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.

I didn't notice that this is actually a special case (Short five lemma) of the five lemma in homological algebra, about which I have almost zero knowledge.
The wikipedia article gives a nice presentation of the proof. 
