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Given an extension $1 \to N \to P \to Q \to 1$ of p-groups. Is it true that $$\dim H^\ast(P,\mathbb{F}_p) = \dim \text{im}(res^P_N) + \dim \text{im}(inf^P_Q)$$ where $\dim$ denotes the Krull dimension, $res$ the restriction and $inf$ the inflation homomorphism ?

Note that the image of the restriction resp. inflation is just the fibre resp. base in the $E_\infty$ term of the Hochschild-Serre spectral sequence of the extension.

This formula holds for abelian p-groups and I also checked it for various extensions of groups of order $p^3$ like Quaternion group and Dihedral group.

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1 Answer 1

up vote 5 down vote accepted

No, this isn't true in general. Let $P=N \ltimes Q$. Then $inf^P_Q$ is injective, so $\dim \text{im}(inf^P_Q) = \dim H^\ast(Q,\mathbb{F}_p)$. By a theorem of Evens, $H^\ast(N,\mathbb{F}_p)$ is finitely generated as module over $\text{im}(res^P_N)$. Hence $\dim \text{im}(res^P_N) = \dim H^\ast(N,\mathbb{F}_p)$. Moreover, by a theorem of Quillen, the Krull dimension of the mod-p cohomology ring of a finite group is equal to the p-rank $r_G$ of $G$. So your claim is in the semi-direct product case: $$r_G = r_N + r_Q.$$ Now a counterexample is given by the extraspecial p-group of order $p^3$: $$P=\langle x,y,c\mid x^p=y^p=c^p=[x,c]=[y,c]=1,[x,y]=c\rangle$$ $$N =\langle x,c\rangle \cong (\mathbb{Z}/p)^2,\quad Q=\langle y\rangle \cong \mathbb{Z}/p$$ where $r_P=2, r_N=2,r_Q=1$.

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