Does $\Gamma$ infinite $p$-group, $M$ (one of) its maximal subgroup imply $\Gamma/M$ has prime order?

Question

If $G$ is finite with a unique maximal subgroup, then $G$ is cyclic.

If $G$ is infinite, not. The example I'm studying is the following: let $Z_{p^{\infty}}$ be the Prufer $p$-group and $C_p$ the cyclic group of order $p$ (obviously $p$ is prime).

Consider now $\Gamma:=Z_{p^{\infty}}\times C_p$. This new group is infinite and not cyclic, in which $M:=Z_{p^{\infty}}\times1$ is clearly maximal.

In order to show that it's unique, we argue by contradiction and suppose that there exists $M'\neq M$ another maximal subgroup of $\Gamma$.

Then my teacher wrote that $\Gamma/M'$ has prime order and $M\cap M'$ is finite: I can't understand why, and I have no idea where to start.

Moreover I'm asking myself if is this true in general for infinite groups.

Thank you all.

Answer 1

Clearly $\Gamma/M'$ is abelian, since $\Gamma$ is abelian. As $M'$ is assumed maximal, $\Gamma/M'$ is simple. Abelian and simple implies cyclic of prime order. (Proof. Good exercise. Hint: lattice correspondence.)

Since $M$ and $M'$ have finite index, their intersection does as well. (Proof. This holds generically, not just in this situation. Let $H$ and $K$ be finite index subgroups of some $G$. The Cartesian product of coset spaces $G/H\times G/K$ is finite and also equipped with a left $G$-action, so the orbit of $H\times K$ is also finite, so its stabilizer $H\cap K$ must have finite index by the orbit-stabilizer theorem.)

This reasoning applies to any abelian group: maximal subgroups, if they exist, must have prime index. Maximal subgroups need not exist (just look at the Prufer $p$-group) and need not be unique, for instance ${\bf Z}_p\times{\bf Z}_p$ (we use $p$-adic integers to get an infinite group, even though this example is perfectly analogous to $C_p\times C_p$). I don't think anything is different for nonabelian groups.