# Examples of p-groups exhibiting isomorphic mod-p cohomology rings.

Hello,

given 2 different finite $p$-groups $G$ and $H$, $|G|=|H|=p^n$. It has been shown by Ian Leary that the mod-p cohomology rings do not determine the groups $G$ and $H$. In fact he gave an example of such groups. Are there more known examples of pairs of non-isomorphic groups that give rise to the same cohomology ring?

• Does the example of Leary covers abelian groups ? For example $\mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p^3\mathbb{Z}$ and $\mathbb{Z}/p^2\mathbb{Z} \oplus \mathbb{Z}/p^2\mathbb{Z}$ have ismomorphic mod-p cohomology rings (for odd p). May 16 '11 at 23:10
• Thanks for your example given in your answer. Concerning your example above. It is not really surprising. For every $n\in\mathbb{N}$ the cohomology of $C_{p^n}$ equals that of $C_p$. As we consider cohomology with coefficients a field the following holds: $H^\ast(C_{p^2}\oplus C_{p^2},\mathbb{F}_p)\cong H^\ast(C_{p^2}, \mathbb{F}_p)\otimes H^\ast(C_{p^2}, \mathbb{F}_p)\cong H^\ast(C_{p},\mathbb{F}_p)\otimes H^\ast(C_{p}, \mathbb{F}_p)$. And the same holds any other direct sum of 2 cyclic p-groups. Much more surprising is the isomorphism for an non-abelian group and a abelian group. May 17 '11 at 10:39
• My example in fact concerns integral cohomology rings of $p$-groups. There are no abelian examples because $H^2(G;\mathbb{Z})\cong \mathrm{Hom}(G,U(1))$ determines $G$ uniquely.
– IJL
Apr 25 '19 at 11:10

Let $\Gamma$ be the kernel of the reduction map $GL_n(\mathbb{Z}/p^3\mathbb{Z}) \to GL_n(\mathbb{Z}/p\mathbb{Z})$. $\Gamma$ is a non-abelian group of order $p^{2n^2}$ and $p$-rank $n^2$. Its mod-$p$ cohomology is $$H^\ast(\Gamma;\mathbb{F}_p) = \mathbb{F}_p[x_1,...,x_{n^2}] \otimes \Lambda(y_1,...,y_{n^2}).$$ In particular the cohomology ring equals those of the abelian group $(\mathbb{Z}/p^2\mathbb{Z})^{n^2}.$ As a reference for the cohomology of $\Gamma$ see this paper of Browder and Pakianathan (Theorem 1 and the example after Prop. 2.5).