# 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?

## 1 Answer

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