# Commutative hexagonal diagram of Abelian groups; proving a certain equality

I'm trying to prove the following lemma by diagram chasing, but I've had no success, so I decided to ask for help here.

Let $$A$$, $$B$$, $$C$$, $$D$$, $$E$$, $$F$$, and $$G$$ be Abelian groups, and let $$a_{1}$$, $$a_{2}$$, $$a_{3}$$, $$g_{1}$$, $$g_{2}$$, $$g_{3}$$, $$c_{1}$$, $$c_{2}$$, $$c_{3}$$, $$e_{1}$$, $$e_{2}$$, $$e_{3}$$ be homomorphisms such that the following diagram commutes (also, $$c_{1}$$ and $$e_{1}$$ are isomorphisms, as indicated in the diagram):

Also, let $$g_{3}a_{2} = 0$$, $$g_{2}c_{2} = 0$$ and $$g_{1}e_{2} = 0$$. Prove that, if one of the "horizontal" diagonals of this hexagon is exact, i.e. if $$\ker{g_{2}} = \mathrm{im}\hspace{0.1cm}c_{2}$$ OR $$\ker{g_{1}} = \mathrm{im}\hspace{0.1cm}e_{2}$$, then $$c_{3}c_{1}^{-1}a_{1} + e_{3}e_{1}^{-1}a_{3} = 0.$$

I tried diagram chasing, but I wasn't able to see a solution. Here's as far as I got: let $$a \in A$$, $$b = a_{1}a$$, $$g = a_{2}a$$, $$f = a_{3}a$$. Then we also have $$b = g_{1}b$$ and $$f = g_{2}g$$. Let $$c = c_{1}^{-1}b$$, $$g' = c_{2}c$$, $$e = e_{1}^{-1}f$$, $$g'' = e_{2}e$$. Then we also have $$g_{1}g' = b$$ and $$g_{2}g'' = f$$. Then, let $$d' = c_{3}c$$ and $$d'' = e_{3}e$$, and we have $$g_{3}g' = d'$$ and $$g_{3}g'' = d''$$.

Ultimately, we have $$d' = c_{3}c_{1}^{-1}a_{1}a$$ and $$d'' = e_{3}e_{1}^{-1}a_{3}a$$, so the point is to prove $$d' = -d''$$.

For example, suppse that the first condition holds, i.e. that $$\ker{g_{2}} = \mathrm{im}\hspace{0.1cm}c_{2}$$. Then, since $$g_{2}g'' = g_{2}g = f$$, we have $$g'' - g \in \ker{g_{2}} = \mathrm{im}\hspace{0.1cm}c_{2}$$, so there exists a $$c' \in C$$ such that $$c_{2}c' = g'' - g$$.

We also have $$c_{3}c' = g_{3}c_{2}c' = g_{3}(g''-g) = g_{3}g'' - g_{3}g = d''$$, because $$g_{3}g = g_{3}a_{2}a = 0$$. If there was a way to prove $$c_{3}(c+c') = 0$$, I'd be done, but I have no idea how to continue.

Am I on the right track? Should I approach this problem differently?

## 1 Answer

Here's a solution from a friend (I was only missing a couple of details):

Note that $$g_{1}(g'+g''-g) = g_{1}(g')+g_{1}(g'')-g_{1}(g) = b + g_{1}e_{2}e - b = b + 0 - b = 0$$, and also $$g_{2}(g'+g''-g) = g_{2}(g')+g_{2}(g'') - g_{2}(g) = g_{2}c_{2}c + f - f = 0$$, so $$g'+g''-g \in \ker{g_{1}} \cap \ker_{g_{2}}$$.

If $$\ker{g_{1}} = \mathrm{im}\hspace{1mm}c_{2}$$, then there exists a $$c'' \in C$$ such that $$c_{2}c'' = g'+g''-g$$ (actually, $$c'' = c+c'$$), so $$c_{1}c'' = g_{1}c_{2}c'' = g_{1}(g'+g''-g) = 0$$, so since $$c_{1}$$ is a bijection, $$c''=0$$, so $$c_{3}(c+c') = 0$$, which is what I wanted to prove.

The other case is analogous to the first one.