1. ## Torsion subgroup

Let G be an abelian group. Prove that $\{g\in G:|g|< \infty\}$ is a subgroup of G. Give an explicit example where this set is not a subgroup when G is non-abelian.

First if G is infinite, then the torsion group would be empty. So suppose G is finite. I am supposed to use the subgroup criterion to show this but not entirely sure with what to do.

Subgroup criterion is a subset H of a group is a subgroup iff H is nonempty and for all x,y in H $xy^{-1}\in H$.

Let $y=x$, $|x|<\infty$, and $x\in H$.
By property 2, $xy^{-1}\in H$. Therefore, $xx^{-1}=e\in H$.
Since e and x are in H, we have $ex^{-1}=x^{-1}\in H$

Correct?

Not sure about the example when G is non-abelian.

2. ## Re: Torsion subgroup

Originally Posted by dwsmith
Let G be an abelian group. Prove that $\{g\in G:|g|< \infty\}$ is a subgroup of G. Give an explicit example where this set is not a subgroup when G is non-abelian.

First if G is infinite, then the torsion group would be empty.
no. (Q-{0},*) is infinite, but its torsion group is not empty (-1 has order 2).

So suppose G is finite.
if G is finite, every element has to have finite order, so for a finite group, Tor(G) = G.

this is really quite simple. suppose a,b are of finite order m and n, respectively.

then $(ab^{-1})^{mn} = a^{mn}(b^{-1})^{mn}$ because G is abelian

$= (a^m)^n(b^{-1})^{nm}= (a^m)^n(b^{-n})^m = (a^m)^n((b^n)^{-1})^m = e^n(e^{-1})^n = ee = e$

so ab must be of finite order.

for the non-abelian case, the only example that comes to mind is the free group on two generators a,b subject to the relations $a^2 = b^2 = e$.

clearly ab is of infinite order.