All possible extensions of $S_3$ by $\mathbb{Z}$? Write $S_3$ for the symmetric group on 3 letters.
The question:
What are the possible extensions of $S_3$ by $\mathbb{Z}$ (up to equivalence)? (To avoid ambiguity, by an extension of $G''$ by $G'$ I mean a short exact sequence of groups $1 \to G' \to G \to G'' \to 1$.)
It is well-known that these are enumerated by the second cohomology group of $S_3$ with $\mathbb{Z}$-coefficients. There are two $S_3$-module structures on $\mathbb{Z}$. I do know that $H^2(S_3;\mathbb{Z}) = \mathbb{Z}/2$; here $\mathbb{Z}$ denotes the trivial $S_3$-module. I am, however, unsure about $H^2(S_3; \tilde{\mathbb{Z}})$ -- what I came up thus far is that it is either $\mathbb{Z}/2$ or $\mathbb{Z}/6$. EDIT: Actually, $H^2(S_3;\tilde{\mathbb{Z}}) = \mathbb{Z}/3$; I did my computations wrong previously.
I can think of only three extensions of $S_3$ by $\mathbb{Z}$: the obvious ones, $\mathbb{Z} \times S_3$ and $\mathbb{Z} \rtimes S_3$, and the infinite dihedral group $\mathbb{Z} \rtimes \mathbb{Z}/2$. EDIT: The fourth one is $\mathbb{Z}/3 \rtimes \mathbb{Z}$, which leaves me with only one group missing. Any ideas?
 A: There are only four such groups.  Consider the preimage of $A_3$ in $G$.  This is a normal subgroup $H$ of index $2$, and $H$ has an infinite cyclic subgroup of index $3$. If $H$ is torsion-free, it is isomorphic to $\mathbb{Z}$, and then since $G$ is not abelian we would have $G\cong D_\infty$.
Otherwise, $H$ has torsion, and so is a semidirect product $\mathbb{Z}\rtimes \mathbb{Z}/3$. Since $\mathbb{Z}/3$ cannot act non-trivially on $\mathbb{Z}$, this is a direct product, and $H\cong \mathbb{Z}\times\mathbb{Z}/3$.
Now similar reasoning holds for $K$, the preimage in $G$ of a transposition in $S_3$.  $K$ is either infinite cyclic, or a semidirect product $\mathbb{Z}\rtimes\mathbb{Z}/2$. Also, for $G$ to have quotient $S_3$, we must act by inversion on $\mathbb{Z}/3$ (which is normal in $G$).
If $K$ is infinite cyclic, we get the group $\mathbb{Z}/3\rtimes\mathbb{Z}$.  If $K$ is $\mathbb{Z}\times\mathbb{Z}/2$, then $G$ is $\mathbb{Z}\times S_3$. And if $K$ is $\mathbb{Z}\rtimes\mathbb{Z}/2$, then $G$ is $\mathbb{Z}\rtimes S_3$.
So, four groups total.
