is it always true that if I have two groups $G,H$, then the character group of the direct product $G \times H$ is (naturally?) isomorphism to $\widehat{G} \times \widehat{H}$ ($\widehat{(X)}$ means the character group of the group $X$). What if we impose the condition that the groups must be finite or (and) abelian? Thank you
characters of group direct product
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All that is written below is referred to the book of Isaacs, "Character Theory of Finite Groups.
Generally the notation $\hat{G}$ is reserved to set of irreducible characters of an abelian group, and with point-wise multiplication is itself a group. If $G$ non abelian then there is no hope to get an irreducible character by multiplying point-wise two characters. Indeed it is not even clear that the product of characters is a character and not just a class function. Fortunately the last sentence turns out to be true: if we have $\chi, \psi$ are characters afforded by $\mathbb{C}[G]$-modules $V,W$, then the multiplication $\chi \psi$ is a character afforded by $V \otimes W$ (Theorem 4.1). There are also some results that guarantee the irreducibility of the product of two characters. For example Gallagher's Theorem 6.17:
Theorem: Let $N$ normal subgroup in $G$ finite group and $\chi \in Irr(G)$ such that the restriction $\theta=\chi_N$ is an irreducible character of $N$. Then the characters $\chi \beta$ are all distinct for $ \beta \in Irr(G/N)$ and are all the irreducible constituents of the induced character $\theta^G$.
The hypothesis $\chi_N \in Irr(N)$ is analyzed in an other Theorem of Gallagher, 8.15 in Isaacs.
An application of what I just wrote is that if we have an abelian group $G$ then two irreducible characters have degree $1$ and their product is a character of degree $1$, so must be irreducible. That's why $\hat{G}$ is closed under pointwise multiplication.
This said, we can characterize the irreducible characters of a direct product in terms of the irreducibles of the direct components. If $\varphi,\theta$ are class functions of two finite groups $H,K$, define the class function of $H\times K$ by $\varphi \times \theta(hk)=\varphi(h)\theta(k)$. We have the:
Theorem (4.21): Let $H,K$ finite groups. Then $Irr(H\times K)=\{\varphi \times \theta \mid \varphi \in Irr(H), \theta \in Irr(K)\}$
The answer is no. Even for finite abelian groups $G$ and $\widehat{G}$ are isomorphic but not naturally so. The reason is because there is a choice on picking an $n^\text{th}$ root of unity.