This post is related to this post.
Denote $\mathbb{S}^1=\{z\in\mathbb{C}:|z|=1\}$ as the multiplicative group on the unit sphere. Given a group $A$ we define its dual group $$\widehat{A}=\text{Hom}(A,\mathbb{S}^1)$$ the group of all homomorphisms from $A$ to $\mathbb{S}^1$, equipped with pointwise multiplication. That is, given $\chi_1,\chi_2\in\widehat{A}$, we define thir product as: $$\forall a\in A: (\chi_1\cdot\chi_2)(a)=\chi_1(a)\cdot\chi_2(a)$$
A standard result, as mentioned in the linked post, states that if $H$ and $K$ are finite abelian groups, then: $$\widehat{H\times K}\cong\widehat{H}\times\widehat{K}$$
While working through the proof of this result, it seems that the assumption that $H$ and $K$ are finite abelian is not used.
Thus, my question is: Does the isomorphism $\widehat{H\times K}\cong\widehat{H}\times\widehat{K}$ hold even if $H$ and $K$ are not assumed to be finite abelian?
Any insights or counterexamples would be greatly appreciated!
Thanks in advance!
