Originally Posted by

**Xyelos** I'm self studying Abstract Algebra Theory and Application and would appreciate a proof check, as I never invoke the fact that $\displaystyle G$ is an abelian group and the solution in the back of the book appears to be taking a different route than me, meaning my proof is likely wrong. If anyone could push me in the right direction I'd appreciate it.

Proposition: Let $\displaystyle G$ be an abelian group. The elements of finite order in $\displaystyle G$ form a subgroup $\displaystyle T$.

Proof. We know $\displaystyle T$ is not empty as the order of the identity of $\displaystyle G$ is one. We now must show that for all elements in $\displaystyle T$ there exists an inverse element. Consider any element $\displaystyle x$ of $\displaystyle T$ and let $\displaystyle m$ be its order. Thus, $\displaystyle x^m = e$, where $\displaystyle e$ is the identity of $\displaystyle G$ and now $\displaystyle T$. Hence, e=x^{m}=x*x^{m-1}=e, implying that x^{m-1} = x^{-1}.

The binary operation on T is associative as it is already associative on G.Quick notation clear up - * represents a binary operation.