## A finite non-abelian group in which every proper subgroup is abelian is not simple

Posted: June 8, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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The symmetric group $S_3$ is an example of a finite non-abelian group in which every proper subgroup is abelian. This group is not simple because its Sylow 3-subgroup is normal. In this post, we’ll show that this is the case for any finite (non-abelian) group all of whose proper subgroups are abelian.

Notation. Let $G$ be a group and let $H$ be a subgroup of $G.$ We will denote by $N(H)$ the normalizer of $H$ in $G.$ We will also define $\mathcal{C}(H)$ to be the union of conjugates of $H,$ i.e. $\mathcal{C}(H)=\bigcup_{g \in G} gHg^{-1}.$

Lemma 1. Let $G$ be a finite group and let $H$ be a subgroup of $G$ with this property that $gHg^{-1} \cap H = \{1\},$ for all $g \notin H.$ Then
1) the intersection of two distinct conjugates of $H$ is the identity element;
2) $|\mathcal{C}(H)|=|G|-[G:H] + 1$ and thus $|G \setminus \mathcal{C}(H)|=[G:H]-1.$

Proof. There is nothing to prove if $H=\{1\}.$ So suppose that $H \neq \{1\}.$ Clearly $H=N(H)$ because if $g \in N(H) \setminus H,$ then $H=gHg^{-1} \cap H = \{1\},$ which is a contradiction. Thus the number of conjugates of $H$ is $[G:N(H)]=[G:H].$
1) Suppose that $g_iHg_i^{-1}, \ i=1,2,$ are two distinct conjugates of $H$ and $g \in g_1Hg_1^{-1} \cap g_2Hg_2^{-1}.$ Then $g=g_1h_1g_1^{-1}=g_2h_2g_2^{-1},$ for some $h_1,h_2 \in H$ and hence $h_2 \in g_2^{-1}g_1 H g_1^{-1}g_2 \cap H.$ So, by hypothesis, either $h_2=1$ or $g_2^{-1}g_1 \in H.$ If $g_2^{-1}g_1 \in H,$ then $g_1Hg_1^{-1}=g_2Hg_2^{-1},$ which is a contradiction. Thus $h_2=1$ and so $g=1.$
2) Since every conjugate of $H$ has $|H|$ elements and, by 1), two distinct conjugates of $H$ have only one element in common , we have

$|\mathcal{C}(H)|=[G:H](|H|-1)+1=|G|-[G:H]+1. \ \Box$

Lemma 2. Let $G$ be a group and let $H_1$ and $H_2$ be two abelian subgroups of $G.$ Let $H$ be the subgroup generated by $H_1$ and $H_2.$ Then $H_1 \cap H_2$ is a normal subgroup of $H.$

Proof. Since $H_1$ and $H_2$ are abelian, $H_1 \cap H_2$ is a normal subgroup of both $H_1$ and $H_2.$ Thus $H_1 \cap H_2$ is also a normal subgroup of $H$ because an element of $H$ is a finite product of elements of $H_1$ and $H_2. \ \Box$

Problem. Let $G$ be a finite non-abelian group in which every proper subgroup is abelian. Prove that $G$ is not simple.

Solution. Suppose, to the contrary, that $G$ is simple and let $H$ be a maximal subgroup of $G.$ Clearly $N(H)=H$ because $H \subseteq N(H)$ and $H$ is a maximal subgroup of $G.$ Let $g \notin H.$ Then $g \notin N(H)$ and so $gHg^{-1} \neq H.$ Hence the subgroup generated by $H$ and $gHg^{-1}$ is $G$ because it contains $H$ strictly. Therefore $gHg^{-1} \cap H$ is a normal subgroup of $G,$ by Lemma 2. Hence $gHg^{-1} \cap H= \{1\}$ and so, by Lemma 1

$|G \setminus \mathcal{C}(H)|=[G:H]-1. \ \ \ \ \ \ \ \ (*)$

In particular $\mathcal{C}(H) \neq G$ and so we can choose $a \notin \mathcal{C}(H).$ Let $K$ be a maximal subgroup of $G$ which contains $a.$ Then again, exactly as we proved for $H,$ we have $gKg^{-1} \cap K = \{1\}$ and so , by Lemma 1

$|\mathcal{C}(K)|=|G|-[G:K]+1. \ \ \ \ \ \ \ \ \ (**)$

Now let $g_1,g_2 \in G.$ Note that $g_2Hg_2^{-1}$ is a maximal subgroup of $G$ because $H$ is a maximal subgroup of $G.$ Since $a \in K$ and $a \notin \mathcal{C}(H),$ we have $g_1Kg_1^{-1} \neq g_2Hg_2^{-1}$ and thus the subgroup generated by $g_1Kg_1^{-1}$ and $g_2Hg_2^{-1}$ is $G.$ Therefore $g_1Kg_1^{-1} \cap g_2Hg_2^{-1}$ is a normal subgroup of $G,$ by Lemma 2. Hence $g_1Kg_1^{-1} \cap g_2Hg_2^{-1}=\{1\}$ and so $\mathcal{C}(H) \cap \mathcal{C}(K)=\{1\}.$ Thus we have proved that

$\mathcal{C}(K) \setminus \{1\} \subseteq G \setminus \mathcal{C}(H).$

It now follows from $(*)$ and $(**)$ and the above inclusion that

$|G| \leq [G:H]+[G:K]-1 \leq |G|/2 + |G|/2 - 1 =|G|-1,$

which is non-sense. This contradiction shows that $G$ is not simple. $\Box$

In fact, the result still holds for finite non-cyclic groups because a simple abelian group is cyclic (of prime order).