# Step in proof that a quotient is isomorphic to cohomology group

The following proof is about the cohomology group. But the fact I do not understand is something like, we have two groups $U \cong V$ and two normal subgroups $N \cong M$ in $U$ respectively $V$, what could be said about the quotients. In general this does not imply $U / N \cong V / M$, as for example $G := \mathbb Z_2 \times \mathbb Z_4$ is a counter-example, it has two normal subgroups isomorphic to $\mathbb Z_2$, but one has $\mathbb Z_4$ as factor group, the other $\mathbb Z_2 \times \mathbb Z_2$. What is valid is, that if the isomorphism is implemented by one mapping, that we have isomorphic quotients, i.e. if $\varphi : G \to H$ is injective, then $G / N \cong \varphi(G) / \varphi(N)$.

In the following proof at the end we have two isomorphisms, hence it is concluded that the quotients are isomorphic. As I do not see why this works here I give all the details of the proof. Unfortunately the definitions are a little bit technical.

Let $G$ be a group. A $G$-module is an abelian group $M$ such that

(1) $m\cdot 1 = m$

(2) $m(gh) = (mg)h$

(3) $(m+n)g = mg + ng$ i.e. $G$ acts on $M$ (condition (1) and (2)) and $M$ is an $G$-operator group (condition (3)), i.e. each element gives rise to an automorphism of $M$.

Define the abelian groups $C^n(G, M)$ by

(1) $C^0(G, M) = M$

(2) $C^n(G, M)$ for $n \ge 1$ is the set of all mappings from the $n$-times cartesian product $G \times \ldots \times G$ to $M$ with pointwise addition.

Then we define mappings $\delta_n \in \mbox{Hom}_{\mathbb Z}(C^n(G, M), C^{n+1}(G, M))$ by

(1) $(f\delta_0)(g) := f - fg$ for $f \in C^0(G, M) = M$,

(2)

\begin{align*} (f\delta_n)(g_1, \ldots, g_{n+1}) & = f(g_2, \ldots, g_{n+1}) \quad + \\ & + \sum_{i=1}^n (-1)^i f(g_1, \ldots, g_{i-1}, g_i\cdot g_{i+1}, g_{i+2}, \ldots, g_{n+1}) \quad + \\ & + (-1)^{n+1} f(g_1,\ldots, g_n) g_{n+1} \end{align*} for $f \in C^n(G, m)$ with $n \ge 1$. For $n \ge 1$ we set $$H^n(G, M) = \mbox{ker}(\delta_n) / \mbox{im}(\delta_{n-1}).$$ These are called the cohomology groups. That they are well defined follows by the fact that the above constructed groups with the mappings (i.e. their so called differentials) form a cochain.

Now the proof concerns an alternative description of these cohomology groups.

Let $\tilde{C}^n(G, m)$ for $n \ge 0$ be the set of mappings from the $(n+1)$-times cartesian product $G \times \ldots \times G$ to $M$ with pointwise addition (i.e. it is an abelian group) and subject to $$f(g_0 g, \ldots, g_n g) = f(g_0, \ldots, g_n)g$$ for all $g_0, \ldots, g_n, g \in G$. Also we define the mappings $\tilde{\delta_n} \in \mbox{Hom}_{\mathbb Z}(\tilde C^n(G, M), \tilde C^{n+1}(G, M))$ by $$(f\tilde{\delta_n})(g_0, \ldots, g_{n+1}) = \sum_{i=0}^{n+1} (-1)^i f(g_0, \ldots, \hat{g_i}, \ldots, g_{n+1});$$ where $\hat{\quad}$ means that the argument is excluded. Then we have $$H^n(G, M) \cong \mbox{ker}(\tilde{\delta_n}) / \mbox{im}(\tilde\delta_{n-1}).$$

Proof: We define mappings $$\mu_n \in \mbox{Hom}_{\mathbb Z}(C^n(G, M), \tilde{C^n}(G, M)) \quad\mbox{and}\quad \tau_n \in \mbox{Hom}_{\mathbb Z}(\tilde{C^n}(G, M), C^n(G,M))$$ by \begin{align*} (f\mu_0)(g_0) & = f g_0 \quad \mbox{for} \quad f \in C^0(G, M) = M; \\ g\tau_0 & = g(1) \quad \mbox{for} \quad g \in \tilde{C^0}(G, M); \end{align*} for $n \ge 1$ and $f \in C^n(G, M)$ let $$(f\mu_n)(g_0, \ldots, g_n) = f(g_0 g_1^{-1}, g_1 g_2^{-1}, \ldots, g_{n-1}g_n^{-1}) g_n;$$ for $n \ge 1$ and $h \in \tilde{C^n}(G, M)$ set $$(h\tau_n)(g_1, \ldots, g_n) = h(g_1 \cdots g_n, g_2 \cdots g_n, \ldots, g_{n-1}g_n, g_n, 1).$$ A trivial computation shows $\mu_n\tau_n = 1$ and $\tau_n \mu_n = 1$. Hence $\mu_n$ is an Isomorphism between the abelian groups $C^n(G,M)$ and $\tilde{C^n}(G, M)$. For $f \in C^n(G, M)$ we have \begin{align*} (f\mu_n \tilde \delta_n)(g_0, \ldots, g_{n+1}) & = \sum_{i=0}^{n+1} (-1)^i (f\mu_n)(g_0, \ldots, \hat{g_i}, \ldots, g_{n+1}) \\ & = \sum_{i=0}^n (-1)^i f(g_0g_1^{-1},\ldots, g_{i-2}g_{i-1}^{-1}, g_{i-1}g_{i+1}^{-1}, g_{i+1}g_{i+2}^{-1}, \ldots, g_n g_{n+1}^{-1}) g_{n+1} \quad + \\ & + (-1)^{n+1} f(g_0 g_1^{-1}, \ldots, g_{n-1} g_n^{-1}) g_n \end{align*} and \begin{align*} (f\delta_n \mu_{n+1})(g_0, \ldots, g_{n+1}) & = (f\delta_n)f(g_0g_1^{-1}, g_1 g_2^{-1}, \ldots, g_n g_{n+1}^{-1}) g_{n+1} \\ & = f(g_1 g_2^{-1},\ldots, g_n g_{n+1}^{-1}) g_{n+1} \quad + \\ & + \sum_{i=1}^n (-1)^i f(g_0 g_1^{-1}, \ldots, g_{i-1}g_{i+1}^{-1}, \ldots, g_n g_{n+1}^{-1}) g_{n+1} \quad + \\ & + (-1)^{n+1} f(g_0 g_1^{-1},\ldots, g_{n-1}g_n^{-1}) g_n. \end{align*} Hence $\mu_n \tilde{\delta_n} = \delta_n \mu_{n+1}$ and so $\mbox{ker}(\tilde{\delta_n}) = (\mbox{ker}\delta_n)\mu_n$ and $\mbox{im}(\tilde{\delta_n}) = (\mbox{im} \delta_n)\mu_{n+1}$, hence $$H^n(G, M) = \mbox{ker}(\delta_n) / \mbox{im}(\delta_{n-1}) \cong \mbox{ker}(\tilde{\delta_n}) / \mbox{im}(\tilde{\delta}_{n-1}).$$

What I do not understand is the isomorphism stated in the last paragraph. As the isomophisms between the kernel and image are realised by different mappings, I do not see that we can conclude that the quotients are isomorphic. See my introductory remarks?

The key difference between the example that you gave at the beginning and what is going on in the proof is that there are no isomorphisms $\alpha$ and $\beta$ such that the following diagram commutes: $$\begin{array}{ccccc} \mathbb{Z}_2&\rightarrow^d&\mathbb{Z}_2\times\mathbb{Z}_4&\rightarrow&0\\ \downarrow_\alpha&&\downarrow_\beta&&\\ \mathbb{Z}_2&\rightarrow^{d'}&\mathbb{Z}_2\times\mathbb{Z}_4&\rightarrow&0 \end{array}$$ Above, $d(1)=(1,0)$ and $d'(1)=(0,2)$.

On the other hand, we have the following:

Proposition: Suppose we have subgroups $H\leq G$ and $H'\leq G'$ and there are homomorphisms $\alpha$ and $\beta$ so that the following diagrams commute: $$\begin{array}{ccccc} H&\hookrightarrow&G&\twoheadrightarrow &G/H\\ \downarrow_\alpha&&\downarrow_\beta&&\\ H'&\hookrightarrow&G'&\twoheadrightarrow&G'/H' \end{array}$$ Then there is a homomorphism $\gamma:G/H\to G'/H'$ such that $$\begin{array}{ccccc} H&\hookrightarrow&G&\twoheadrightarrow &G/H\\ \downarrow_\alpha&&\downarrow_\beta&&\downarrow_\gamma\\ H'&\hookrightarrow&G'&\twoheadrightarrow&G'/H' \end{array}$$ commutes.

proof: It is easy to check that $\gamma(gH)=\beta(g)H'$ is well-defined and gives the desired result.

Proposition: If $\alpha$ is surjective and $\beta$ is injective, then $\gamma$ is injective.

proof: Suppose $gH\in\ker\gamma$, so that $\gamma(gH)=\beta(g)H'=H'$. We need to show $gH=H$. Well, $\beta(g)\in H'$ and so $\beta(g)=\alpha(h)$ for some $h\in H$ since $\alpha$ is surjective. But, the diagram commutes, so $\alpha(h)=\beta(h)$. Thus, $g=h\in H$ because $\beta(g)=\beta(h)$ and $\beta$ is injective. We now conclude that $gH=hH=H$.

Proposition: If $\beta$ is surjective, so is $\gamma$. In particular, if both $\alpha$ and $\beta$ are isomorphisms, then so is $\gamma$.

proof: pretty obvious.

remark: I stated this result in terms of groups, but this is still true in larger generality. In particular, it works in the category of $G$-modules (or $R$-modules), where $\alpha$ and $\beta$ should be module maps.

Now, in the proof in your question, you have $\mu_n$ and $\mu_{n-1}$ are isomorphisms, and the following diagram commutes: $$\begin{array}{ccccc} \mathrm{im}(\delta_{n-1})&\hookrightarrow&\ker(\delta_n)&\twoheadrightarrow &\ker(\delta_n)/\mathrm{im}(\delta_{n-1})\\ \downarrow_{\mu_{n-1}}&&\downarrow_{\mu_n}&&\\ \mathrm{im}(\tilde{\delta}_{n-1}) &\hookrightarrow&\ker(\tilde{\delta}_n)&\twoheadrightarrow&\ker(\tilde{\delta}_n)/\mathrm{im}(\tilde{\delta}_{n-1}) \end{array}$$

• Thanks, your answer really enlightened me! But at first sight the relation $\mu_{n-1}\tilde \delta_{n-1} = \delta_{n-1}\mu_n$ seems to fit with your last diagram, but then I noticed that I did myself an error. We do not have that $\mu_{n-1}$ maps $\mbox{im}(\delta_{n-1})$ onto $\mbox{im}(\tilde \delta_{n-1})$, but it is also $\mu_n$ which maps these sets onto each other, but your explanation is the right one, the fact noted at the beginning, i.e. if $\varphi : G \to G'$ is an isomorphism, then $G / N \cong \varphi(G) / \varphi(N)$ is a special case of your propositions. – StefanH Jan 7 '16 at 16:15
• I would prefer if you do not alter your answer, as I find the explanation good anyway. I add a diagramm of the situation: $$\begin{array}{ccccc} C^n(G, M) & \rightarrow_{\delta_{n-1}} & \mathrm{im}(\delta_{n-1})&\subseteq& \ker(\delta_n) \\ \downarrow_{\mu_{n-1}}&&\downarrow_{\mu_n}&& \downarrow_{\mu_n}&\\ \tilde C^n(G,M) & \rightarrow_{\tilde\delta_{n-1}} & \mathrm{im}(\tilde{\delta}_{n-1}) &\subseteq&\ker(\tilde{\delta}_n)& \end{array}$$ Here the relation $\mu_{n-1}\tilde \delta_{n-1} = \delta_{n-1}\mu_n$ is expressed in the leftmost rectangle of the diagram. (continue) – StefanH Jan 7 '16 at 16:26
• (continuation) The map $\mu_n$ is injective, and as $\mu_{n-1}$ is surjective we can "hit" every element of $\mbox{im}(\tilde\delta_{n-1})$ by going the path $\mu_{n-1}\tilde \delta_{n-1}$ at the bottom, hence $\mu_{n-1}$ is injective and surjective between the image sets. It is left to show that it is also surjective between the kernels. So if $f\tilde\delta_{n-1} = 0$ by surjectivity of $\mu_n$ we have $g$ such that $f = g\mu_n$ and $0 = f\tilde\delta_n = g\delta_n\mu_{n+1}$ and by injectivity of $\mu_{n+1}$ this implies $g\delta_n = 0$, hence $g \in \mbox{ker}(\delta_n)$. – StefanH Jan 7 '16 at 16:26