A finite group of order $n$, having a subgroup of order $k$ for each divisor $k$ of $n$, is not simple?

Question

I was asked to prove that, if a finite group $G$ of order $n$ has a subgroup of order $k$ for each divisor $k$ of $n$, then $G$ is not simple.

I tried to do this but I could not. Can anyone please give me a hint?

Answer 1

A theorem of Hall says that any group having a $p$-complement for every $p$ is solvable. So you can reduce to $p$-groups. These always have non-trivial center by the class equation, so are simple only when they are of prime order (and the statement permits these as counterexamples, but is easily modified to exclude them).

You can find this theorem in most graduate group theory texts. In Rotman it is in the Normal Series chapter.

Answer 2

I assume you're using a definition of "simple" that excludes cyclic groups of prime order, because if you don't exclude them then they are counterexamples to the claim. So now here's an "overkill" solution. If the order of $G$ is even, then by assumption it has a subgroup of index $2$, and such subgroups are always normal, so $G$ isn't simple. If, on the other hand, the order of $G$ is odd, then by the Feit-Thompson theorem $G$ is solvable and thus not simple. (OK, it's Overkill with a capital O.)

Answer 3

The claim is false, if $G$ is cyclic of prime order. Those are simple groups. Apparently you are to exclude that case, so we assume that $|G|$ is not a prime number.

Let $p$ be the smallest prime divisor of $|G|$. Your assumption implies that there exists a subgroup $H\le G$ such that $[G:H]=p$. The group $G$ acts on the cosets of $H$ transitively, so this gives a non-trivial homomorphism $f$ from $G$ to $S_p$.

Because $\gcd(p!,|G|)=p$, we see that the image of $f$ must be of order $p$. Because $\ker f\subseteq H$ this implies that $H=\ker f$. Hence $H\unlhd G$. Thus $G$ is not simple.

Answer 4

The question of Petar refers to the class of finite groups known as CLT groups, where CLT stands for Converse Lagrange Theorem:

$G$ is a CLT group if for each positive integer $d$ dividing $|G|$, $G$ has at least one subgroup of order $d$.

These groups have been studied extensively. It turns out for example that all supersolvable groups are CLT, and all CLT groups are solvable. See also one of the early papers of Henry G. Bray, Pac. J. Math 27 (1968).