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For the symmetric group $S_3$, there is an inclusion $i:\mathbb{Z}/3\mathbb{Z}\hookrightarrow S_3$. How can I assert that the induced homomorphism $$i^{\ast}:H^{n}(S_3,\mathbb{Z})\rightarrow H^{n}(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})$$ is non-trivial? I am interested in the cases $n=4k$, $k>0$, where $i^{\ast}:\mathbb{Z}/6\mathbb{Z}\rightarrow\mathbb{Z}/3\mathbb{Z}$.

This should be a simple question, but I haven't found an answer.

Also, it should be useful that $H^{\ast}(S_3,\mathbb{Z})$ can be computed with the Lyndon-Hochschild-Serre spectral sequence $$E_{2}^{p,q}=H^{p}(\mathbb{Z}/2\mathbb{Z}\,,\,H^{q}(\mathbb{Z}/3\mathbb{Z}\,,\,\mathbb{Z})),$$ where there is a non-trivial (edge) morphism $$H^{4k}(S_3,\mathbb{Z})\rightarrow E_{\infty}^{0,4k}=H^{0}(\mathbb{Z}/2\mathbb{Z}\,,\,H^{4k}(\mathbb{Z}/3\mathbb{Z}\,,\,\mathbb{Z}))=H^{4k}(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})=\mathbb{Z}/3\mathbb{Z}.$$ But I can't affirm that this is $i^{\ast}$.

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    $\begingroup$ There is a map $H^*(\Bbb Z/3;\Bbb Z) \to H^*(S_3;\Bbb Z)$ called the transfer map. Its composite with $i^*$ (note that $i^*$ is often called restriction) is multiplication by $2$. In particular, the composite is an isomorphism on $H^*(\Bbb Z/3;\Bbb Z)$ for $*>0$. So you conclude that $i^*$ is surjective. $\endgroup$
    – mme
    Commented Nov 3, 2018 at 1:02
  • $\begingroup$ @MikeMiller But how do I obtain this transfer map and how I know that the composition with $i^\ast$ is that? Can you explain or give a specific reference? Thanks. $\endgroup$
    – mrde05
    Commented Nov 3, 2018 at 2:26
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    $\begingroup$ Look at Ken Brown's book on group cohomology, which is a very good reference. In general, if $H$ is a subgroup of $G$, there is a transfer map from $H$-cohomology to $G$-cohomology, whose composite with the restriction is $[G:H]$. In this case that's 2. $\endgroup$
    – mme
    Commented Nov 3, 2018 at 2:31

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