# Cohomology of Symmetric Group 3 using Lyndon-Hochschild-Serre spectral sequence

For the symmetric group $$S_{3}$$ we have the short exact sequence $$0\rightarrow C_{3}\rightarrow S_{3}\rightarrow C_{2}\rightarrow 0,$$ where $$C_{n}$$ is the cyclic group of order $$n$$. Using the Lyndon-Hochschild-Serre spectral sequence we obtain $$E_{2}^{p,q}=H^{p}(C_{2},H^{q}(C_{3},\mathbb{Z})),$$ where we would have $$0$$ for $$q$$ odd (right?).

So my doubt is that I'm not sure of how to obtain the non-trivial action of $$C_{2}$$ on $$\mathbb{Z}$$ or $$C_{3}$$ when $$q$$ is even.

A trivial action doesn't lead to the correct result, since after adding the diagonals in $$E_{\infty}$$ it should be the cohomology of $$S_{3}$$, which is $$H^{n}(S_{3},\mathbb{Z})=\begin{cases} \mathbb{Z} & n=0 \\ 0 & n \hbox{ odd} \\ C_{2} & n\equiv2 \hbox{ mod 4} \\ C_{6} & n\equiv0 \hbox{ mod 4.} \end{cases}$$

## 1 Answer

Every short exact sequence of groups induces an action of the quotient on the normal subgroup by conjugation. Here, the action is inversion in $$C_3$$; to see this explicitly, observe that $$(12)(123)(12)^{-1} = (132).$$ Anyway, you want to know what the map $$i^*: H^*(C_3; \Bbb Z) \to H^*(C_3;\Bbb Z)$$ induced by inversion is. The first useful fact is that because inversion is a group homomorphism, this map is an algebra homomorphism.

Because $$H^*(C_3;\Bbb Z) = \Bbb Z[c]/(3c)$$, where $$|c| = 2$$, we see that we just need to determine what this map does in $$H^2(C_3;\Bbb Z)$$; it will either be the identity, or multiplication by $$-1$$; in the former case the induced map on $$H^*(C_3;\Bbb Z)$$ is the identity. In the latter case, $$c^k \mapsto (-c)^k = (-1)^k c^k$$, and so $$i^*: H^{4n+2}(C_3;\Bbb Z)$$ is multiplication by $$-1$$, and $$i^*: H^{4n}(C_3;\Bbb Z)$$ is the identity.

To pin this down, recall the universal coefficient theorem gives a natural isomorphism for any finite group $$G$$ $$H^2(G;\Bbb Z) \cong \text{Ext}(H_1(G;\Bbb Z), \Bbb Z),$$ because group cohomology in positive degrees is torsion, and in particular $$\text{Hom}(H_2(G;\Bbb Z), \Bbb Z) = 0.$$

Lastly, conclude with the fact that $$H_1(G;\Bbb Z) \cong G^{\text{ab}}$$ is a natural isomorphism, and $$\text{Ext}(A, \Bbb Z)$$ is naturally equivalent to $$\text{Hom}(A, S^1)$$ for any finite group $$A$$; essentially, this is saying that $$\text{Ext}(A, \Bbb Z)$$ is non-canonically isomorphic to $$A$$ itself, but that the induced maps are dualized; they go in the opposite direction.

In any case, applying all this to $$G = C_3$$, we find that the induced map of inversion on $$H^2(C_3;\Bbb Z) = \text{Ext}(\Bbb Z/3, \Bbb Z)$$ is multiplication by $$-1$$.

Now you need to know the group cohomology of $$H^*(C_2;\Bbb Z/3)$$, where $$\Bbb Z/3$$ is either given the trivial action or the negation action. Using the calculation here, as well as the fact that $$H^0(G;M) = M^G$$ by definition, we see that $$H^*(C_2;\Bbb Z/3) = 0$$ in positive degrees, and that $$H^0(C_2;\Bbb Z/3) = \Bbb Z/3$$ for the trivial action, but $$H^0(C_2;\Bbb Z/3) = 0$$ with the negation action.

(A high-tech calculation: if $$G$$ is a finite group and $$M$$ is a $$G$$-module in which multiplication by $$|G|$$ is an isomorphism, the existence of the transfer map implies that $$H^*(G;M) = 0$$ in positive degrees. In particular, this applies to $$G = \Bbb Z/2$$ and $$M = \Bbb Z/3$$ with any action.)

Now run your spectral sequence. The $$q = 0$$ line is $$\Bbb Z, 0, \Bbb Z/2, 0, \Bbb Z/2, 0, \cdots$$ while the $$p = 0$$ line is $$\Bbb Z, 0, 0, 0, \Bbb Z/3, 0, 0, 0, \Bbb Z/3, 0, \cdots$$

That is, there is a $$\Bbb Z/3$$ on every $$E^{0,4n}_2$$, and it is otherwise zero.

The spectral sequence is completely supported on these axes. Because everything is supported in even bidegree, there can be no nontrivial differentials, and $$E_2 = E_\infty$$. Along with the observation that there are no nontrivial extension problems (because $$\Bbb Z/2$$ and $$\Bbb Z/3$$ only give rise to a single abelian group as extension), we get the desired calculation.

If you also want to know what the product structure is, observe that the map $$H^*(C_2;\Bbb Z) \to H^*(\Bbb S_3; \Bbb Z)$$ sends the degree $$2$$ generator to the degree $$2$$ generator (you can see this at the level of the $$E^2$$ page of your spectral sequence), and is a ring homomorphism.

Similarly, one sees that the homomorphism $$H^*(S_3; \Bbb Z) \to H^*(C_3; \Bbb Z)$$ sends a 3-torsion generator in degree $$4$$ to the generator of $$H^4(C_3;\Bbb Z)$$.

These two facts, combined with the description of the underlying cohomology groups above, gives an isomorphism $$H^*(S_3;\Bbb Z) \cong \Bbb Z[c_1, c_2]/(2c_1, c_2-3c_1^2),$$ where $$|c_i| = 2i$$; this means that $$c_1$$ is $$2$$-torsion and $$c_2 = 3c_1^2$$ is $$6$$-torsion.

• This computation extends without difficulty to any dihedral group $D_{2n+1}$ of order $4n+2$. For $D_{2n}$, you need to pay attention to the extension problems. – user98602 Nov 11 '18 at 2:30