The kernel of a homomorphism at the end of an exact sequence

Question

Suppose that $A,B,$ and $C$ are groups and there is an exact sequence $A \to B \to C \to 1 $, and $A_1,B_1,$ and $C_1$ are also groups and there is an exact sequence $A_1 \to B_1 \to C_1 \to 1$. So $B \to C$ and $B_1 \to C_1$ are surjective but $A \to B$ and $A_1 \to B_1$ are not necessarily injective. Suppose there are also homomorphisms $\alpha \colon A \to A_1$ and $\beta \colon B \to B_1$ and $\gamma \colon C \to C_1$ such that the resulting diagram commutes. If the kernels of $\alpha$ and $\beta$ are finite, must the kernel of $\gamma$ be finite?

I don't see why it would be true, but I don't know any counterexample. I thought about it for a while and can't come up with any simple proof. My guess is that it's not true.

Answer 1

Isn't this a counterexample, with the natural embeddings, modding out, and projections?

$$\newcommand\twoheaduparrow{\mathrel{\rotatebox{90}{$\twoheadrightarrow$}}} \begin{array} A & \mathbb{Z}_2 & {\hookrightarrow} & \mathbb{Z}_2 &\oplus& \mathbb{Z} &{\twoheadrightarrow} & \mathbb{Z} & \ \\ & \downarrow{} & & &\downarrow{}& &&\downarrow{}\\ & 2\mathbb{Z} & \stackrel{}{\hookrightarrow} && \mathbb{Z} & \stackrel{}{\twoheadrightarrow} &&\mathbb{Z}_2 & & \end{array}$$

Here $\ker{\gamma} = 2\mathbb{Z}$, so, infinite. Note the first downward map is just the $0$ map which commutes in that square since on the right side of the square we embed then project, so $\mathbb{Z}_2$ goes to $0$ as well.