is maximal, then for all g \in G, g^{-1}Mg \leq G is maximal. Proof: Suppose to the contrary that M is maximal and that g^{-1}Mg is not; let g^{-1}Mg \leq H \leq G. Then M \leq gHg^{-1} \leq G, with all containments proper, a contradiction. \square

Let G be a finite nonabelian group in which every proper subgroup is abelian. Suppose that G is simple.

Let M \leq G be a maximal subgroup. By this previous exercise, G \neq \bigcup_{g \in G} g^{-1}Mg; let x \in G such that x is not in a conjugate of M. Now x \in N for some maximal subgroup N \leq G. Note that N is not conjugate to M, so that the conjugates of N and M are pairwise distinct. Moreover, by this previous exercise, the conjugates of N and M intersect trivially pairwise. Thus the nonidentity elements in conjugates of N and M form disjoint subsets of G. By this previous exercise, the number of nonidentity elements in conjugates of M is (|M|-1)[G:M] and in conjugates of N is (|N|-1)[G:N], so that G has at least (|M|-1)[G:M] + (|N|-1)[G:N] nonidentity elements. Now note the following.

|G| > (|M|-1)[G:M] + (|N|-1)[G:N] (strictness since G also contains 1)
 =  (|M|-1)\frac{|G|}{|M|} + (|N|-1)\frac{|G|}{|N|} (since G is finite)
 =  |G| - \frac{|G|}{|M|} + |G| - \frac{|G|}{|N|}
 =  |G|(2 - (\frac{1}{|M|} + \frac{1}{|N|})), so that
|G| > |G|(2 - \frac{|M|+|N|}{|M||N|}) and hence
1 > 2 - \frac{|M|+|N|}{|M||N|}, and thus
|M|+|N| > |M||N|.

Then (without loss of generality) we have |M| \in \{1,2\}, since for natural numbers a and b with a \geq 3, ab \geq a+b.

Suppose |M| = 1. If |G| is composite, then via Cauchy’s theorem we have a contradiction (as there exist proper nontrivial subgroups in G). Then |G| is prime, but then G \cong \mathbb{Z}/(p) is abelian, a contradiction. In particular, we may assume that |M|, |N| \geq 2.

Suppose |M| = 2. Then since M is maximal, M is in fact a Sylow 2-subgroup of G. Hence |G| = 2k where k is odd, and by this previous exercise, G is not simple, a contradiction.

Thus no such group G exists.

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Comments

  • Lennon  On February 28, 2011 at 7:48 am

    in the paragraph starting with “Let M in G be a maximal subgroup”, why do we have |G|=|MN| since MN may not be a subgroup(if it is, then we are done but what if not?)

    • nbloomf  On February 28, 2011 at 10:01 am

      Good point.

      I think that the proof still works using the weaker (and certainly true) inequality |G| \geq |MN|.

    • nhanttruong  On July 6, 2011 at 6:36 pm

      We’re assuming M and N to be distinct maximal subgroups of G finite simple where every subgroup is abelian. So by the previous problem, M\cap N = 1 and G = MN.

  • nhanttruong  On July 6, 2011 at 6:53 pm

    A (just) slightly different proof starting with G having at least (|M| - 1)|G:M| + (|N| - 1)|G:N| nonidentity elements, so that G has at least (|M| - 1)|G:M| + (|N| - 1)|G:N| + 1 = 1 + 2|G| - \frac{|G|}{|M|} - \frac{|G|}{|N|} elements. We want to show that this is > |G|. Rearranging, we want to show 1+|M||N| = 1 + |G| > \frac{|M|+|N|}{|M||N|}|G| = |M| + |N|. The first and last equalities follow because M and N are distinct maximal subgroups of G with trivial intersection. Then the inequality follows since for a,b\geq 2, 1+ab>a+b.

    • nhanttruong  On July 8, 2011 at 2:26 am

      This proof is wrong. The first and last inequalities don’t follow because G needs not be MN.

  • Gobi Ree  On December 26, 2011 at 3:45 am

    I can’t understand why G has at least |G|+|M||N|-|M|-|N| elements.

    • nbloomf  On January 18, 2012 at 1:26 pm

      I can’t either… I came up with another argument which (I think) gets around that problem.

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