Let $G$ be an abelian group. Prove that the set of elements of finite order in $G$, that is $T(G) :=\bigg{\{} g \in G \space \bigg{|} \space|g| < \infty \bigg{\}}$ is a subgroup of $G$.
Proof
Since $1 \in G$ and $|1| <\infty$, $1 \in T(G)$ therefore $T(G)$ is non empty.
Let take any $a,b \in T(G)$ say $|a|=n<\infty$ and $|b|=m<\infty$.
Consider, $(ab^{-1})^{nm}=a^{mn}(b^{-1})^{mn}=(a^n)^{m}((b^{-1})^{m})^n=1$.
Therefore $|ab^{-1}| \leq nm$ that is $|ab^{-1}|<\infty$. So $ab^{-1} \in T(G)$. By the subgroup criterion $T(g)$ is a subgroup of group $G$
Can anyone verify my answer?
