Is it true that if $\phi_1$ and $\phi_2$ are characters then $\phi_1\phi_2$ is a character of a representation?

I think this is true, say for instance if $R_1$ is a matrix representation with character $\phi_1$ of $G$ and $R_2$ is a matrix representation of $H$ with character $\phi_2$, then we can say $\phi_1\phi_2$ as the character of the matrix representation of $G\times H$ if we define $\phi_1\phi_2(g, h)=\phi_1(g)\phi_2(h)$. Will this suffice to prove our claim?

  • $\begingroup$ The product of two characters is indeed a character, but the proof is a lot more subtle. The proofs I've seen involving using modules and tensor products. You shouldn't need an auxiliary group $H$ (and indeed, a representation of $G \times H$ would give you characters of $G \times H$) $\endgroup$
    – pjs36
    May 6, 2015 at 3:21
  • $\begingroup$ @pjs36: Could you please elaborate on the tensor product idea? $\endgroup$
    – user104221
    May 6, 2015 at 15:08

1 Answer 1


Suppose $G$ acts on a vector space $V$ with character $\phi_1$, and $H$ acts on a vector space $W$ with character $\phi_2$. Then $G \times H$ acts on the vector space $V \otimes W$ with character $\Phi(g,h) = \phi_1(g)\phi_2(h)$. Moreover every irreducible complex representation of $G \times H$ is a tensor product of an irreducible representation of $G$ with one of $H$. This is what is often called the external tensor product of representations. It is external since it takes in representations of two groups and spits out a representation of a third.

There is also an internal tensor product of representations which takes in two representations of $G$ and spits out a third. It is related to the external tensor product as follows: We take two representations of $G$ and get a representation of $G \times G$ via the external tensor product, then we restrict this representation to the diagonal copy of $G$ in $G \times G$. Understanding how these internal tensor products decompose is a much subtler problem, which is referred to in general as the Clebsch-Gordan problem.


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