I encountered a group $G =\langle(1,3,2,4),(3,5,4,6)\rangle\subseteq S\_6$ in my study, but I do not know its name.


Let $f=(1,3,2,4)$ and $g=(3,5,4,6)$. We have $g^2=fg^2f$, and thus $\langle f,g^2\rangle\simeq D\_4$. I have proved that $\langle f,g\rangle=\langle f,g^2\rangle \cup \:g\langle f,g^2\rangle \cup\:fg \langle f,g^2 \rangle$, so $G$ is of order 24.


I think this specific group must have a name. Thanks a lot if you tell me its name.



The name of a group of order 24
It is an easy exercise to show that $G$ is isomorphic to $S\_4$ in its action on $6=\binom{4}{2}$ pairs of the four points of its natural action.


