Let $r$ be a positive integer, and $T$ be a finite group. Let $A,B$ be abelian groups which satisfies $A\cong \Bbb{Z}^r\times T$ and $B\cong \Bbb{Z}^r\times T$.
Suppose $0\to \Bbb{Z}/2\Bbb{Z}\to A\to B\to X\to 0$ is an exact sequence.
From these ingredients, can we conclude that $X\cong \Bbb{Z}/2\Bbb{Z}$ ?
In the case $r=0$, by applying isomorphism theorem to $A\to B$, we can definitely say order of $X$ is $2$, so the above claim is true.
But in general case, I don't have a confident to say order of $X$ is $2$.
The background of this question came when I was studying elliptic curves. p$350$ of Silverman's book 'The arithmetic of elliptic curves'. Let $E/K$ be an elliptic curve over a number field and $A=E(K), B=E'(K)$, middle map in the exact sequence is $\phi :A\to B$, which is an isogeny of degree 2. $r=rank(E/K)$ and $T\cong \Bbb{Z}/2\Bbb{Z}$.
$T$ can be identified uniquely inside $A$ as the subgroup of torsion elements of $A$ (i.e. finite order.) Let's call it $T_A$. In the short exact sequence $$0\to T_A\to A\to A/T_A \to 0$$ the group $A/T_A\cong \Bbb{Z}^r$ is a free group. It follows that the short exact sequence splits, and we can find a direct summand $A_0\subseteq A$ such that $A = A_0 \oplus T_A$ and $A_0\cong \Bbb{Z}^r$. (Note: $A_0$ unlike $T$ is not uniquely identified inside $A$.)
Let $\sigma$ be the middle map. Torsion elements must map to torsion elements, so $\sigma(T_A)\subseteq T_B$ where $T_B$ is the torsion subgroup of $B$ (which is also isomorphic to $T$).
Now we consider $\sigma\rvert_{A_0}$. Since $\ker \sigma$ has $2$ elements and $A_0$ is torsion-free, it follows that $\ker \sigma\rvert_{A_0} = 0$, so $\sigma$ maps $A_0$ injectively into $B$. Moreover, $\sigma(A_0)$ is torsion-free, so $\sigma(A_0)\cap \sigma(T_A) = 0$. The subgroup $\sigma(A_0)$ of $B$ is a free subgroup of rank $r$. The subgroup $\sigma(T_A)$ of $T_B$ must have index $2$, because $|T_A| = |T_B| = |T|$ and $\ker \sigma\rvert_{T_A}$ has $2$ elements.
So now, the only thing remaining is whether $\sigma(A_0)\oplus T_B = B$. If they're equal then $X\cong T(B)/\sigma(T_A)$ so $|X|=2$. If they're not equal then $(B/T_B) / ((\sigma(A_0)\oplus T_B)/T_B)$ is some finite group of size $N>1$ and $|X| = 2N$.
It is possible of course for $\sigma(A_0)\oplus T_B$ to be a proper subgroup of $B$. If we start out with map $\sigma$ such that $\sigma(A_0)\oplus T_B = B$, we can take a basis for $A_0$, and then for some basis element $e$, alter $\sigma$ so that it maps $e$ to $2\sigma(e)$ instead of $\sigma(e)$.