Example of a group where $o(a)$ and $o(b)$ are finite but $o(ab)$ is infinite Let G be a group and $a,b \in G$. It is given that $o(a)$ and $o(b)$ are finite. Can you give an example of a group where $o(ab)$ is infinite?
 A: If $\circ(a)$ is the order of $a$, usually denoted $|a|$, looking at permutations of $\mathbb{N}$ are an easy way to find an example.
For example consider the permutations:
$$ a = (12)(34)(56)...$$
$$ b = (23)(45)(67)...$$
It isn't hard to see that both $a$ and $b$ have order 2, but consider the image of $1$ under $(ab)^n$ to see that the order of $ab$ is infinite.
A: Every direct plane rotation (of angle $2\theta$) is the composite of two orthogonal reflections with fixed lines making an angle of $\theta$. If $2\theta/2\pi$ is irrational, then the rotation of angle $2\theta$ has infinite order.
A: Probably you have seen this effect: mirrors on two opposite walls of a room.  
 
Take two parallel hyperplanes in $\mathbb R^n$.  Reflection in each of them is an isometry of order 2.  But their composition is a translation of infinite order.
A: If there's any example at all, then
$$ G = \langle a,b \mid a^m , b^n \rangle $$
is an example for some positive integers $m,n$. I assert that every choice with $m,n > 1$ is an example.
A: The standard example is the infinite dihedral group.
Consider the group of maps on $\mathbf{Z}$ 
$$
D_{\infty} = \{ x \mapsto \pm x + b : b \in \mathbf{Z}  \}.
$$
Consider the maps
$$
\sigma: x \mapsto -x,
\qquad
\tau: x \mapsto -x + 1,
$$
both of order $2$. Their composition
$$
\tau \circ \sigma (x) = \tau(\sigma(x)) : x \mapsto x + 1
$$
has infinite order.
A: Consider the matrices $$A=\left(\begin{array}[cc] .1 & -1\\ 0 & -1\end{array}\right)\quad\text{and}\quad B=\left(\begin{array}[cc].1&0\\ 0 & -1\end{array}\right).$$ You can check that each have order $2$, but their product gives $$(AB)^n=\left(\begin{array}[cc] .1 & n\\ 0 & 1\end{array}\right),$$a matrix without finite order.
