Proof that every element of A_5 is an involution or a product of two involutions? It can be verified with brute force that the alternating group on 5 elements ($A_5$) has the property that every member is either an involution or can be written as the product of two involutions.  Is there a simple proof of this fact?
 A: The elements of $\,A_5\,$ are of the form (using the unique decomposition (up to order) of a permutation into disjoint cycles):$$(i\,j)(k\,m)\,,\,(i\,j\,k)\,,\,(i\,j\,k\,m\,n)$$The first one above is an involution, and now we prove the other two are products of involutions:$$(i\,j\,k)=(i\,k)(i\,j)\,\,,\,\,(i\,j\,k\,m\,n)=(i\,n)(i\,m)(i\,k)(i\,j)$$
Added m.k. proposed in her/his comment below that every element in $\,A_5\,$ should\could be expressed as product of involution within $\,A_5\,$ itself. Well:$$(i\,j\,k)=\left[(j\,k)(i\,m)\right]\left[(m\,k)(i\,m)\right]$$$$(i\,j\,k\,m\,n)=(i\,m\,n)(i\,j\,k)\,,\,\text{and now apply the line above}$$
A: In $A_5$, we have four different cycle structures: the identity $(1)$, the $3$-cycles, the $5$-cycles and products of two disjoint transpositions.
The identity is the product of an any involution with itself. Products of two disjoint transpositions are involutions.
For $3$-cycles: $(123) = (13)(12) = (13)(45)(12)(45)$.
For $5$-cycles: $(12345) = (13)(45)(12)(35)$.
Just in case it isn't obvious, the general case $(abc)$ and $(abcde)$ follows by replacing $1, 2, 3, 4$ and $5$ by $a, b, c, d$ and $e$ respectively.
A: Two involutions generate a dihedral group, so we'd like to show that the alternating group on five symbols is a union of dihedral groups.
One conceptual way to do this is to identify the alternating group with PSL(2,4) and use the very easy subgroup structure of such groups $\operatorname{PSL}(2,p^k)$: every element is either a p-element (and so not a product of involutions unless $p^k \mod 4$ is 0 or 1) or a member of one of two conjugacy classes of dihedral subgroups of orders $p^k\pm 1$ (the normalizers of the split and the non split tori).
In our case $p^k=4$, so the only elements not handled by the two tori are the involutions themselves.
