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

  • $\begingroup$ yes, it is true $\endgroup$ – Loreno Heer Nov 16 '14 at 12:30
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    $\begingroup$ @sanjab: some proof or something ? $\endgroup$ – user123733 Nov 16 '14 at 12:46

$\color{red}{\left[\right.}$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 $\;\Bbb Z/p\Bbb Z\;$

  • $\begingroup$ Excuse my ignorance, but I am confused. For example, in $(\mathbb{Q}, +)$, $1\not\in\langle2\rangle$, so surely $1$ is not contained in some maximal subgroup (by your first paragraph)? And the same trick can always be repeated, $n\not\in\langle 2n\rangle$. So...what am I missing? It seems that what you have said - what the theorem says! - is that a divisible group has no proper, non-trivial subgroups (which is obviously false!). So...help? $\endgroup$ – user1729 Nov 17 '14 at 10:01
  • $\begingroup$ (Also, is your adding in \; around your MathJax deliberate? Or are you using a program?) $\endgroup$ – user1729 Nov 17 '14 at 10:02
  • $\begingroup$ @user1729 Ok, clearly I didn't make myself clear: $\;1\notin\langle 2\rangle\implies \;$ there exists a maximal subgroup wrt not containing $\;1\;$ , not maximal in itself. Yet I see this can probably lead to some problems in my proposed answer, so let me check this and I will add, if necessary, an explanation...or I'll delete my answer. Thank you very much. $\endgroup$ – Timbuc Nov 17 '14 at 12:00
  • $\begingroup$ @user1729. I'd appreciate any insight/comment on the edition I added to my answer. Thank you. $\endgroup$ – Timbuc Nov 17 '14 at 14:21

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