Pontryagin dual for non abelian $\widehat G := \operatorname{Hom}(G, T) $? For a locally compact abelian group  $G$ , the Pontryagin dual  is the group  $\widehat G$  of continuous  group homomorphisms  from $G$  to the circle group  $T$ . That is,
:
$$\widehat G := \operatorname{Hom}(G, T). $$
The Pontryagin dual $\widehat G$ is usually endowed with the  topology  given by uniform convergence on  compact sets (that is, the topology induced by the  compact-open topology  on the space of all continuous functions from $G$ to $T$).
For example,
$$\widehat {\mathbb Z} = T,\  \widehat {\mathbb R} = \mathbb R,\ \widehat T = \mathbb Z.$$
questions

*

*do we have a relation that
$$ \operatorname{Hom}(G_1, G_2)= \operatorname{Hom}(G_2, G_1)?$$
for any groups? I think the answer is no. If no, can you give the reasonings behind why and for counter examples? (for abelian groups examples? for non-abelian groups examples?)


*How do we define $\widehat G$ for non-abelian $G?$ Is that $\widehat SU(2)=SO(3)$ and $\widehat SO(3)=SU(2)$?
 A: For abelian groups $G_1$  and $G_2$, the map
$$
  \varphi  \in  \text{Hom}(G_1, G_2) \mapsto  \hat \varphi \in \text{Hom}(\hat G_2, \hat G_1),
  $$
given by
$$
  \hat \varphi (\gamma ) = \gamma \circ \varphi , \quad \forall \gamma \in  \hat G_2,
  $$
is a bijection.  Now, for every finite abelian group $G$ one has that  $\hat G$ is isomorphic to $G$.  So, if $G_1$ and $G_2$ are
finite abelian, one gets a bijection
$$
  \text{Hom}(G_1, G_2) \buildrel \widehat{} \over \longrightarrow \text{Hom}(\hat G_2, \hat G_1) = \text{Hom}(G_2, G_1).
  $$
Besides finite abelian groups, things fail miserably.
Without finiteness: Notice that $\text{Hom}(\mathbb T, \mathbb Z_2)$ has only one element,  namely the trivial homomorphism,  but in
$\text{Hom}(\mathbb Z_2, \mathbb T)$ one may find the trivial homomorphism in addition to the map sending the generator to
$-1$.
Without commutativity: Since the alternating group $A_n$ is simple for $n\geq 5$, one has that
$\text{Hom}(A_n, \mathbb Z_2)$ only has  the trivial homomorphism, while
$\text{Hom}(\mathbb Z_2, A_n)$ has as many elements as order-two even permutations.
Regarding the dual of non-abelian groups,  one usually defines $\hat G$ as the set of equivalence classes of irreducible unitary
representations on Hilbert spaces.  Here $\hat G$ does not have a group  structure  and the theory is simultaneously  a lot more
sophisticated and a lot less powerful than Pontryagin duality.
If $G$ is abelian, Schur's Lemma implies that every unitary irreducible representation
$$
  π:G→U(H)
  $$
is one-dimensional, that is,  $H$ is a one-dimensional Hilbert space.   In this case $U(H)=\mathbb T$, so $π$ is
actually a  homomorphism from $G$ to $\mathbb T$.  In other words,  the dual  reduces to the Pontryagin dual.
However,
for lots of non-abelian groups, e.g. $A_n$,  there are no nontrivial
one-dimensional representations, so
$
  \text{Hom}(G, \mathbb T)
  $
may be empty!
