Order of the products of two order-2 elements in a finite group 
Let $G$ be a finite group and $g,h \in G$ both have order 2. Determine
  the possible orders of $gh$.

So I first think of this in terms of symmetric groups. Obviously $g$ and $h$ could be the same transposition in $S_n$ and thus $gh$ is the identity (order 1). They could also be two disjoint transpositions and have order 2. And if they are two non-disjoint transpositions they would have order 3. 
For order 4 or higher I couldn't really construct more examples, but I think that is mainly because I am limiting myself to the symmetric groups. Perhaps there are finite groups that are more complex that would help me arrive at an answer faster... Any thoughts or hints?
 A: Consider the dihedral group $D_{2n} = \langle a,b\mid a^n = b^2 = 1 = (ab)^2\rangle$ of order $2n$. We have also $D_{2n} = \langle ab,b\rangle$ with both $ab,b$ elements of order $2$, and $a = (ab)b$ has order $n$. This holds for all $n\in\mathbb{N}^*$, $n\ge 3$. Therefore, every positive integer can be a possible order (it is easy to find examples for $n = 1,2$).
A: Keith Conrad gives this example (4.4) in one of his expository papers:
Let
$$ A = \begin{pmatrix} -1 & 1 \\ 0 & 1 \end{pmatrix}, B = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}.$$
Then you can check that
$$ A^2 = I, B^2 = I, AB = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$$
Notice that
$$ (AB)^m = \begin{pmatrix} 1 & m \\ 0 & 1 \end{pmatrix}.$$
So if we look at everything mod $m$, then $A$ and $B$ have order $2$ and $AB$ has order $m$. The group here is $\operatorname{GL}_2(\mathbf{Z}/m)$. The one exception is when $m = 2$ in which case $B \equiv I \pmod 2$ (which isn't order $2$). So this provides an example for all $m \ge 3$.
And, of course, if you don't look mod $m$, this provides an example where $A$ and $B$ have order $2$ and their product has infinite order.
