Show that $\langle a,b,c;a^3,b^2,ab=ba^2 , c^2, ac=ca,bc=cb \rangle$has order 12 and find the permutation group isomorphic to it!
I know that $S_3$ is presented by $\langle a,b;a^3,b^2,ab=ba^2\rangle$ where $a \to (xyz)$ and $b\to (xz)$ and I think there's some connection between these two groups
There is a nice fact in which you can see why @Don did that nice comment first. This fact tells us if $G=\langle X\mid R\rangle$ and $H=\langle Y\mid S\rangle$ then $$G\times H=\langle X,Y\mid R,S,T\rangle$$ where $T=\{aba^{-1}b^{-1}\mid a\in X, b\in S\}$. Now, as we know the presentation of $S_3$: $$S_3=\langle a,b\mid a^3=b^2=1,ab=ba^2\rangle$$ so we need to consider $H=\mathbb Z_2 =\langle c\rangle$ to have a presentation for new group $S_3\times\mathbb Z_2$ of order $12$.