Pontryagin dual of 2-part of abelian groups

Question

Let $A$ be an abelian group. Let $A[2]$ be a 2-part of $A$, which is a subgroup of $A$ killed by 2.

Let $A'=Hom_{\text{conti}}(A,\Bbb{Q}/\Bbb{Z})$ be a Pontryagin dual of $A$.

Is there a known formula that expresses $(A[2])'$ in terms of $A'$?

If $A$ is finite, $(A[2])'=A'[2]$ is clear, but I don't have a confident that such relation holds for arbitrary (not necessarily finite) abelian groups.

If $(A[2])' = A'[2]$ is true for a general Abelian group $A$, I would like to know the proof. If not, is there an alternative formula that can be considered?

Thank you in advance.

Answer 1

For locally compact abelian groups (which is the usual context in which Pontryagin duality is considered), $A'[2]$ is canonically isomorphic to $(A/2A)'$. Indeed, Pontryagin duality is an exact contravariant functor, which means that dualising the exact sequence $$ 0\to A \stackrel{\times 2}{\to} A\to A/2A \to 0 $$ gives the exact sequence $$ 0\to (A/2A)'\to A' \stackrel{\times 2}{\to} A'\to 0. $$ From this you see that $(A/2A)'$ is the kernel of the multiplication-by-$2$ map on $A'$, which is just the $2$-torsion of $A'$.

Note that even for finite groups, it is not true that $A'[2]=(A[2])'$. It just so happens that for a finite group $B$ the groups $B[2]$ and $B/2B$ are abstractly isomorphic, but not canonically so, so you should not write that they are equal. And as Lukas Heger's example demonstrates, for general locally compact groups these need not even be abstractly isomorphic.

Answer 2

First, a formula relating $A$ and $A'[2]$: Note that $A[2]=\mathrm{Hom}_{conti}(\Bbb Z/2\Bbb Z,A)$, where $\Bbb Z/2\Bbb Z$ carries the discrete topology. Applying the fact that Pontryagin duality is a self-duality (i.e. a contravariant auto-equivalence of categories), we get that

$$A'[2]=\mathrm{Hom}_{conti}(\Bbb Z/2\Bbb Z,A')=\mathrm{Hom}_{conti}((\Bbb Z/2\Bbb Z)',A')=\mathrm{Hom}_{conti}(A,\Bbb Z/2\Bbb Z)$$

Consider for example $A=\Bbb Z$, we have $A[2]=0$, thus $(A[2])'=0$. But $A'=\Bbb R/\Bbb Z$ has $2$-torsion isomorphic to $\Bbb Z/2\Bbb Z$, so $A'[2]$ is non-zero.