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?


Thanks in advance.



Examples of p-groups exhibiting isomorphic mod-p cohomology rings.
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](http://hopf.math.purdue.edu/Browder-Pakianathan/pakianat.pdf) of Browder and Pakianathan (Theorem 1 and the example after Prop. 2.5). 


Remark 1: There is a classification of finite groups those mod-p cohomology ring equals the cohomology of an abelian p-group. This can be used to obtain further examples. As reference see Theorem B in the following [paper](http://www.ams.org/journals/tran/2000-352-09/S0002-9947-99-02385-5/S0002-9947-99-02385-5.pdf) of Weigel. 


Remark 2: In the examples above only the cohomology rings as abstract rings agree. I can remember of having seen somewhere an example of non-isomorphic 2-groups of the same order those cohomology rings even agreed as Steenrod algebras. 


