MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is set on a finite $2$-group $G$ and a subgroup $H$ of index $2$ (but perhaps the question could be answered for arbitrary orders/indexes).

It was asked here on MO whether $|\text{Ker}(res^G_H)|\le 2\cdot|H_2(G,\mathbb{Z})|$, where $res^G_H:H_1(G,\mathbb{Z})\rightarrow H_1(H,\mathbb{Z})$ is the restriction map in dimension $1$ (also known as the Verlagerung transfer map $Ver$). Using the integral coefficients module $\tilde{\mathbb{Z}}$ with the twisted $G$-action $g\cdot z=-z$ for $g\notin H$, I was able to establish the relation $|\text{Ker}(res^G_H)|\le |H_2(G,\tilde{\mathbb{Z}})|$.

Is there any feasible way to obtain $H_2(G,\tilde{\mathbb{Z}})$ from $H_2(G,\mathbb{Z})$, or even just relations of orders?

There are the long exact sequences $\cdots\rightarrow H_2(G,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow} H_1(G)\stackrel{res}{\rightarrow}H_1(H)\rightarrow H_1(G,\tilde{\mathbb{Z}})$ and $\cdots\rightarrow H_2(G,\tilde{\mathbb{Z}})\rightarrow H_2(H)\stackrel{cor}{\rightarrow} H_2(G)\stackrel{\partial}{\rightarrow} H_1(G,\tilde{\mathbb{Z}})$ which arise from the short exact sequences of modules $\mathbb{Z}\hookrightarrow Ind^G_H\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ with the twisted-action on either the first or last $\mathbb{Z}$, respectively. But this ultimately leaves you with a relation based on images and kernels that doesn't give sufficient information. There is also the Serre spectral sequence $E_2^{p,q}=H_p(G/H,H_q(H,\tilde{\mathbb{Z}}))\Rightarrow H_{p+q}(G,\tilde{\mathbb{Z}})$, but it's too tough to crack open, and doesn't use the untwisted-homologies of $G$.

Generally, is there a way to obtain $H_n(G,\tilde{M})$ from $H_n(G,M)$ in dimension $n\ge 1$ where $G$ is a $p$-group for arbitrary prime $p$? Here $\tilde{M}$ just refers to some twisting that you can put on some $M$.

share|cite|improve this question
Last sentence: how do you define the $G$-action on $\tilde{\mathbf{Z}}$ if $G$ is a $p$-group and $p$ isn't 2? – Kevin Buzzard May 22 '11 at 18:32
PS for what it's worth, my gut feeling is that there is basically no relation at all between the cohomology of a module and the cohomology of a twist of it. Geometrically you're asking something like this: say we have a vector bundle on a manifold and we know something about its cohomology. Now if I twist the bundle what can we say about the cohomology? Even in degree 0, surely the simplest case, the answer is "nothing much", because if there are no global sections then who knows whether one will appear after twisting. – Kevin Buzzard May 22 '11 at 19:21
You can get a little info from the sequences $0\to \mathbb Z\to \mathbb Z\to \mathbb Z/2\to 0$ and $0\to\tilde{\mathbb Z}\to\tilde{\mathbb Z}\to \mathbb Z/2\to 0$. The resulting homology exact sequences imply that the rank (i.e. number of cyclic summands) of $H_n(G,\tilde{\mathbb Z})$ is one more than the rank of $H_n(G,\mathbb Z)$ if $n>0$ is even, one less if $n$ odd. – Tom Goodwillie May 23 '11 at 2:18
On the other hand, you can make examples where $H_2(G,\tilde{\mathbb Z})=\mathbb Z/2\times\mathbb Z/2$ and $H_2(G,\mathbb Z)$ has arbitrarily large order. – Tom Goodwillie May 23 '11 at 16:19
I deleted my first attempt to give such an example, because it was wrong. Here's an example where the twisted $H_2$ is big (roughly $\mathbb Z/2^k$ and the untwisted $H_2$ is small. $H=\mathbb Z/2^k\times \mathbb Z/2^k$ and $G$ is the semidirect product for the action of $\mathbb Z/2$ that switches the two factors. – Tom Goodwillie May 23 '11 at 22:10

Your Answer


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

Browse other questions tagged or ask your own question.