$2$ part of $Ext(\Bbb{Z}/2\Bbb{Z}, A)$ when $A$ is finite

Question

Let $A$ be a finite group. Let denote $Ext^1$ by $Ext$ functor in homological algebra.

Is $(Ext^1(\Bbb{Z}/2\Bbb{Z}, A))[2]$ finite?

When $A$ is finite cyclic group $\Bbb{Z}/m\Bbb{Z}$, $\#Ext^1(\Bbb{Z}/2\Bbb{Z}, A)=gcd(2,\#A)=gcd(2,m)$. Thus $(Ext^1(\Bbb{Z}/2\Bbb{Z}, A))[2]=1 \text{or} 2$.

But is $(Ext^1(\Bbb{Z}/2\Bbb{Z}, A))[2]$ finite in the case $A$ is not cyclic ?

Answer 1

Yes. Actually, the whole Ext group is finite, even without limiting to the $2$-torsion. This group classifies extensions $G$ of $\mathbb{Z}/2\mathbb{Z}$ by $A$, i.e. exact sequences $$ 0 \to A \to G \to \mathbb{Z}/2\mathbb{Z} \to 0 $$ There are only finitely many things $G$ can be (only finitely many group structures on a finite set) and only finitely many index two subgroups such a $G$ can have.