TITLE: In an abelian group, the elements of finite order form a subgroup.
QUESTION [7 upvotes]: I need to show that elements of finite order in an abelian group form a subgroup of that group. Where do i start ?

REPLY [11 votes]: Let $U$ be the set of all elements of finite order.

The neutral element $1$ has order 1. So $1\in U$.
Let $g,h\in U$. Then there are positive integers $n,m \geq 1$ with $g^n = 1$ and $h^m = 1$. So
$$(gh)^{nm} \overset{gh = hg}{=} g^{nm} h^{nm} = (g^n)^m (h^m)^n = 1^m 1^n = 1.$$
Hence $\operatorname{ord}(gh) \le nm$ and therefore $gh \in U$.
Let $g\in U$. Then there is a positive integer $n$ with $g^n = 1$.
Multiplication with $(g^{-1})^n$ yields $$\underbrace{(g^{-1})^n g^n}_{=1} = (g^{-1})^n.$$ So $\operatorname{ord}(g^{-1})\le n$ and $g^{-1}\in U$.

Therefore, $U$ is a subgroup.