Find groups that contain elements $a$ and $b$ such that $|a|=|b|= 2$ and $|ab|=5$ Find groups that contain elements $a$ and $b$ such that $|a|=|b|= 2$ and $|ab|=5$

My thoughts:
$|a|=|b|=2\implies a^2=e$ and $b^2=e$
I see that the group cannot be abelian as the order wont be greater than $2$ :  $(ab)^2 = a^2b^2=e$
Not really sure how to proceed further. Greatly appreciate any help. Thanks!

Here's one example: in the symmetric group $S_5$
$$(1\ 2\ 3\ 4\ 5) = (1\ 2)(3\ 5)\cdot(2\ 5)(3\ 4),$$
so take $a=(1\ 2)(3\ 5)$ and $b=(2\ 5)(3\ 4).$
You can get many more examples like this using the fact that any permutation  can be expressed as the product of two involutions. So, for another concrete example,
$$(1\ 2\ 3\ 4\ 5)(6\ 7\ 8\ 9\ 10)=(1\ 2)(3\ 5)(6\ 7)(8\ 10)\cdot(2\ 5)(3\ 4)(7\ 10)(8\ 9).$$
