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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?

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I think this is it. You can think of it this way: fix a normal copy of $\mathbb{Z}$ in your extension with quotient $S_3$. Consider the following two properties of the extension: $C_2\leq S_3$ acts either trivially or non-trivially on $\mathbb{Z}$; and the extension is either split or non-split. All four combinations can occur, as your four examples show, and I think that these two properties determine the extension in this particular case. –  Alex B. Mar 30 '12 at 13:22
    
$H^2$ does not classify the extensions up to isomorphism! –  user641 Mar 30 '12 at 16:04

1 Answer 1

up vote 4 down vote accepted

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.

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