Suppose that $G$ has exactly one nonlinear irreducible character. Show that $G'$ is an elementary abelian $p$-group.

Question

I know that the number of linear characters is equal to $|G/G'|$, so $|\text{Irr}(G)|=1+|G/G'|$. I know that if $\phi$ is a character of $G'$ then for $x\in G$, $\phi^x$ is a character of $G'$. I'm not sure how to proceed with the problem though.

Answer 1

For any finite group, each coset of $G'$ is a union of conjugacy classes of $G$. If $G$ is not abelian (equivalently, if $G$ has a nonlinear irreducible character), the coset $1\cdot G'$ contains at least $2$ conjugacy classes, since $G'$ properly contains the conjugacy class $\{1\}$. Thus, any finite nonabelian group $G$ has at least $[G:G']+1$ conjugacy classes.

However, if $G$ has exactly one nonlinear irreducible character, then the number of irreducible characters is exactly $[G:G']+1$. Since the number of irreducible characters equals the number of conjugacy classes in $G$, a group with exactly one nonlinear irreducible character sharply attains the lower bound mentioned in the preceding paragraph. This forces each nonidentity coset of $G'$ to equal a single conjugacy class, and it forces $G'$ to decompose into exactly two conjugacy classes, namely $\{1\}$ and $G'\setminus \{1\}$.

We have established that if $G$ has exactly one nonlinear irreducible character, then all nonidentity elements of $G'$ are conjugate. It follows that $G'$ is an elementary abelian $p$-group for some prime $p$. [To see this, choose any element $g\in G'$ of prime order. Use the fact that $G'\setminus\{1\}$ consists of the conjugates of $g$ to deduce that $G'$ is a $p$-group. Adjust the choice of $g$ so that it belongs to $Z(G')$. Again use the fact that $G'\setminus\{1\}$ consists of the conjugates of $g$ to show that $G'=Z(G')$ and that this group satisfies the identity $x^p=1$.]