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?

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  • $\begingroup$ When you say "an even counterexample", you mean a finite group $G$ of even order for which the three bulleted items do hold (for some $m$ and $p$)? Is it required that $m \ne p$? $\endgroup$ – LSpice Mar 5 '19 at 22:11
  • $\begingroup$ I am not finding the question clear. How can you have an even counterexample to a statement about groups of odd order? So you are looking for a group of even order that satisfies the three conditions in the bullet points, is that right? Also you have not said what $m$ is. $\endgroup$ – Derek Holt Mar 5 '19 at 22:12
  • $\begingroup$ Thank you for the comments. I have edited the question to answer your comments. $\endgroup$ – Joakim Færgeman Mar 5 '19 at 22:22
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    $\begingroup$ So $m = 1$ would be all right? (I don't have any examples; I just want to make sure I understand what you're asking.) $\endgroup$ – LSpice Mar 6 '19 at 2:55
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    $\begingroup$ If $|G|$ is even, then the third condition implies that $p=2$. $\endgroup$ – Derek Holt Mar 6 '19 at 7:59

$\DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\PGL}{PGL} \DeclareMathOperator{\mcd}{mcd} \newcommand{\C}{\mathbb{C}}$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.

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  • $\begingroup$ You are absolutely right, of course! Thank you for this argument. $\endgroup$ – Joakim Færgeman Mar 7 '19 at 15:22

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