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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.

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    $\begingroup$ Sea turtles answer tells why it works. If G is not abelian, indeed if it is not nilpotent, then maximal subgroups need not have prime index. Examples of nonnilpotent p groups are tarski monsters and McClain char simple p groups. $\endgroup$ May 1, 2014 at 23:32

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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.

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  • $\begingroup$ Thanks a lot! $M\cap M'$ has finite index and I agree. However I asked why it has finite order too. I'm arguing about this, but I can't see the light! $\endgroup$
    – Joe
    May 2, 2014 at 17:26
  • $\begingroup$ @Joe $M$ is the only proper subgroup of $\Gamma$ that isn't finite. How to prove this? In any direct product $G=H\times K$, if $\pi_H$ and $\pi_K$ are the projection maps and $M\le G$ is any subgroup, we have the containment $M\le \pi_H(M)\times\pi_K(M)$. The Prufer $p$-group's proper subgroups are all cyclic of prime power order, in particular they are finite. $\endgroup$
    – blue
    May 2, 2014 at 17:34
  • $\begingroup$ Thanks a lot! The finitness of $M\cap M'$ is trivial, I've seen this after your comment answer. However your answer was really clear and useful, thank you! :-) $\endgroup$
    – Joe
    May 2, 2014 at 19:18

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