I'm self studying Abstract Algebra Theory and Application and would appreciate a proof check, as I never invoke the fact that 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 be an abelian group. The elements of finite order in form a subgroup .
Proof. We know is not empty as the order of the identity of is one. We now must show that for all elements in there exists an inverse element. Consider any element of and let be its order. Thus, , where is the identity of and now . Hence, e=xm=x*xm-1=e, implying that xm-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.