The lower bound of a group with characters of special degrees

Question

Is there any lower bound for the order of a group with an irreducible character of degree $p$, where $p$ is a prime.

Is there any similar result for $p^2$ or $p^3$ instead of $p$?

Thanks for your helps

Answer 1

Let me try to answer in some detail the case $p$. So $G$ is a finite group of minimal possible order subject to having a complex irreducible character $\chi$ of degree $p$, where $p$ is a chosen prime. Note that $\chi$ is necessarily faithful by minimality of $|G|$. As noted in my comment, by elementary character theory, we necessarily have $|G| >p^{2}$, while also have $p | |G|$, so certainly $|G| \geq p^{2}+p.$ And as noted there, at least when $ p =2$ or a Mersenne prime, there is a Frobenius group $G$ of order $p(p+1)$ with an irreducible character of degree $p$.

Let me go further, and try to find the smallest such group for other odd primes $p$ which are not Mersenne. There is always a $p$-group $P$ of order $p^{3}$ with an irreducible character of degree $p$, so we need only concern ourselves with the case$|G| < p^{3}.$ Since $p > 3$ now, we know that ${\rm PSL}(2,p)$ has a faithful irreducible character of degree $p$, and that $|{\rm PSL}(2,p)| = \frac{p(p^{2}-1)}{2}.$

Hence we need to consider whether there are groups $G$ for which these bounds can be improved. This unavoidably involves some number-theoretic issues. If $mp + 1 = q^{r}$ is a prime power for some positive integer $m$ and prime $q$, then there is a Frobenius group $H$ with Frobenius kernel of order $q^{r}$ and complement of order $p$, and $H$ has an irreducible character of degree $p$. For an arbitrary prime $p$, it is not a priori obvious what the smallest such integer $m$ is. Since $p$ is odd and not Mersenne, $m \geq 2$. We are interested in cases when $mp+1 < \frac{p^{2}-1}{2}.$

Let us return to our group $G$. Let $P \in {\rm Syl}_{p}(G)$. If $P$ is non-Abelian, then $\chi$ restricts irreducibly to $P$, and $|P| \geq p^{3}.$ Hence we may suppose that $P$ is Abelian.

Consider first the case that $G = G^{\prime}.$ Then $P \cap Z(G) = P \cap G^{\prime} \cap Z(G) \leq P^{\prime} = 1$ by an elementary transfer argument. Hence $Z(G) = 1$ in this case, using Schur's Lemma, since any non-identity scalar matrix of order prime to $p$ in $G$ would have determinant different from $1$, and would lie outside $G^{\prime}.$ By a Lemma of Burnside (used in the proof of his $p^{a}q^{b}$-theorem), we have $\chi(x) = 0$ for each non-identity $x \in C_{G}(P).$ Thus $\frac{\chi(1)}{|C_{G}(P)|} = \langle {\rm Res}^{G}_{C_{G}(P)}(\chi), 1 \rangle \in \mathbb{Z}$ and $C_{G}(P) = P$ has order $p$. We have already noted that if $N \lhd G$ and $1 \neq N \neq G,$ then $N$ is Abelian, so now $N$ must have order prime to $p$ ( for otherwise $p = \chi(1)$ divides $[G:N]$ by a Theorem of Ito, whereas $[G:N]$ is prime to $p$ when $p$ divides $|N|$, as $|P| = p$). Now $PN$ is non-Abelian, and using Ito's theorem again, we see that the degrees of its irreducible characters divide $p$, so $\chi$ must restrict irreducibly to $PN$, and we must have $PN = G$ by the minimality of $|G|$. But this contradicts $G = G^{\prime}$, since if $G = PN$, we have $G^{\prime} \leq N$ and $G$ is solvable. Our conclusion when $G = G^{\prime}$ is that $G$ is non-Abelian simple and that $p^{2}$ does not divide $|G|$. But then a Theorem of R. Brauer tells us that if $|G| <p^{3}$ for some prime $p$ and $G$ is non-Abelian simple with Sylow $p$-subgroup of order $p >3$, we have $G \cong {\rm PSL}(2,p).$

Hence we may suppose that $G \neq G^{\prime}.$ But then $G^{\prime}$ is Abelian and $G$ is solvable. Now as $|G| < p^{3},$ we have $G \neq F(G)$, so that $F(G)$ is Abelian, and in fact $F(G)$ must be the unique maximal normal subgroup of $G$, as all proper normal subgroups of $G$ are Abelian. Hence $[G:F(G)] = p$ ( for $[G:F(G)]$ must be a prime $q$, while all irreducible characters of $G$ have degree dividing $[G:F(G)]$, again using Ito's theorem, so $q=p).$ Thus $N = O_{p^{\prime}}(F(G))$ is not central in $G$. Hence there is a Sylow $r$-subgroup $R$ of $N$ such that $RP$ is non-Abelian and (again using Ito) has an irreducible character of degree $p$. We have $G = RP$ by the minimality of $|G|$. By standard facts about coprime automorphisms, and using minimality, we have $R = [R,P]$ and $C_{R}(P) = 1$ as $R$ is Abelian. Now $N_{R}(P) = C_{R}(P) =1$, so that $N_{G}(P) = P$ and Sylow's Theorem implies that $|R| \equiv 1$ (mod $p$).

We may summarize as follows:

Let $G$ be a finite group with a complex irreducible character of prime degree $p$. Then if there is positive integer $m < \frac{p^{2}-3}{2p}$ such that $mp+1$ is a power of a prime, we have $|G| \geq p(mp+1)$ for the smallest such integer $m$. If there is no such positive integer $m$, then we have $|G| \geq \frac{p(p^{2}-1)}{2}$. Furthermore, in either case, there is a finite group $G$ with an irreducible character of degree $p$ which attains that bound.

An interesting example is provided by $p = 19$. The smallest positive integer $m$ such that $19m+1$ is a prime power is $m = 10.$ Hence the smallest solvable group with an irreducible character of degree $19$ has order $19 \times 191$. However ${\rm PSL}(2,19)$ has order $180 \times 19$, so that ${\rm PSL}(2,19)$ is the smallest group which has a complex irreducible character of degree $19$.