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.

• 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.