Let $A,B,C,D$ be abelian groups and $A$ is finite. Suppose $D[2]$ is finite.
Is the following true ?
Suppose $0\to A\to B\to C\to D\to 0$ is exact. Let $0\to A[2]\to B[2]\to C[2]\to X\to 0$ be an exact sequence(Every morphism of this sequence is restriction of the first exact sequence). Then group structure of $X$ is uniquely determined. I want to prove $X [2]$ is finite.
$A[2]=Hom(\Bbb{Z}/2\Bbb{Z},A)$, so we want to apply Ext functor to a short exact sequence, but our sequence is five term. So I'm at a loss.
I tried to apply Ext functor to $0\to B/A \to C \to D \to 0$, then there is an exact sequence $0 \to (B/Z)[2]\to C[2] \to D[2] \to Ext^1(\Bbb{Z}/2\Bbb{Z},B/Z)$. If $Ext^1(\Bbb{Z}/2\Bbb{Z},B/Z)$ is finite, we are done, but there is no reason unless we give more information about $B/A$.
This is a bit ugly but works anyway.
Consider the following diagram:
$\require{AMScd}$ \begin{CD} @.B[2] @>>> C[2] @>>> X @>>> 0\\ @. @V{f}VV @V{g}VV @V{h}VV \\ 0 @>>>B/A @>>> C @>>> D \end{CD}
We know the image of $h$ lands on $D[2]$ which is finite, so to show $X$ is finite, it suffices to show $\ker h$ is finite.
By the snake lemma, there is an exact sequence $$\ker g\rightarrow \ker h\rightarrow \text{coker} f \rightarrow \text{coker}g$$
Note that $\ker g = 0$, so we have $$0\rightarrow \ker h \rightarrow B/(A+B[2])\xrightarrow {\bar{\rho}} C/C[2]$$ where $\rho:B\rightarrow C$ and $\bar \rho$ is the induced map. And we may identify $\ker h$ and $\ker{\bar{\rho}}$.
$x+A+B[2]\in \ker h$ is equivalen to $\bar{\rho}(x+A+B[2])=0$, then to $$\rho(x+A+B[2])\in C[2]\Leftrightarrow 2\rho(x+A+B[2])=0\Leftrightarrow \rho(2x+2A+2B[2])=0$$
finally to
$$\rho(2x+2A)=0\Leftrightarrow 2x\in A$$
Now if $|\ker h|>|A|$, then there are distinct $x+A+B[2], y+A+B[2]$ from $\ker h$, such that $2x=2y$, therefore $x-y \in B[2]$, which contradicts $x-y\not\in A+B[2]$.
Actually, there is a straightforward diagram chasing on:
\begin{CD} 0 @>>> A[2] @>>> B[2] @>>> C[2] @>{x'\mapsto x}>> X @>>> 0\\ @. @VVV @VVV @VVV @V{x\mapsto 0}VV\\ 0 @>>> A @>>{x'''\mapsto 2x''}> B @>{x''\mapsto x'}>{2x''\mapsto 0}> C @>{x'\mapsto 0}>> D @>>> 0 \end{CD}
Starting with $x\in \ker(X\rightarrow D)$, we can find $x', x''$ as shown above. If $B[2]\rightarrow B$ is surjective, we would be able to show $x=0$, and we're done (the classical four lemma, used in the proof of the five lemma), but it's not the case. However, note that as $x'\in C[2]$, $2x'=0$, therefore $2x''\in B\mapsto 0\in C$, hence we can further chase to find $x'''\in A$ such that $x'''\mapsto 2x''\in B$. As the lift from $B$ to $A$ (if exists) is unique, we may identify $x'''=2x''$.
Now we show the map $x\mapsto 2x''$ is well-defined and injective. Indeed if $x$ can be lifted to both $x''$ and $y''$ in B, then $x''-y''$ is mapped to $0$ in $X$, hence $x''-y''$ can be lifted to $B[2]$, that is $2x''=2y''$. If $x$ is lifted to $x''$, and $y$ to $y''$, such that $2x''=2y''$, then $x''-y''\in B[2]$, therefore $x=y$ following the first row.
(I first posted the following wrong example where $A$ is not finite.)
Let $B=(\mathbb Z/4\mathbb Z)^\infty, C=2B\simeq(\mathbb Z/2\mathbb Z)^\infty$, where the map $\rho:B\rightarrow C$ is given by the surjective map $\rho(x)=2x$.
Now we can work out that $A\simeq \ker(\rho)=C, D\simeq\text{coker}(\rho)=0$.
And, $A[2]=A, B[2]=A, C[2] = C = A, D[2]=0$. That is, $$0\rightarrow A\xrightarrow{id}A \xrightarrow{0} A\rightarrow X\rightarrow 0$$
Therefore $X\simeq A$ which is not finite.