I have these two permutations of $S_{12}$:
$\alpha =(1\;3\;5\;7\;9)(2\;4\;6\;8\;10)(11\;12)$
$\beta=(1\;6\;8\;10)(2\;3\;5\;7)(4\;9)(11\;12)$
I need to prove that if $G$ is a subgroup of $S_{12}$ and $\alpha,\beta\in G$ then exist a permutation $\gamma \in G$ such that $\gamma(1) = 4$
Any hint? I have no idea how to demonstrate it.
HINT: In $\alpha$ you see that $1$ and $9$ are in a cycle together, and in $\beta$ you see that $9$ and $4$ are in a cycle together, so...