 Solution

For $1\to H\to G\to Q\to 1$, filter a bar resolution of $G$ by the number of quotient variables, or equivalently use the double complex obtained from projective resolutions over $\mathbb{Z}H$ and $\mathbb{Z}Q$. Taking cohomology first in the $H$-direction and then in the $Q$-direction produces the Lyndon–Hochschild–Serre spectral sequenceThe quotient acts on $H^q(H,\mathbb{Z})$ through conjugation. Differentials have bidegree $(r,1-r)$; after determining the $E_2$-page, one follows these differentials and then resolves the filtration extensions on each total degree.

For the dihedral group of order ten, write $G=D_{10}=C_5⋊C_2$, with $H=C_5$ and $Q=C_2$. The integral cohomology of a finite cyclic group isIf $u$ generates $H^2(C_5,\mathbb{Z})$, inversion acts by $u\mapsto -u$, and hence by $(-1{)}^r$ on $u^r\in H^{2r}(C_5,\mathbb{Z})$. Since multiplication by $2$ is invertible on $\mathbb{Z}/5$, every positive-degree cohomology group of $C_2$ with coefficients in this module vanishes. Its invariants are $\mathbb{Z}/5$ when $q$ is divisible by four and zero when $q\equiv 2(\mathrm{mod}4)$.

Thus the only nonzero terms aretogether with the bottom rowDegree considerations leave no possible nonzero differential, so the spectral sequence collapses. In total degrees divisible by four, the $\mathbb{Z}/2$ and $\mathbb{Z}/5$ filtration factors combine uniquely as $\mathbb{Z}/10$ because their orders are coprime. Therefore the integral cohomology of the dihedral group of order ten is