The order of an element a of an abelian group G is the smallest positive int such that ma = 0, if such exists; otherwise, the order of a is said to be infinite. The order of a thus equals the order of the subgroup generated by a. (a) Show the elements of finite order in G form a subgroup of G, called its torsion subgroup. (b) Show that if G is free abelian, it has no elements of finite order. (c) Show the additive group of rationals has no elements of finite order, but is not free abelian.
We have to show that the elements of finite order in form a subgroup of .
It is enough to show that the subgroup generated by finite order elements in is closed under sums and inverses.
Let be the subgroup generated by some finite order element in .
Let with orders respectively. Then we have . Note that
It follows that the order of the element is less than or equal to and so it is finite. Hence .
Now note that
It follows that .
Therefore, is a subgroup of .