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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.

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  • $\begingroup$ Where did this problem arise? (This looks rather like an exercise) $\endgroup$
    – Yemon Choi
    Apr 12, 2015 at 11:38
  • $\begingroup$ @YemonChoi I was calculating the group of all linear fractional transformations that fix n distinct points on the extended complex plane. When n=6 and the the six points are 0, \infty, 1, i, -1, -i, the result is the group above. $\endgroup$
    – user150248
    Apr 12, 2015 at 12:03
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    $\begingroup$ According to GAP, this is $S_4$. $\endgroup$ Apr 12, 2015 at 12:03
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    $\begingroup$ The description in the comment gives the orientation-preserving symmetries of the octahedron. $\endgroup$
    – Sam Nead
    Apr 12, 2015 at 12:52

1 Answer 1

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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.

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