All you gotta do to show something is a subgroup is show it is closed under the group operation and closed under inverses (the rest is inherited from the mother group).

This is the torsion subgroup. Let . Then such that . We wish to show .

This is particularly easy because G is assumed to be abelian.

Thus as desired.

Let , suppose .

Thus as desired

T is a subgroup of G.

You just modded out by everything of finite order by definition of T. G/T cannot have anything other than the coset 1T itself of finite order, think about it. G is abelian so every subgroup is normal so this is a well defined group.

Just a note, if G were finite everything is torsion by LaGrange, so G/T is trivial. The only interesting things are for infinite abelian groups.