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I am revising for my Representations of Groups exam and am stuck on the following question.

Let $\ G,H \ $ be finite groups. Prove that if $ \ \chi \ $ is a character of $G$ and $ \ \phi \ $ is a character of $H$ then

\begin{align*} \chi \ \ast \ \phi:& G \times H \longrightarrow \mathbb{C} \ & (g,h) \longmapsto \chi(g) \ \phi(h) \end{align*}

is a character of $G \times H$.

Any help would be very much appreciated!.

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Where are you stuck? – azarel Mar 31 '12 at 1:45
We can help you best if you tell us what you have tried. – Mariano Suárez-Alvarez Mar 31 '12 at 1:51
So I know that given a presentation $\rho: G \longrightarrow GL(n,C)$ then we have $\chi(g) = tr(\rho(g))$ for all $g \in G$ and given a presentation $\sigma: H \longrightarrow GL(m,C)$ then we have $\phi(g) = tr(\sigma(h))$ for all $h \in H$ but where do you go from here. This is just spelling out the basic definitions but I am really stuck from here. – Alex Kite Mar 31 '12 at 1:55
I have defined a representation $\rho \times \sigma: G \times H \longrightarrow GL(mn,C)$ by $(\rho \times \sigma)(g,h) = \rho(g)\sigma(h)$ but this wrong. – Alex Kite Mar 31 '12 at 2:03
You should consider the tensor product of the two representations. – Mariano Suárez-Alvarez Mar 31 '12 at 2:07
up vote 2 down vote accepted

$\def\tr{\mathrm{tr}\;}\def\GL{\mathrm{GL}}\def\CC{\mathbb C}$ Let $\rho:G\to\GL_n(\CC)$ and $\sigma:H\to\GL_m(\CC)$ be two group homomorphisms. Let us name the components of the matrices in the image of $\rho$ in the obvious way so that we may write $\rho(g)=(\rho_{i,j}(g))_{1\leq i,j\leq n}$, and similarly for $\sigma$.

On the other hand, let $I=\{(i,j):1\leq i\leq n, 1\leq j\leq m\}$ and let us index the coefficients of a matrix $A\in\GL_{nm}(\CC)$ using $I$, so that we may write $A=(a_{(i,j),(k,l)})_{(i,j),(k,l)\in I}$. This makes sense, of course, because the matrices in $\GL_{mn}(\CC)$ have $mn=|I|$ rows and columns.

Consider the function $\lambda:G\times H\to\GL_{mn}(\CC)$ such that $\lambda(g,h)=(\lambda_{(i,j),(k,l)}(g,h))_{(i,j),(k,l)\in I}$ with $$\lambda_{(i,j),(k,l)}(g,h)=\rho_{i,k}(g)\sigma_{j,l}(h)$$ for all $(i,j)$, $(k,l)\in I$.

With some work —which is best done in private— one can show that $\lambda$ is an homomorphism of groups. A little computation, then, shows that for each $(g,h)\in G\times H$ we have $$\tr\lambda(g,h)=\tr\rho(g)\cdot\tr\sigma(h).$$ This means that the character of $\lambda$ is has the relation you want with the characters of $\rho$ and $\sigma$.

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Carrying out the details of this computation should be enough justification to find a textbook which introduces tensor products and does this more sanely! – Mariano Suárez-Alvarez Mar 31 '12 at 5:48
Basically this Kronecker product is the matrix of the tensor product. – anon Mar 31 '12 at 5:50
Thank you very much for the answer! I do have a textbook which introduces tensor products, so I will read definitely read it. – Alex Kite Mar 31 '12 at 11:53

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