# $\phi \times \theta$ is faithful iff $(|Z(H)| ,|Z(K)|)=1$ for faithful characters $\phi \in Irr(H)$ and $\theta \in Irr(K)$ .

Let $G = H \times K$. Let $\phi \in \operatorname{Irr}(H)$ and $\theta \in \operatorname{Irr}(K)$ be faithful. Show that $\phi \times \theta$ is faithful iff $(|Z(H)| ,|Z(K)|)=1$.

Problem 4.3 of Isaac's character theory.

## 1 Answer

This is a duplicate, but there do not seem to be any answers to the earlier question, so here are some hints.

Since the characters are irreducible, the centres of $H$ and $K$ are mapped by the associated (complex) representations onto scalar matrices. So if $(|Z(H)|,|Z(K)|) \ne 1$, then $Z(H)$ and $Z(K)$ both have elements of the same prime order, and you can use those elements to find an element in the kernel of $\phi \times \theta$.

For the converse, note that an element $h \in H$ with $h \not\in Z(H)$ does not map onto a scalar matrix, and so $|\phi(h)| < |\phi(1)| = {\rm deg}\ \phi$. So $(\phi \times \theta)(hk)=(\phi \times \theta)(1)$ is only possible if $h \in Z(H)$ and (similarly) $k \in Z(K)$. Now you can prove that this is only possible if $h$ and $k$ have the same orders.