Let $G$ be a finite group and let $\text{Irr}(G)$ be the set of irreducible complex characters of $G$. A character $\chi\in\text{Irr}(G)$ is monomial if there exists a subgroup $H\leq G$ and a linear character $\lambda\in\text{Lin}(H)$ such that $\chi=\lambda^G$. Let $\text{Irr}_m(G)$ be the set of irreducible monomial characters of $G$. Then a group $G$ is monomial if $\text{Irr}(G)=\text{Irr}_m(G)$. I'm interested in solvable groups even if probably this is not necessary.

My question is: given two finite solvable groups $G$ and $H$ is it true that $$\text{Irr}_m(G\times H)=\{\varphi\times \psi\mid \varphi\in\text{Irr}_m(G),\psi\in\text{Irr}_m(H)\}$$

Using GAP I checked some (really small) cases and this always works.

What is this useful for? I'm interested in this question because a positive answer will give an easy way to construct a family of solvable groups $(G_n)_{n\in\mathbb{N}}$ such that

$$\lim\limits_{n\to \infty} \frac{|\text{Irr}_m(G_n)|}{|\text{Irr}(G_n)|}=0$$ To construct such a family fix a solvable nonmonomial group $G$ (e.g. $\text{SL}_2(3)$) and define $G_n$ to be the direct product of $n$-copies of $ G$.

Update: The above question has a negative answer, in general, by work of van der Waall "Direct products and monomial characters". However, a positive answer is given when at least one of the factors, say $G$, satisfies the following property: either $G=1$ or every maximal subgroup $M$ of $G$ has a normal subgroup $N\unlhd M$ with only abelian Sylow subgroups (for every prime) and such that $M/N$ is nilpotent. In particular, the above family can be constructed considering $G=\text{SL}_2(3)$.

  • $\begingroup$ Out of curiosity: How large orders did you check? It seems that comparing the numbers of monomial characters might be the quickest, but I am not sure how quick TestMonomial is. $\endgroup$ Jun 1, 2018 at 18:52
  • $\begingroup$ @TobiasKildetoft: With GAP I checked the cases $G=SL_2(3)$, $H\in \{S_3,D_8,Q_8,D_{10}, A_4, D_{12}, \text{SL}_2(3), \text{GL}_2(3), \text{SmallGroup}(48,28), \text{SmallGroup}(48,32), \text{SmallGroup}(48,33), \text{SmallGroup}(72,3), \text{SmallGroup}(72,25)\}$, where $\text{SL}_2(3), \text{GL}_2(3), \text{SmallGroup}(48,28), \text{SmallGroup}(48,32), \text{SmallGroup}(48,33), \text{SmallGroup}(72,3), \text{SmallGroup}(72,25)$ are the smallest solvable nonmonomial groups. To find the monomial characters of $G$ I do Filtered(Irr(G), x-IsMonomial(x)). This method is not so fast however. $\endgroup$
    – DRossi
    Jun 2, 2018 at 15:59

1 Answer 1


It is true that the product of two M-groups is an M-group. This is Lemma 2.2.12 in John McHugh, Monomial Characters of Finite Groups.

Actually, $G$ and $H$ need not be solvable.

  • $\begingroup$ This doesn't answer the question. $\endgroup$
    – DRossi
    Oct 26, 2018 at 18:56
  • 1
    $\begingroup$ I know - I am simply giving a case of where it is true. I think (but I am unsure of this) that one can confirm your question by following the proof of Lemma 2.2.12 $\endgroup$ Oct 26, 2018 at 19:08

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