The elements of finite order in an abelian group form a subgroup: proof check If G is an abelian group, show that the set of elements of finite order is a subgroup of G.
Proof:
Let G be an abelian group and H be the set of elements of finite order.
(1) nonempty
Now e ∈ H, since $a^n$ = e, by definition of order, and |e| = 1  ∈ H. Thus, H is nonempty.
(2) Closure
Let a, b  ∈ H, where |a| = k and |b| = m for all k, m ∈ G. Then $a^k$ = e and $b^m$ = e. So $a^{km} = (a^k)^m = e$ and $b^{km} = (b^k)^m = e$. Then $(ab)^{km} = e$ and $ab$ ∈ H with a finite order at most $km$. Hence, H has closure.
(3) Inverse
Let a ∈ H with |a| = k for all k ∈ G. Then $(aa^{-1})^k = e$. So $a^k(a^-1)^k = e$. Since $a^k = e$, then $(a^{-1})^k = e$. Thus $|a^{-1}| = k$ and $a^{-1}$ ∈ H. Hence, H has an inverse.
Therefore, H is a subgroup of G. 
I think I messed up on the inverse part of my proof.

Inverse: It is sufficient to show $ka=0$ iff $k(-a)=0$ (and thus $|a|=|-a|$ ) .
But inversion is an invertible function so it is sufficient to show $ka=0$ implies $k(-a)=0$ . Assume $ka=0$.  $ka=2ka+(k)(-a)$ by repeated application of the identity axiom. But this is $0=2*0+k(-a), k(-a)=0 $.   
