The symmetric group 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 be a group and let be a subgroup of We will denote by the normalizer of in We will also define to be the union of conjugates of i.e.

**Lemma 1**. Let be a finite group and let be a subgroup of with this property that for all Then

1) the intersection of two distinct conjugates of is the identity element;

2) and thus

*Proof*. There is nothing to prove if So suppose that Clearly because if then which is a contradiction. Thus the number of conjugates of is

1) Suppose that are two distinct conjugates of and Then for some and hence So, by hypothesis, either or If then which is a contradiction. Thus and so

2) Since every conjugate of has elements and, by 1), two distinct conjugates of have only one element in common , we have

**Lemma 2**. Let be a group and let and be two abelian subgroups of Let be the subgroup generated by and Then is a normal subgroup of

*Proof*. Since and are abelian, is a normal subgroup of both and Thus is also a normal subgroup of because an element of is a finite product of elements of and

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

**Solution**. Suppose, to the contrary, that is simple and let be a maximal subgroup of Clearly because and is a maximal subgroup of Let Then and so Hence the subgroup generated by and is because it contains strictly. Therefore is a normal subgroup of by Lemma 2. Hence and so, by Lemma 1

In particular and so we can choose Let be a maximal subgroup of which contains Then again, exactly as we proved for we have and so , by Lemma 1

Now let Note that is a maximal subgroup of because is a maximal subgroup of Since and we have and thus the subgroup generated by and is Therefore is a normal subgroup of by Lemma 2. Hence and so Thus we have proved that

It now follows from and and the above inclusion that

which is non-sense. This contradiction shows that is not simple.

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