Exercise:
α) Show that in an abelian group G the set T of its elements with finite order is a subgroup of G (torsion subgroup).
β) Consider if T is a normal subgroup of group G.
I am sorry that I don't appose any attempt but I don't know even how to start. Any help or thorough solution would be precious for me as I need to clear my mind on this type of problems.
Hint!
For α), if $g$ has order $r$, $h$ has order $s$, consider $(gh)^{rs}$.
For β), if the group is commutative, any subgroup is normal, so there's nothing to prove. However, there exist non-commutative groups for which the torsion elements are a subgroup, viz. nilpotent groups, but it is very technical.
Added:
However, if you know that, for some reason, the set $T$ of torsion elements in a (not necessarily abelian) group is a subgroup, then it is a normal subgroup.
Actually, it is even more: recall that a normal subgroup is a subgroup which is invariant by inner automorphisms of the group. Actually, $T$ is invariant by every endomorphism of the group, i.e. it is a fully characteristic subgroup.
Indeed, if $g\in T$, there exists a natural number $n$ such that $g^n=e$. Let $u$ be any endomorphism of $G$, we have: $$\bigl(u(g)\bigr)^n=u(g^n)= e^n=e,\;\text{ so }\enspace u(g)\in T.$$