application of the five lemma suppose we are given a short exact sequence of $\mathbb{Z}G$-modules $$0\to K\to F\to A\to 0$$ where $F$ is free. and we form a diagram with that first row and with a second row $0\to L\to M\to N\to 0$ short exact but with no special condition and vertical arrows $\alpha\colon K\to L$, $\beta\colon F\to M$ and $\gamma\colon A\to N$. (Sorry about the horrible way of writing the diagram but I don't know how to write it without the xy package!) Suppose also that $\beta$ is an isomorphism. Could we say that $\gamma$ is an isomorphism as well? Could we use the five lemma to say so, or should we necessarily assume that $\alpha$ is an isomorphism too?
Thank you in advance, bye!
 A: As I indicated in a comment, the first question has a negative answer, as shown by a silly example. The second one has a positive answer, though, as is easy to check (and it suffices to assume that $\alpha$ is epi and there's no need to assume that $F$ is free).
For the fun of it, let me mention an overkill way of doing it:
The sharp four lemma states:
If the rows of 
$$\begin{array}{ccccccc}
A_1 & \to & A_2 & \to & A_3 & \to & A_4 \\\
\downarrow & & \downarrow & & \downarrow & & \downarrow \\\
B_1 & \to & B_2 & \to & B_3 & \to & B_4
\end{array}$$
are exact, $A_1 \to B_1$ is epi and $A_2 \to B_2$ and $A_4 \to B_4$ are isos then $A_3 \to B_3$ is mono. The proof is a simple diagram chase.
From this and its dual — or by a direct diagram chase — you can easily conclude the sharp five lemma:
If the rows of
$$\begin{array}{ccccccccc}
A_1 & \to & A_2 & \to & A_3 & \to & A_4 & \to & A_5\\\
\downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow\\\
B_1 & \to & B_2 & \to & B_3 & \to & B_4 & \to & B_5
\end{array}$$
are exact, $A_1 \to B_1$ is epi, $A_5 \to B_5$ is mono and $A_2 \to B_2$ and $A_4 \to B_4$ are isos then $A_3 \to B_3$ is an iso.
Now let $A_1 \to B_1$ be $\alpha$, $A_2 \to B_2$ be $\beta$ and let $A_4 = A_5 = B_4 = B_5 = 0$ to get the desired conclusion that $\gamma$ is an isomorphism.
