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.

Solution

Verified$\textsf{\color{#c34632}Answer for (a)}$ We have to show that the elements of finite order in $G$ form a subgroup of $G$.

It is enough to show that the subgroup generated by finite order elements in $G$ is closed under sums and inverses.

Let $T(G)$ be the subgroup generated by some finite order element in $G$.

Let $a,b \in T(G)$ with orders $m,n$ respectively. Then we have $ma=0=nb$. Note that

$mn(a+b) = n(ma) + m(nb) = 0$

It follows that the order of the element $a+b$ is less than or equal to $mn$ and so it is finite. Hence $a+b \in T(G)$.

Now note that

$m(-a) = -(ma) = 0$

It follows that $-a \in T(G)$.

Therefore, $T(G)$ is a subgroup of $G$.