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Torsion elements in an abelian group form a subgroup

Let $G$ be a group. An element $x \in G$ is called torsion if it has finite order. The set of all torsion elements in $G$ is denoted $T(G)$. Prove that if $G$ is abelian, then $T(G)$ is a subgroup of $G$. Show by example that $T(G)$ need not be a subgroup if $G$ is not abelian.

Solution: Note that $T(G)$ is nonempty since the identity has order 1. Now if $x,y \in T(G)$, we have $x^a = y^b = 1$ for some positive integers $a$ and $b$. Then $$(xy^{-1})^{ab} = (x^a)^b (y^b)^{-a} = 1,$$ so that $xy^{-1} \in T$. By the Subgroup Criterion, $T(G) \leq G$ is a subgroup.

Now consider the infinite dihedral group, $D_\infty$. Both $sr^2$ and $sr$ have order 2, but $srsr^2 = r$ has infinite order; in this case, $T$ is not closed under the group operator.


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