Even Counterexample to Statement About the Non Existence of Certain Groups with Two Irreducible Monomial Character Degrees Let $\textrm{cd}(G)=\lbrace \chi(1)\,|\, \chi\in\textrm{Irr}(G)\rbrace$ denote the set of character degrees of a finite group $G$. Similarly, denote by $\textrm{mcd}(G)$ the set of monomial character degrees. I have proved the following statement:
There exists no finite group $G$ of odd order such that:


*

*$\textrm{mcd}(G)=\lbrace 1,m\rbrace$

*$\textrm{cd}(G)=\lbrace 1,m,p\rbrace $ where $p$ is a prime

*$\textrm{gcd}(|G|,p^2-1)=1$
where $m$ is any positive integer and $m\neq p$. I have managed to find such groups if you remove the condition that $\textrm{gcd}(|G|,p^2-1)=1$ (with the help of Professor Mark Lewis). I would like to find an example of a group of even order satisfying the three bulleted conditions but have not suceeded.
Where would be a good place to look?
 A: $\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\PGL}{PGL}
\DeclareMathOperator{\mcd}{mcd}
\newcommand{\C}{\mathbb{C}}$Such groups do not exist. Indeed, suppose that $G$ has even order and satisfies the three conditions. The third condition forces $p$ to be $2$. Let $\chi$ be an irreducible character of $G$ of degree $2$. Then $\chi$ is a faithful irreducible degree $2$ character of $\tilde{G}=G/\ker \chi$, and $\tilde{G}$ is realised as a finite subgroup of $\GL_2(\C)$. By the usual classification, its image in $\PGL_2(\C)$ is either cyclic or dihedral or isomorphic to one of $A_4$, $S_4$, $A_5$. Let's exclude all these possibilities.
If $\tilde{G}$ is cyclic modulo the centre, then an easy exercise shows that it is abelian, so has no faithful character of degree $2$ – a contradiction.
If the image of $\tilde{G}$ in $\PGL_2(\C)$ is dihedral, then it has a monomial irreducible character of degree $2$, therefore so does $G$ (the lift of a monomial character from a quotient is easily seen to be monomial), contradicting your first condition, which demands that $2\not\in \mcd(G)$.
If the image of $\tilde{G}$ is isomorphic to one of $A_4$, $S_4$, $A_5$, then its order, and therefore also the order of $G$, is divisible by $p^2-1=3$, contradicting the third condition.
