I have been struggling with this question for quite a time now:
Let $f,g$ be mappings of a set $S$ into itself. Assume that $f^2 = g^2 = I$ and that $f \circ g = g \circ f$. Prove that $(f \circ g)^2=I$. Prove that $(f \circ g)^3=I$.
I have proved $(f \circ g)^2 = I$: $$ (f \circ g) \circ (f \circ g)= (f \circ g)^2 = (f^2 \circ g^2)=I \circ g^2= I \circ I = I. $$
But when trying to prove that $(f \circ g)^3=I$ it seems impossible I have tried all I can. Is it even true?
Consider the following counterexample, with transposition permutations $f = (1,2)$ and $g = (3,4)$ in $S_4$ (these are bijections on $\{1,2,3,4\}$ that just swap the pair of elements specified and leave other elements unchanged). Then $f^2 = g^2 = I$ and $f\circ g = g\circ f$ but what is $(f\circ g)^3$?