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Added a sentence to answer the Op's concern.
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user43326
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Let's take $M$ to be $Z/3$ with trivial $S _3$ action for concreteness' sake. Then we have $$H^*(C_3,M)\cong Z/3[x_2]\otimes \Lambda (e_1).$$ Now, $S_3/C_3\cong F_3 ^*$ acts on this ring just by multiplication by constant on $x_2$. So as $a^2=1$ for all $a \in F_3 ^*$ by Fermat's small Fermat's theorem, invariants are $ Z/3[x_2^2]\otimes \Lambda (e_1).$ Thus we pass from period 2 to period 4. In other words, the $S_3$ action differs on dim $4n$ and $4n+2$.

Let's take $M$ to be $Z/3$ with trivial $S _3$ action for concreteness' sake. Then we have $$H^*(C_3,M)\cong Z/3[x_2]\otimes \Lambda (e_1).$$ Now, $S_3/C_3\cong F_3 ^*$ acts on this ring just by multiplication by constant on $x_2$. So as $a^2=1$ for all $a \in F_3 ^*$ by small Fermat's theorem, invariants are $ Z/3[x_2^2]\otimes \Lambda (e_1).$ Thus we pass from period 2 to period 4.

Let's take $M$ to be $Z/3$ with trivial $S _3$ action for concreteness' sake. Then we have $$H^*(C_3,M)\cong Z/3[x_2]\otimes \Lambda (e_1).$$ Now, $S_3/C_3\cong F_3 ^*$ acts on this ring just by multiplication by constant on $x_2$. So as $a^2=1$ for all $a \in F_3 ^*$ by Fermat's small theorem, invariants are $ Z/3[x_2^2]\otimes \Lambda (e_1).$ Thus we pass from period 2 to period 4. In other words, the $S_3$ action differs on dim $4n$ and $4n+2$.

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user43326
  • 3.1k
  • 17
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Let's take $M$ to be $Z/3$ with trivial $S _3$ action for concreteness' sake. Then we have $$H^*(C_3,M)\cong Z/3[x_2]\otimes \Lambda (e_1).$$ Now, $S_3/C_3\cong F_3 ^*$ acts on this ring just by multiplication by constant on $x_2$. So as $a^2=1$ for all $a \in F_3 ^*$ by small Fermat's theorem, invariants are $ Z/3[x_2^2]\otimes \Lambda (e_1).$ Thus we pass from period 2 to period 4.