Let $\alpha = (1\ 2\ 4\ 5\ 6) (23\ 24\ 26\ 7\ 9) (3) (17) (13\ 20\ 21\ 8\ 22\ 19\ 25) (9\ 11\ 12\ 14\ 15\ 16\ 10)$ be a permutation.
Express the permutation as a product of two involutions without fixed elements in two different ways
I have computed that $\alpha = (9\ 11\ 12\ 14\ 15\ 16\ 10\ 23\ 24\ 26\ 7)(13\ 20\ 21\ 8\ 22\ 19\ 25) (1\ 2\ 4\ 5\ 6)(3)(17)(18)$. Hence, the cycle type of $\alpha$ is $(11,7,5,1,1,1)$.
We know that if two permutations $\beta$ and $\delta$ are involutions without fixed elements, then each number of the cycle type of $\beta \delta$ appears an even number of times.
Since in the type of $\alpha$ no number appears an even number of times, that means we cannot express $\alpha$ as a product of two involutions without fixed elements? Am I doing something wrong?