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I've got a question concerning the cohomology of $S_3$, the symmetric group on three letters. According to my calculations, its first cohomology group with coefficients in arbitrary $\mathbb{Z}[S_3]$-module is equal to (notation below) $$ H^1(S_3;M)={}_{N_2}\left(\frac{{}_{N_3}M}{\omega-1}\right)\oplus\frac{{}_{N_2}M^{C_3}}{\gamma-1}. $$ Here I treat $S_3$ as the semi-direct product $C_3\rtimes C_2$. The notation is:

  • $\omega$ is the generator of $C_3$
  • $\gamma$ is the generator of $C_2$
  • $N_3=1+\omega+\omega^2$
  • $N_2=1+\gamma$
  • $_{x}M$ denotes elements of $M$ such that $xm=0$, i.e., they are $x$-torsion.

The question I have is: what is the example of a $\mathbb{Z}[S_3]$-module such that both summands are non-zero?

I can think of some examples where either one is non-zero - if we think of $\mathbb{F}_3$ as a $\mathbb{Z}[S_3]$-module with the action given by the sign of a permutation, then the first summand is equal to $\mathbb{F}_3$ and the second is zero - this is because division by $\gamma-1$ in the second summand makes the elements of ${}_{N_2}\mathbb{F_3}^{C_3}$ 2-torsion, and since they were initially 3-torsion, whole module becomes zero. On the other hand, if we take $\mathbb{F}_2$ with trivial action, the first summand becomes zero. But I cannot come up with an example where both summands are zero.

Additionally, if anybody knows how to simplify the expression of $H^1(S_3;M)$, or wants to dispute my calculations, this will also be very welcome :)

Thanks!

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    $\begingroup$ I haven't checked your calculations, but if you have one module $M$ for which the first summand is nonzero, and another module $M'$ for which the second summand is nonzero, then won't both summands be nonzero for $M\oplus M'$? $\endgroup$ Commented Jun 17 at 16:45
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    $\begingroup$ An common and simple example is $M = \mathbb{Z}/6\mathbb{Z}$, with the $S_3$ action defined as follows:ω·m=m,γ·m＝-m. $\endgroup$ Commented Jun 17 at 18:17
  • $\begingroup$ So the sign action on $\mathbb{Z}/6$. Seems about right, thanks! $\endgroup$ Commented Jun 17 at 18:23

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