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At quick glance my following solution seems fine, and will edit it otherwise:

Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma. As I remarked in the comment attached to this post, the coefficient module from the latter $\mathbb{Z}$ in the short exact sequence has nontrivial $\Gamma$-action (I will denote this coefficient by $\tilde{\mathbb{Z}}$, where the action is multiplication by $-1$ via elements of the nontrivial coset of $\Gamma/G$). Keeping that in mind, we have in particular:

$H_2(\Gamma,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\stackrel{res}{\rightarrow}H_1\Gamma$.

Exactness implies $Ker(tr)=Im(\delta)=H_2(\Gamma,\tilde{\mathbb{Z}})/Ker(\delta)$, so that $|Ker(tr)|\le |H_2(\Gamma,\tilde{\mathbb{Z}})|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

I want to claim that $|H_2(\Gamma,\tilde{\mathbb{Z}})|=2\cdot|H_2\Gamma|$ (the latter homology has $\mathbb{Z}$-coefficient with trivial action), but at this moment I am unsure how to prove it. I will come back to this soon.

• Chris Gerig

Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma. As I remarked in the comment attached to this post, the coefficient module from the latter $\mathbb{Z}$ in the short exact sequence has nontrivial $\Gamma$-action (I will denote this coefficient by $\tilde{\mathbb{Z}}$, where the action is multiplication by $-1$ via elements of the nontrivial coset of $\Gamma/G$). Keeping that in mind, we have in particular:

$H_2(\Gamma,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\rightarrow H_1(\Gamma,\tilde{\mathbb{Z}})$.

Exactness implies $Ker(tr)=Im(\delta)=H_2(\Gamma,\tilde{\mathbb{Z}})/Ker(\delta)$, so that $|Ker(tr)|\le |H_2(\Gamma,\tilde{\mathbb{Z}})|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

I want to claim that $|H_2(\Gamma,\tilde{\mathbb{Z}})|\le 2\cdot|H_2\Gamma|$ (the latter homology has $\mathbb{Z}$-coefficient with trivial action), but at this moment I am unsure how to prove it. Hatcher's Algebraic Topology textbook gives a long exact sequence for general coefficient systems (pg330), with $H_3G\stackrel{res}{\rightarrow}H_3\Gamma\rightarrow H_2(\Gamma,\tilde{\mathbb{Z}})\rightarrow H_2G\stackrel{res}{\rightarrow}H_2\Gamma$, so this could be of use.

• Chris Gerig

At quick glance my following solution seems fine, and will edit it otherwise:

Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma.

In particular: $H_2\Gamma\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\stackrel{res}{\rightarrow}H_1\Gamma$.

Exactness implies $Ker(tr)=Im(\delta)=H_2\Gamma/Ker(\delta)$, so that $|Ker(tr)|\le |H_2\Gamma|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

• Chris Gerig

At quick glance my following solution seems fine, and will edit it otherwise:

Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma. As I remarked in the comment attached to this post, the coefficient module from the latter $\mathbb{Z}$ in the short exact sequence has nontrivial $\Gamma$-action (I will denote this coefficient by $\tilde{\mathbb{Z}}$, where the action is multiplication by $-1$ via elements of the nontrivial coset of $\Gamma/G$). Keeping that in mind, we have in particular:

$H_2(\Gamma,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\stackrel{res}{\rightarrow}H_1\Gamma$.

Exactness implies $Ker(tr)=Im(\delta)=H_2(\Gamma,\tilde{\mathbb{Z}})/Ker(\delta)$, so that $|Ker(tr)|\le |H_2(\Gamma,\tilde{\mathbb{Z}})|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

I want to claim that $|H_2(\Gamma,\tilde{\mathbb{Z}})|=2\cdot|H_2\Gamma|$ (the latter homology has $\mathbb{Z}$-coefficient with trivial action), but at this moment I am unsure how to prove it. I will come back to this soon.

• Chris Gerig

At quick glance my following solution seems fine, and will edit it otherwise:

Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, in particular:

$H_2\Gamma\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\stackrel{res}{\rightarrow}H_1\Gamma$.

Exactness implies $Ker(tr)=Im(\delta)=H_2\Gamma/Ker(\delta)$, so that $|Ker(tr)|\le |H_2\Gamma|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

• Chris Gerig

At quick glance my following solution seems fine, and will edit it otherwise:

Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma.

In particular: $H_2\Gamma\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\stackrel{res}{\rightarrow}H_1\Gamma$.

Exactness implies $Ker(tr)=Im(\delta)=H_2\Gamma/Ker(\delta)$, so that $|Ker(tr)|\le |H_2\Gamma|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

• Chris Gerig