2
$\begingroup$

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 $S_5$.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Derek Holt Jan 23 '13 at 9:32
  • $\begingroup$ 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). $\endgroup$ – Derek Holt Jan 24 '13 at 9:10
1
$\begingroup$

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 $n=p$.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ So it fails for solvable groups. I'd imagine then that nilpotent groups are the most interesting class. $\endgroup$ – JSchlather Jan 23 '13 at 6:18
  • $\begingroup$ In fact $A_n$ has no subgroup of index $p$ for $p\not= n\geq 5$ (though it is not solvable). $\endgroup$ – Alexander Gruber Jan 23 '13 at 6:20
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.