Let $G_1$ and $G_2$ be finite groups and let $G = G_1 \times G_2$. Suppose $\rho: G_1 \to GL_m(\mathbb{C})$ and $\phi: G_2 \to GL_n(\mathbb{C})$ are representations. Let $V =M_{mn}(\mathbb{C})$. Define $\tau : G \to GL(V)$ by the rule $\tau(g_1,g_2)(A) = \rho_{g_1}A\phi_{g_2}^T $

Prove that if $\rho,\phi$ are irreducible representations then so is $\tau$.


A TLDR disclaimer: the representation you are looking at is called the external tensor product of the two individual representations so if you've seen this before, irreducibility should be immediate. Instead of doing the general case, I will prove the problem in this specific case.

We want to use character theory here. Let's see what this representation does to the standard basis $\{E_{ij}\}$ of $M_{mn}.$

$$E_{ij} \phi_{g_{2}}^{T}$$

basically takes the $j$th column of $\phi_{g_{2}}$ and places it into the $i$th row of the new $m \times n $ matrix and puts $0$ everywhere else.

So, the $k, l$th entries of

$$\tau(g_{1}, g_{2})(E_{ij})$$

is $(\rho_{g_{1}})_{k,i} (\phi_{g_{2}})_{l, j}.$ Since we only care about the trace of $\tau(g_{1}, g_{@})$, we only care about the $i, j$th coefficient here, which is

$$(\rho_{g_{1}})_{i, i}(\phi_{g_{2}})_{j, j}.$$

So now, we see that the character $$\chi_{\tau}(g_{1}, g_{2}) = \sum_{i, j} (\rho_{g_{1}})_{i, i}^{2}(\phi_{g_{2}})_{j, j}^{2} = \sum_{i} (\rho_{g_{1}})_{i, i}^{2} \sum_{j} (\phi_{g_{2}})_{j, j}^{2} = \chi_{\rho}(g_{1})\chi_{\phi}(g_{2}).$$

Hence, we have

$$\frac{1}{|G_{1} \times G_{2}|} \sum_{g \in G_{1} \times G_{2}} |\chi_{\tau}(g_{1}, g_{2})|^{2} = \left[\frac{1}{|G_{1}|} \sum_{g \in G_{1}} |\chi_{\rho}(g_{1})|^{2}\right] \left[\frac{1}{|G_{2}|}\sum_{g \in G_{2}} |\chi_{\phi}(g_{2})|^{2}\right] = 1.$$

Hence, the new representation is irreducible.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.