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When I was attempting (and ultimately not succeeding) to answer this question I wanted something in group theory to be true, which I didn't think was. Namely if $G$ is a finite group and $p$ was a prime such that $p$ divided the order of $G$ then $G$ had a subgroup of index $p$. The proposition is clearly true for nilpotent groups, but beyond that I couldn't decide anymore. In particular I'm curious about whether or not its true for solvable groups and if it's easy to construct a counterexample for any prime $p$. For $p=3$ we have for instance $S_5$.

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It's true for supersolvable groups (apply induction to $G/N$ where $N$ is a cyclic normal subgroup of prime order, together with Schur-Zassenhaus), but it's also true in $S_4$, so that is not a necessary condition. – Derek Holt Jan 23 '13 at 9:32
A closely related question is whether there is any characterization of finite groups that have subgroups of all possible orders (i.e. possible by Lagrange's Theorem). – Derek Holt Jan 24 '13 at 9:10
up vote 1 down vote accepted

If $M\leqslant G$ has index $m$, then $G/\text{core}_G(M)$ embeds into $\text{Sym}(G;M)\cong S_m$. If $G$ is simple, then $M$ is core-free, so we have that $G$ injects into $\text{Sym}(G;M)$ which is of order $m!$. Thus whenever $G$ is a finite simple group and $M$ a subgroup of prime index $p$, $p$ must be the largest prime divisor of $|G|$ and $p^2\not\mid |G|$. In particular, $|A_n|=n!/2$ only divides $p!$ when $n=p$.

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I'm a little confused. Doesn't $A_5$ have a subgroup of index $5$? How does this fit into your argument? – JSchlather Jan 23 '13 at 6:33
Yeah let me fix this – Alexander Gruber Jan 23 '13 at 6:34
@JacobSchlather That should do it. – Alexander Gruber Jan 23 '13 at 7:36
I am going to try to figure out an argument for solvable groups as well, I'll edit if I come up with anything. – Alexander Gruber Jan 23 '13 at 22:10

The alternating group $A_4$ has no subgroup of index 2.

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So it fails for solvable groups. I'd imagine then that nilpotent groups are the most interesting class. – JSchlather Jan 23 '13 at 6:18
In fact $A_n$ has no subgroup of index $p$ for $p\not= n\geq 5$ (though it is not solvable). – Alexander Gruber Jan 23 '13 at 6:20

As the example given by Gerry shows, even if $G$ is solvable, it does not necessarily have a subgroup of prime index $p$ for every $p$ dividing $|G|$.

However, there is a partial result for solvable groups. If $G$ is a nontrivial solvable group, you can show that $G$ has at least one subgroup of prime index. Note that this might not be true for nonsolvable groups. The alternating group $A_6$ is one example of a finite group with no subgroups of prime index.

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