A diagram of group homomorphisms 
Consider the commutative diagram of group homomorphisms:
  $$\require{AMScd}
\begin{CD}
A @>{f}>> B\\
@V{h}VV @V{k}VV \\
C @>{g}>> D
\end{CD}$$
  If both $f$ and $g$ are injective and $h$ is an isomorphism, what can we say about $k$?


I guess it might be true that $k$ is also injective. By definition $k\circ f=g\circ h$. I can see that $\ker k\circ f=1$. But I don't see any information about $\ker k$. How should I go on?

[Added:]
The question is based on the following problem:
$$\require{AMScd}
\begin{CD}
1@>>> A @>{f_1}>> B@>{f_2}>>C@>>>1\\
\   @V{h}VV @V{k}VV @VV{j}V \\
1@>>> D @>{g_1}>> E@>{g_2}>>F@>>>1
\end{CD}$$
The two lines in the commutative diagram are assumed to be short exact sequences and $h$ and $j$ are isomorphisms. Show that $k$ must be isomorphism too. 
I thought I could divide the diagram into two parts and show that $k$ is 1-1 and onto.
Thanks to the comment, it seems that this does not work at all.

Let's show $k$ is onto.  Given $y \in E$, since $g_2\circ k = j\circ f_2$ is onto, there exists $x \in B$ such that $g_2(k(x)) = g_2(y)$.  Therefore $k(x) - y \in \text{ker}(g_2) = g_1(E)$.
Since $h$ is onto, there exists $z \in A$ such that $g_1(h(z)) = k(x) - y$.  But $g_1(h(z)) = k(f_1(z))$, and hence $k(x - f_1(z)) = y$.
Now let's show $k$ is 1-1.  Suppose there are $x_1,x_2 \in B$ such that $k(x_1) = k(x_2)$.  Then $j(f_2(x_1)) = g_2(k(x_1)) = g_2(k(x_2)) = j(f_2(x_2))$.  Hence $f_2(x_1) = f_2(x_2)$.  Therefore $x_1 - x_2 \in \text{ker}(f_2) = f_1(A)$.  Therefore $x_1 - x_2 = f_1(z)$ for some $z \in A$.  Therefore $0 = k(x_1 - x_2) = k(f_1(z)) = g_1(h(z))$.  Since $g_1\circ h$ is 1-1, we have $z = 0$.
