The group $\langle a,b,c \ | a^3,b^2, ab=ba^2, c^2, ac=ca, bc=cb \rangle$?

Question

The group $\langle a,b,c \ | a^3,b^2, ab=ba^2, c^2, ac=ca, bc=cb \rangle$ is isomorphic to which permutation group.

I have calculated its order and it is $12$, so my guess was $A_4$ but it is not working out as I tried to satisfy relations, which group it is isomorphic to?

Answer 1

Note that $c$ commutes with $a$ and $b$, so we have immediately that

$G=\langle a,b,c \ | a^3,b^2, ab=ba^2, c^2, ac=ca, bc=cb\rangle = H \oplus \langle c \ | c^2=1 \rangle = H \oplus \, \mathbb{Z}_2$

for some group $H$ of order 6. So how many groups of order 6 are there...?

Answer 2

Let $G = \langle a,b,c |a^3,b^2,ab=ba^2,c^2,ac=ca,bc=cb\rangle$. The subgroup $H$ generated by $\{a,b\}$ is isomorphic to $S_3$. The element $c$ commutes with $H$, and so lies in the center of $G$, which intersects $H$ trivially (since $H$ has trivial center). So $G$ is the direct product $H \times \langle c\rangle$.

This group is not isomorphic to a subgroup of $S_4$, since in $S_4$ there are no pairs of commuting elements $a,c$ such that $a$ has order 3 and $c$ has order 2.

This group is isomorphic to a subgroup of $S_5$. Let $H$ be the subgroup which fixes the last two points, and let $c$ be the transposition which interchanges the last two points.