A group is divisible if and only if it has no maximal subgroup ?

Question

Is it true that a group is divisible if and only if it has no maximal subgroup ?

Answer 1

$\color{red}{\left[\right.}$To begin with let us make the following general observation: if in some group $\;G\;$ there exists an element $\;g\in G\;$ and a proper (obviously) subgroup $\;H\le G\;$ s.t. $\;g\notin H\;$, then there exists a maximal proper subgroup $\;M\le G\;\;s.t.\;\;g\notin M\;$ . This is a rather straightforward (not necessarily trivial!) application of Zorn's Lemma.

Suppose now $\;G\;$ is a group without maximal (proper) subgroups (note that btw $\;G\;$ is infinite) , and let us take the subgroup

$$G^n:=\left\langle\;x^n\;:\;\;x\in G\;\right\rangle$$

If $\;G\;$ isn't divisible then there exist $\;g\in G\;,\;\;n\in\Bbb N\;$ s.t. $\;g\notin G^n\;$ , which by the above leads to a maximal proper subgroup not containing $\;g\;$ . Thus, for any $\;n\in\Bbb N\;,\;\; G^n=G\;$ and G is thus divisible.$\color{red}{\left.\right]}$

The other direction is simpler (for abelian groups, which are the groups for which "divisible" is more usually applied): if $\;G \;$ is divisible and nevertheless it has a proper maximal subgroup $\;M\;$ then $\;G/M\;$ is a group without any non-trivial subgroups and thus finite and of order a prime. yet it is easy to show that any homomorphic image of a divisible group is divisible, and since non-trivial groups cannot be divisible (why? Take $\;n=$ order of the group...) , we're done.

Edit: Thanks to user1729's comment below, one serious flaw in the the first part of this answer (between red square parentheses) has been found: it's not that I meant $\;M\;$ is a maximal subgroup of $\;G\;$ (though this is what is being understood from my lousy redaction!), but rather that $\;M\;$ is maximal with respect to not containing $\;g\;$ . What follows then doesn't work as the "contradiction" gotten is no contradiction at all: what was just proved is that there is a maximal subgroup wrt not containing $\;g\;$...

Let's change that part to the following: if (the abelian) group $\;G\;$ has no maximal subgroups, then it can't have finite homomorphic images, since

$$G/H\;\;\text{finite}\;\implies\;\exists \;\text{maximal}\;\;M/H\le G/H\implies \;M\le G\;\;\text{maximal}$$

Now, if $\;G\;$ isn't divisible then there exists a prime $\;p\;$ s.t. $\;pG\lneq G\;$ , so that $\;G/pG\neq 0\;$ , but then

$$\;p\left(G/pG\right)=0\implies\;\;\exp\left(G/pG\right)=p\implies\;\exists \;\text{epimorphism}\;\;G/pG\to \Bbb Z/p\Bbb Z$$

as then $\;G/pG\;$ is elementary abelian and thus a direct (finite, infinite: it doesn't matter) sum of copies of $\;\Bbb Z/p\Bbb Z\;$