# If a single integer power kills every element of abelian group $G$, then $G$ is finite (UPDATE after shown false)

Is it true that for any abelian group $G$, if there is some integer $k$ satisfying $g^k = e$ for every $g \in G$, then $G$ is finite?

This is true if $G$ is finitely generated.

I believe that it is true in the general case but am struggling to prove this. Any help is appreciated.

EDIT: Alan Wang points out a simple counterexample to the proposition.

My original idea for this question came from Keith Conrad's algebra notes, which has the following exercise.

Show a finitely generated $A$-module $M$ is a torsion module if and only if there is some $a \ne 0$ in $A$ such that $aM = 0$. (This is false without a hypothesis of finite generatedness even for $A = \mathbb{Z}$, since infinite torsion abelian groups exist.)

It's the parenthetical which now has me confused. How is an infinite torsion abelian group an automatic counterexample?

• What about infinite direct product of $\Bbb{Z}_2$? – Alan Wang Jul 21 '18 at 6:43
• It is false, take $C_p^{\mathbb{N}}$ for example. – jgon Jul 21 '18 at 6:43
• Why couldn't the group have an infinite number of elements all of finite order? – fleablood Jul 21 '18 at 6:44
• "This is true if G is finitely generated." And it'd be false if it where infinitely generated. So the question is: is it possible to have an infinitely generated group where every element has a finite order? And ... why the heck not? As Alan Wang suggests why not $\mathbb Z_2^{\mathbb N}$ or, more generally, $\times_{i\in \mathbb N} \mathbb Z_{k_i}$ – fleablood Jul 21 '18 at 6:49
• For nonabelian groups you might want to explore the Burnside Problem. It is now known that there are infinite groups which are finitely generated and in which all the elements have bounded order. – Mark Bennet Jul 21 '18 at 7:03

However, consider e.g. $\Bbb Q/\Bbb Z$. Here each element $a/b$ is killed by an integer ($b$), but no integer kills the whole group.