Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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 $r_P=2, r_N=2,r_Q=1$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.