$\chi(1)>1$, then $|\chi|^2$ is not an irreducible character.

Question

Ex 2.6 Character Theory, Isaac: Let $\chi$ be a charcter of $G$, such that $\chi(1)>1$. Show that $\chi \bar{\chi} \notin Irr(G)$. $Irr(G)$ denotes the set of characters induced from irreducible representations.

What I can deduce so far...

(i) If $\chi$ were a linear character, and $\chi \in Irr(G)$, then I can prove that the product is in fact in $Irr(G)$.

(ii) If $\chi \bar{\chi} = |\chi|^2 \in Irr(G)$, then by orthogonality relations, $[|\chi|^2, |\chi|^2] =1$, i.e. $\frac{1}{|G|} \sum_{g \in G} |\chi(g)|^4 =1.$

Answer 1

Note that $|\chi|^2$ is not the trivial character since $\chi(1)>1$. Hence if $|\chi|^2$ is the character of an irreducible representation, taking the inner product with the trivial character yields $$ \sum_{g\in G}|\chi(g)|^2=0 $$ But this is impossible because every term in the sum is non-negative, and at least one term is positive.

Answer 2

Another approach: $|\chi|^2=\chi\overline{\chi}$, so it is the product of two irreducible characters and hence a character. But $[|\chi|^2,1_G]=[\chi,\chi]=1$ and since $\chi$ is non-linear, this show that the trivial character $1_G$ is always an irreducible constituent with multiplicity $1$ and there must be at least an other irreducible constituent. Example, if $\lambda $ is the non-trivial linear character of $G=S_3$ and $\chi$ is its irreducible character of degree $2$, then $|\chi|^2=1_G+\lambda+\chi$.