Order of elements in non-abelian groups Now I've in this algebra class I had to prove that 

for all abelian groups, if $x$, $y$ have finite order then $xy$ also has a finite order. 

Fair enough. However I was wondering: what happens to this property for non-abelian groups?
 A: To give some concrete example: The two matrices
$$\begin{pmatrix}0&2\\\frac{1}{2}&0\end{pmatrix} \text { and } \begin{pmatrix}0&1\\1&0\end{pmatrix}$$
have order $2$ in $\operatorname{GL}_2(\mathbb Q)$, but their product is equal to $\begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}$, which has infinite order.
A: $G = \langle x,y \mid x^2=y^2=1 \rangle $ is an example of a group generated by elements $x, y$ and the (only) relation $x^2=y^2=1$. It is non-abelian and infinite. In fact, it is a free product, as it is called, of a cyclic group of order 2 with itself.
A: $\newcommand{\Z}{\mathbb{Z}}$Just as a variation on the previous answers, let $A = \Z / n \Z$, for $n \ge 0$ (where for $n = 0$ we understand $A = \Z$).
Then the two matrices with coefficients in $A$ 
$$
x
=
\begin{bmatrix}
-1 & 1\\
0 & 1\\
\end{bmatrix},
\quad
y
=
\begin{bmatrix}
-1 & 0\\
0 & 1\\
\end{bmatrix},
$$
have order $2$, but their product
$$
x y
=
\begin{bmatrix}
1 & 1\\
0 & 1\\
\end{bmatrix}
$$
has order $n$ if $n > 0$, and infinity if $n = 0$.
So even when the product of two elements of order $2$ is finite, it can be anything.
