Torsion Subgroups

October 30, 2019October 30, 2019 bs8277

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

A solution: Let $T(G)= \{g \in G \text{ }|\text{ } |g| < \infty \}$ . Since the identity is of order 1, it lies in $T(G)$ and therefore $T(G)$ is not empty. Now if $a,b \in T(G)$ , we have $a^m = b^n = 1$ for some positive integers $m$ and $n$ . Then $(ab^{-1})^{mn} = (a^m)^n (y^n)^{-m} = 1$ , so that $ab^{-1} \in T(G)$ . Therefore, by the Subgroup Criterion, $T(G) \leq G$ is a subgroup.

An explicit example when $T(G)$ is not a group is the infinite dihedral group, $D_\infty$ . In $D_\infty$ , both $sr^2$ and $sr$ have order 2, but $srsr^2 = r$ has infinite order. Thus, $T(G)$ is not closed under the group operation.

Torsion Subgroups

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