Intuition behind Pontryagin duality We have that for a locally compact abelian group,
$$A \mapsto \operatorname{Hom}(A, \mathbb{R}/\mathbb{Z})$$
is a biduality map when endowing $\operatorname{Hom}$ with the compact-open topology.
My question is why $\mathbb{R}/\mathbb{Z}$? This sounds like a rather counter-intuitive result. I can see one why one would give $\operatorname{Hom}$ the compact-open topology (there are many categorical and topological reasons to do this) but it looks rather ad-hoc that we use the circle group.
How did this come up? Are there other bidualities using different groups?
 A: The fundamental notion behind the theory of groups is the concept of symmetries and the best place to visualize symmetries is Euclidean space and, by extension, Hilbert's space.
By a symmetry on a Hilbert space $H$ one means any map
$$
  U:H\to H
  $$
that preserves distance, and sends the origin to itself.  In fact any such map is necessarily a unitary operator.
The group $\mathscr U(H)$, formed by all unitary operators on Hilbert's space  is, according to this, the archetype of symmetry!
Given any group $G$, it therefore makes a lot of sense  to try to  model $G$ via  $\mathscr U(H)$,  and this is usually done by considering
group homomorphisms
$$
  u:G\to \mathscr U(H),
  $$
often called  group representations.
If $u_1$ and $u_2$ are representations of $G$,  one may define a direct sum representation
$$
  u=u_1\oplus u_2
  \tag 1
  $$
on $H\oplus H$,  but
everything there is to know about $u$ is already present in either $u_1$ or $u_2$.
On the other hand, should we be
interested in studying a given representation $u$, we might ask whether or not it has an expression such as (1), in
which case   it
would be better to study $u_1$ and $u_2$ separately.
This means that one should concentrate in the study of irreducible group representations, namely those  which cannot
be split as in (1).
For abelian groups, Schur's Lemma says that all irreducible unitary representations are necessarily one-dimensional,
meaning that the underlying  Hilbert space is $\mathbb C$, and since
$$
  \mathscr U(\mathbb C) = \mathbb T,
  $$
we are inexorably led to  studying homomorphisms from $G$ to $\mathbb T$!
