Suppose that $G$ has exactly one nonlinear irreducible character. Show that $G'$ is an elementary abelian $p$-group. 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.
 A: 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 $x^p=1$.]
