Elements of finite order in an abelian group form a subgroup Let $G$ be an abelian group.  
Show that $\{x\in{G} | |x| < \infty\}$ is a subgroup of $G$.  Give an example of a non-abelian group where this fails to be a subgroup.
 A: Hint: Here is one of many ways of constructing an example. Let $G$ be the group of permutations of the integers.  Let $f$ be the permutation that takes any integer $x$ to $-x$, and $g$ the permutation that takes any integer $x$ to $1-x$. 
Both $f$ and $g$ have order $2$.  Now consider the permutation $gf$, meaning $f$, followed by $g$. Show that $gf$ does not have finite order. 
If you prefer matrices, let $A=\begin{pmatrix}-1 &0\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}-1 &1\\0&1\end{pmatrix}$.
Then $A^2$ and $B^2$ are both the identity matrix. But $BA$ has infinite order. To see this, check what $BA$ does to the vector $\begin{pmatrix}n \\1\end{pmatrix}$  
Remark: For the Abelian case, we need to show closure under product and inverse. For product, note that if $a^m=e$ and $b^n=e$, then $(ab)^{mn}=a^{mn}b^{mn}=e$. Inverse is easier, since in any group, Abelian or not, the inverse of $a$ has the same order as $a$. 
A: There is another counter example which I think you may find it interesting. Please see Exercise 2.17 of J.J.Rotman's well-known book. There; he gave us $G=GL(2,\mathbb Q)$ such that $tG$ is not a subgroup.. 
A: I assume that by $\lvert g \lvert$ you mean the order of $g$, so that the subset you are considering is the subset of all elements with finite order. Call this subset $X$.
Without actually doing the problem for you, here are a few things that might help.
To check that a subset of a group is a subgroup, all you need to do is to check that the subset itself satisfies the axioms for a group. First, then, you need to figure out whether if $g,h\in X$ then $gh\in X$. In your case this translates to the question about whether the product of two elements with finite order has finite order. Since your group $G$ is abelian, then this shouldn't be to hard to prove.
Then you need to check that all elements in $X$ have an inverse in $X$. An element $g\in G$ obviously has an inverse in $G$, but is this inverse actually in $X$? Well, if $g$ has finite orderm it shouldn't be too hard to see that $-g$ (writing the group additively because it is abelian) also has finite order (Hint: $-0 = 0$).
Without listing all the axioms, you also have to check that the addition of to elements of finite order has finite order.
Googeling or looking in a book will give you other things that you need to check.
As for the second question about an example of $G$ non-abelian where $X$ is not a subgroup, you can take a look at this qusetion: $T(G)$ may not be a subgroup? (As mentioned in the comments by @anon). Note, though, that in this question the group is written multiplicative. 
Hopefully this was helpful.
