TITLE: Dicyclic group as subgroup of $S_6$?
QUESTION [5 upvotes]: I have the dicyclic group $G$ of order 12 generated by $x,y$ satisfying $x^4 = y^3 = 1$ and $xyx^{-1} = y^2$, and am trying to determine whether the symmetric group $S_6$ contains a subgroup isomorphic to it. 
So far I've tried looking for an appropriate set of 6 elements for $G$ to act on, and hoping that the permutation representation $ \phi: G \to S_6$ is injective, but haven't had any luck: $ \phi$ is not injective for the action of $G$ conjugating its set of 6 elements of order 4, nor is it injective for the action of $G$ translating its set of 6 cosets of a subgroup of order 2.
Is it even true that $S_6$ does contain such a subgroup isomorphic to $G$, and if so how would I construct the isomorphism?

REPLY [3 votes]: If $y$ is of order $3$, it must have cycle type $(abc)$ or $(abc)(def)$. If $x$ is of order $4$, it must have cycle type $(abcd)$ or $(abcd)(ef)$. Also, you know how $x$ acts on $y$ by conjugation...