2
$\begingroup$

Let $G$ be an infinite abelian group with at least one element of infinite order. Prove that $G$ is generated by it's elements of infinite order.

As a hint, I was given that it may be useful to recall some facts about elements of finite order in an abelian group. The main thing that this makes me think is that these elements form a subgroup, the torsion subgroup.

As far as further ideas go, I was thinking of taking the factor group $\frac{G}{T(G)}$ which exists since $G$ is abelian and so all subgroups are normal. Since $G$ has an element of infinite order, this factor group must be infinite. But now I am stuck.

$\endgroup$
  • 2
    $\begingroup$ Each finite order element is the difference of two infinite order elements. $\endgroup$ – Somos Apr 10 at 22:09
  • $\begingroup$ @Somos, does this mean that the rotational motions of the plan generated by some infinite order rigid motion? i.e how are elements of finite order in Special linear group generated by elements of infinite group in general linear group over the plane? $\endgroup$ – mathpadawan Apr 10 at 22:14
  • $\begingroup$ Okay, the rotational group about a single point in the plane is abelian, but why are you asking about Special linear group which is not abelian? $\endgroup$ – Somos Apr 10 at 22:15
  • 1
    $\begingroup$ The interesting question is whether this result is true without assuming that the group is abelian. $\endgroup$ – Derek Holt Apr 10 at 22:22
9
$\begingroup$

Let $a$ be any element of infinite order, that is, $na$ is different for each $n\in\Bbb Z$.
Then if element $b$ has order $k$, then $nk(a+b) =nka$ are all different for $n$, hence $a+b$ has infinite order.
So, $b$ is generated by $a$ and $a+b$, both of infinite order.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.