The task: prove that every bijection of a set into itself is representable as a composition of two symmetries (on this set). There is no presumption on finiteness or infinity of the set.   Proof (where I am at now and looking for insights):  1) Writing down the task into math language  Let $f$ be bijection of the set $X$ into itself: $f: X\leftrightarrow X$.  Let $h$ be a symmetry on the set $X$ that means: $h: X\leftrightarrow X$ and $h^{-1} = h$.  Let $g$ be a symmetry on the set $X$ that means: $g: X\leftrightarrow X$ and $g^{-1} = g$.  2) Need to prove that $$\forall f:X\leftrightarrow X \exists (h:X\leftrightarrow X, h^{-1}=h \; and\; g:X\leftrightarrow X, g^{-1}=g):f= g\circ h.$$  3) Proof  Prove that there are $h$ and $g$ so that $f=g\circ h$.  Here is where I am getting into the problem. While   I know that composition of two bijections is bijection. I feel intuitively that every bijection can be represented as a composition of two bijections (by permutation in the middle - not sure that permutation thinking is valid for infinite sets).   I am struggling to get to the proof of that for every bijection $f:X\leftrightarrow X$ among many of the compositions of bijections $X\leftrightarrow X$ that equals $f$ - there are at least one composition that consists of two symmetries.
How can we prove that any bijection on any set is  a composition of 2 involutions ?  Since involutions are bijections mapping elements of a set to elements of the same set, I find it weird that this applies to any bijection.  Thanks for your help !
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