# 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?

• 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 $m \ne p$? – LSpice Mar 5 '19 at 22:11
• 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. – Derek Holt Mar 5 '19 at 22:12
• Thank you for the comments. I have edited the question to answer your comments. – Joakim Færgeman Mar 5 '19 at 22:22
• 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.) – LSpice Mar 6 '19 at 2:55
• If $|G|$ is even, then the third condition implies that $p=2$. – 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.