For basic definition of group cohomology, one can read it the book of Lorenz Falco (Algebra II). We will now treat group (co)homology in terms of derived functors, as presented in the lecture notes of Milne. In the first section, we will introduce about induced modules. Roughly speaking, it defines an exact functor from the category of $H$-modules to the category of $G$-modules, where $H$ is a subgroup of $G$. Especially, when $H=\{1\}$, then we obtain an exact functor from the category of abelian groups to the category of $G$-modules.
Definition. Let $G$ be a group, an abelian group $M$ is said to be a $G$-module if there exists an action from $G$ to $M$, and $g(m_1+m_2)=gm_1+gm_2$, for all $g\in G, m_1,m_2\in M$. Let $M,N$ be $G$-modules, a map $\varphi: M\to N$ is said to be a $G$-module homomorphism if $\varphi$ is a homomorphism of abelian groups, and $\varphi(gm)=g\varphi(m)$, for all $g\in G,m\in M$. The set of all $G$-module homomorphisms between $M$ and $N$ is denoted by $Hom_G(M,N)$. We also denote $Mod_G$ the category with objects consisting $G$-modules, and morphisms are $G$-modules homomorphisms.
Let $G$ be a group and $H$ a subgroup of $G$, and $M$ is an $H$-module, we denote:
$$Ind^G_H(M):=\{\varphi: G\to M|\varphi(hg)=h\varphi(g), \forall h\in H,g\in G\}$$
One can equip $Ind^G_H(M)$ a structure of $G$-module by defining $(\varphi_1+\varphi_2)(g)=\varphi_1(g)+\varphi_2(g)$, and $(g_1\varphi)(g)=\varphi(gg_1)$.
Lemma 2.1. The map $\phi_M: Ind^G_H(M)\to M$ defined by $\varphi\mapsto \varphi(1_G)$ is an $H$-module homomorphism.
Proof. It can be seen that $\phi$ is a homomorphism of abelian groups, and
$$\phi_M(h\varphi)=(h\varphi)(1_G)=\varphi(1_Gh)=\varphi(h1_G)=h\varphi(1_G)=h\phi_M(\varphi)$$
(Q.E.D)
Lemma 2.2. With the assumption in Lemma 2.1, when $H=G$, then $\phi_M$ is an isomorphism.
Proof. We have $Ind^G_G(M)=\{\varphi:G\to M|\varphi(g_1g_2)=g_1\varphi(g_2),\forall g_1,g_2\in G\}$, and any $\varphi$ in this set is completely determined by $\varphi(1)$. And the map $\phi_M$ above is injective and surjective. (Q.E.D)
Proposition 2.3. Let $G$ be a group, $H$ a subgroup of $G$, $M$ a $G$-module, and $N$ an $H$-module, then there exists a canonical isomorphism $Hom_G(M, Ind^G_H(N))\cong Hom_H(M,N)$.
Proof. Let $\phi$ be a $G$-module homomorphism from $M$ to $Ind^G_H(M)$, then the map $\phi':=\phi_N\circ \phi: M\to N$, where $\phi_N$ is defined in Lemma 2.1 is an $H$-module homomorphism.
Conversely, if we begin with $\phi':M\to N$ an $H$-module homomorphism, then for any $\alpha\in Ind^G_G(M)$, we have $\phi'\circ \alpha\in Ind^G_H(M)$, since for all $h\in H,g\in G$, we have
$$(\phi'\circ \alpha)(hg)=\phi'(h\alpha(g))=h(\phi'\circ \alpha(g))$$
And we hence obtain a map $\phi: Ind^G_G(M)\to Ind^G_H(N)$ defined by $\alpha\mapsto \phi'\circ \alpha$, and it can be seen that for all $g,g_1\in M$, we have
$$\phi(g\alpha)(g_1)=\phi'\circ (g\alpha)(g_1)=\phi'\circ\alpha(g_1g)=(g(\phi'\circ\alpha))(g_1)=g\phi(\alpha)(g_1)$$
Hence, $\phi$ is a homomorphism of $G$-modules. Due to Lemma 2.2, we have $M\cong Ind^G_G(M)$, and hence, one obtains $Hom_G(M,Ind^G_H(N))\cong Hom_H(M,N)$. (Q.E.D)
Via this theorem, we obtain the universal property of induced modules.
Corollary 2.4. With the assumption in Proposition 2.3, then for any $H$-module homomorphism $\phi'$ from $M\to N$, there exists one and only one $G$-module homomorphism $\phi$ from $G$ to $Ind^G_H(N)$, such that $\phi_N\circ \phi=\phi'$.
Via the proof of Proposition 2.3, if $M,N$ are $H$-modules, and $\phi'\in Hom_H(M,N)$, then one can construct $\phi: Ind^G_H(M)\to Ind^G_H(N)$ defined by $\alpha\mapsto \phi'\circ\alpha$, and $\phi$ is a $G$-module homomorphism. We hence obtain a functor from $Mod_H$ to $Mod_G$ defined by $M\mapsto Ind^G_H(M)$.
Proposition 2.5. The functor $Ind^G_H$ defined above is exact.
Proof. Let
$$0\to M\xrightarrow{i} N\xrightarrow{p} P\to 0$$
be an exact sequence of $H$-modules, then via the functor $Ind^G_H$, the induced map from $Ind^G_H(M)$ to $Ind^G_H(N)$ is defined to be $\alpha\mapsto i\circ \alpha$, and the induced map $Ind^G_H(N)\to Ind^G_H(P)$ is defined to be $\beta\mapsto p\circ\beta$, and this yields the following complex of $G$-modules
$$0\to Ind^G_H(M)\to Ind^G_H(N)\to Ind^G_H(P)\to 0$$
The exactness at $Ind^G_H(M)$ is clear. We will prove next the exactness at $Ind^G_H(P)$. Taking any $\gamma\in Ind^G_H(P)$, one can represent $G=\bigcup_{i\in I}Hs_i$, where $\{s_i\}_{i\in I}$ are representatives of right coset of $H$ in $G$. One has $\gamma(hs_i)=h\gamma(s_i)$. Let $n_i\in N$, such that $p(n_i)=\gamma(s_i)$. If we defined $\beta(hs_i):=hn_i$, then the map $\beta: G\to N$ is well-defined, and $p\circ \beta=\gamma$. Moreover, for any $g\in G$ there exists a unique $s_i$ such that $g\in Hs_i$, and hence, for all $h\in H$, there exists $h'$ such that $hg=h's_i$, i.e. $g=h^{-1}h's_i$, and
$$\beta(g)=\beta(h^{-1}h's_i)=h^{-1}(h'n_i)=h^{-1}\beta(h's_i)=h^{-1}\beta(hg)$$
And this yields $h\beta(g)=\beta(hg)$, and $\beta\in Ind^G_H(N)$. This yields the surjectivity of the map $Ind^G_H(N)\to Ind^G_H(P)$.
To complete the proof, we have to show that this complex is also exact at $Ind^G_H(N)$. Let $\beta\in Ind^G_H(N)$, such that $p\circ \beta=0$. From the original exact sequence, for any $g\in G$, there exists a unique $m_g\in M$ such that $\beta(g)=i(m_g)$. And one can define the map $\alpha: G\to M$ defined by $g\mapsto m_g$, and $\alpha(hg)=m_{hg}$, and $i(m_{hg})=\beta(hg)=h\beta(g)=hi(m_g)=i(hm_g)$. And since $i$ is injective, we have $m_{hg}=hm_g$, and this yields $\alpha(hg)=h\alpha(g)$, for all $h\in H,g\in G$. From this, $\alpha\in Ind^G_H(M)$ with $i\circ \alpha=\beta$. And it follows that $Ind^G_H$ is an exact functor. (Q.E.D)
When $H=\{1\}$, an $H$-module is just an abelian group. Let $M_0$ be an abelian group, we have $Ind^G(M_0):=Ind^G_{\{1\}}(M_0)=\{\alpha:G\to M_0 \text{ as sets}\}$. We said that a $G$-module $M$ is induced if there exists an abelian group $M_0$ such that $M\cong Ind^G(M_0)$ as $G$-modules.
If we denote
$$\mathbb{Z}[G]:=\{\sum_{g\in G}n_gg|n_g\in \mathbb{Z}\text{ and }n_g\ne 0\text{ for finitely many } g\in G\}$$
then $\mathbb{Z}[G]$ becomes a $G$-module under the action $g'(\sum_{g\in G}n_gg):=\sum_{g\in G}n_g(g'g)$. And this follows easily that $Ind^G(M_0)=Hom_{Ab}(\mathbb{Z}[G],M_0)$. And so, when $G$ is finite, we get
$$Ind^G(M_0)\cong Hom_{Ab}(\mathbb{Z}[G],M)\cong \mathbb{Z}[G]\otimes_\mathbb{Z}M_0$$
Proposition 2.6. Let $G$ be a group, and $M$ is a $G$-module.
(i) Let $M_0$ be the underlying abelian group structure of $M$, then there exists an embedding of $G$-modules $\phi: M\to Ind^G(M_0)$ defined by $m\mapsto \varphi_m$, where $\varphi_m(g)=gm$, for all $g\in G$.
(ii) When $G$ is finite, $M$ is a quotient of an induced module.
Proof.
(i) The map $\phi$ is obvious injective, and $\phi(g'm)=\varphi_{g'm}$, where $\varphi_{g'm}(g)=gg'm$, and $(g'\phi(m))(g)=(g'\varphi_m)(g)=\varphi_m(gg')=gg'm$. Hence, $\phi(g'm)=g'\phi(m)$, for all $g'\in G,m\in M$. This yields $\phi$ is an embedding of $G$-modules.
(ii) Let $M_0$ be the underlying abelian group structure of $M$, we can have a map $\phi$ from $Ind^G(M_0)$ to $M$ defined by $\varphi\mapsto \sum_{g\in G}g\varphi(g^{-1})$. It can be seen that $\phi$ is a surjective homomorphism of abelian groups. Moreover, we have
$$\phi(g_1\varphi)=\sum_{g\in G}g(g_1\varphi)(g^{-1})=\sum_{g\in G}g\varphi(g^{-1}g_1)$$
And $g_1\phi(\varphi)=\sum_{g\in G}g_1g\varphi(g^{-1})$. Let $g'^{-1}=g^{-1}g_1$, then $g'=g_1^{-1}g$, and $g_1g'=g$. And this yields
$$\sum_{g\in G}g\varphi(g^{-1}g_1)=\sum_{g'\in G}(g_1g')\varphi(g'^{-1})=\sum_{g\in G}g_1g\varphi(g^{-1})$$
So $\phi(g_1\varphi)=g_1\phi(\varphi)$. Hence, the map $\phi$ above is a $G$-module homomorphism. (Q.E.D)
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