Representation dimension of some finite groups

∎

• (i)

  Let $\delta(G)=p$. In view of Theorem 3.1, $\delta(G)$ equals $\delta_{irr}(G)$. Consequently, $\delta_{irr}(G)=p$. Hence, the derived length of $G$ is at most $2$, due to Theorem 22.25 of [4]. This implies that $G$ is a metabelian group. Since metabelian groups are normally monomial, it follows as a consequence of Remark 3.2 that the set of all possible irreducible character degrees is $\{1,p\}$. Now, suppose that $\operatorname{cd}(G)=\{1,p\}$. Since $G$ is a finite $p$-group with cyclic center, it follows from Theorem 3.1 that $\delta_{irr}(G)$ exists and $\delta(G)=\delta_{irr}(G)$. Consequently, $\delta(G)=\delta_{irr}(G)=p$.

• (ii)

  If $\operatorname{cd}(G)=\{1,p^{a}\}$, then it follows from Theorem 3.1 that $\delta(G)=\delta_{irr}(G)=p$. If $\operatorname{cd}(G)=\{1,p,p^{a}\}$, then it immediately follows from Lemma 21 of [33] that $\delta(G)=p$.

∎

We first prove that in each of the following cases, $\delta(G_{i})=\delta_{irr}(G_{i}),$ for $i=1,2$.
Case (1). $G_{1}$ and $G_{2}$ are simple groups.
Since $G_{i}$ is a simple group, for $i=1,2$, we have that any nonlinear character must be faithful. Hence $\delta(G_{i})=\delta_{irr}(G_{i}).$
Case (2). $G_{1}$ and $G_{2}$ are $p$-groups with cyclic center.
It immediately follows from Theorem 3.1(1) that $\delta(G_{i})=\delta_{irr}(G_{i}),$ for $i=1,2$.
Case (3). $G_{1}$ and $G_{2}$ are monolithic groups.
Let $G$ be a monolithic group and let $\chi$ be a faithful character of $G$ with $\deg(\chi)=\delta(G)$. We write $\chi=\sum\eta_{i}\psi_{i}$, where $\psi_{i}$ are irreducible characters of $G$. Now, $\{1\}=\operatorname{ker}(\chi)=\cap\operatorname{ker}(\psi_{i}),$ where $\operatorname{ker}(\chi)$ denotes the kernel of the character $\chi$. We observe that $\operatorname{ker}(\psi_{j})=\{1\}$ for some $j$, else $\operatorname{ker}(\chi)$ would contain the minimal normal subgroup of $G$, contradicting that $\chi$ is faithful. Thus, $\delta(G)\leq\deg(\psi_{j})\leq\deg(\chi)=\delta(G).$ Hence, $\chi$ must be a faithful irreducible character of $G$. Consequently, $\delta(G_{i})=\delta_{irr}(G_{i})$, for $i=1,2$.
Case (4). $G_{1}$ and $G_{2}$ have unique nonlinear irreducible character.
In view of Lemma 1.4 of [19], $\delta(G_{i}),$ for $i=1,2$, is the degree of the unique nonlinear irreducible character and hence the result follows.

In all the above cases, it is proved that $\delta(G_{i})=\delta_{irr}(G_{i}),$ for $i=1,2$. Defining a representation on $G_{1}\times G_{2}$ by block diagonal matrices yields that $\delta(G_{1}\times G_{2})\leq\delta(G_{1})+\delta(G_{2})$. If the orders of $\mathcal{Z}(G_{1})$ and $\mathcal{Z}(G_{2})$ are relatively coprime, then $\delta_{irr}(G_{1}\times G_{2})=\delta_{irr}(G_{1})\delta_{irr}(G_{2})$, due to Problem 4.3 of [20].

Suppose that $G_{1}$ and $G_{2}$ are of coprime order. If $\chi$ is a minimal faithful character of $G_{1}\times G_{2}$, then $\chi=\chi_{1}+\chi_{2}+\cdots+\chi_{n}$, where $\chi_{i}\in\operatorname{Irr}(G_{1}\times G_{2})$, for $1\leq i\leq n$. Let $\chi_{i}=\tau_{i}\psi_{i}$, for $1\leq i\leq n$, where $\tau_{i}\in\operatorname{Irr}(G_{1})$ and $\psi_{i}\in\operatorname{Irr}(G_{2})$. Since $G_{1}$ and $G_{2}$ are of coprime order, $\langle(g_{1},g_{2})~|~g_{1}\in\underset{1\leq i\leq n}{\cap}\operatorname{ker}(\tau_{i}),g_{2}\in\underset{1\leq i\leq n}{\cap}\operatorname{ker}(\psi_{i})\rangle\leq\underset{1\leq i\leq n}{\cap}\operatorname{ker}(\tau_{i}\psi_{i})=\operatorname{ker}(\chi)=\{1\}$. This implies that $\langle(g_{1},g_{2})~|~g_{1}\in\underset{1\leq i\leq n}{\cap}\operatorname{ker}(\tau_{i}),g_{2}\in\underset{1\leq i\leq n}{\cap}\operatorname{ker}(\psi_{i})\rangle=\{1\}$. Hence $\tau_{1}+\tau_{2}+\cdots+\tau_{n}$ is a faithful character of $G_{1}$ and $\psi_{1}+\psi_{2}+\cdots+\psi_{n}$ is a faithful character of $G_{2}$. If $\tau\in\operatorname{Irr}(G_{1})$ and $\psi\in\operatorname{Irr}(G_{2})$ such that $\operatorname{deg}(\tau)=\delta_{irr}(G_{1})=\delta(G_{1})$ and $\operatorname{deg}(\psi)=\delta_{irr}(G_{2})=\delta(G_{2})$, then $\tau\psi\in\operatorname{Irr}(G_{1}\times G_{2})$ is a minimal faithful character of $G_{1}\times G_{2}$. Therefore, $\delta(G_{1}\times G_{2})=\delta_{irr}(G_{1})\delta_{irr}(G_{2})$.∎

Since $G$ is a nilpotent group, it is a direct product of Sylow $p$-subgroups. Due to Theorem 4.1, it follows that $\delta(G)=\delta_{irr}(G)$. ∎

From Theorem 1 of [10], a non abelian group $G$ of odd order which has exactly two non linear irreducible characters of each degree, is one of the following:

• (i)

  an extraspecial $3$-group of order $3^{2r+1}$, for some integer $r\geq 1$; or

• (ii)

  a Frobenius group of odd order $\frac{(p^{n}-1)p^{n}}{2}$ for some odd prime $p$ and some integer $n$, with abelian Frobenius kernel $K$ of order $p^{n}$ and cyclic Frobenius complement.

Clearly, (i) follows from Theorem 3.1(6). For (ii), first consider the case when the Frobenius kernel $K$ of the Frobenius group $G$ is cyclic, and so is its Frobenius complement. We denote the Frobenius complement by $H$. It follows from Theorem 37.5.1 of [25] that the irreducible characters of $G$ are $\chi_{1}^{G},~\chi_{2}^{G},\cdots,~\chi_{r}^{G},\lambda_{1},~\lambda_{2},\cdots,~\lambda_{s}$, where $\{\chi_{1},\chi_{2},\cdots,\chi_{r}\}$ is a complete set of representatives of $G$-conjugacy classes of non principal irreducible characters of $K$, $\chi_{i}^{G}$ are their induced representations on $G$, for $1\leq i\leq n$, and $\lambda_{1},\lambda_{2},\cdots,\lambda_{s}$ are extensions of irreducible characters of $H$ to $G$. Clearly, $\lambda_{i}$’s are linear non faithful characters, for $1\leq i\leq s$. Note that the degree of each nonlinear irreducible character of $G$ is $[G:K]$. Since $K$ is cyclic, $G$ has a nonlinear irreducible character which is faithful of degree $[G:K]$, i.e., $\frac{p^{n}-1}{2}$. Since it is the minimal nonlinear irreducible character degree, it immediately follows that $\delta(G)=\delta_{irr}(G)=\frac{p^{n}-1}{2}$.

Now, suppose that the Frobenius kernel $K$ of $G$ is abelian but non cyclic and its complement $H$ is cyclic. Since $K$ is abelian, it is a direct product of cyclic groups. Hence, each subgroup of $K$ is normal in $G$. Consequently, for $1\leq i\leq n$, $\operatorname{ker}(\chi_{i}^{G})=\operatorname{core}_{G}(\operatorname{ker}(\chi_{i}))=\operatorname{ker}(\chi_{i}),$ where $\operatorname{core}_{G}(\operatorname{ker}(\chi_{i}))$ denotes the largest normal subgroup of $G$ contained in $\operatorname{ker}(\chi_{i})$. As $K$ is non cyclic, the Frobenius kernel $K$ has no faithful irreducible character. Hence, the group $G$ does not have a faithful irreducible character. Further, $K\leq\operatorname{ker}(\lambda_{i})$, for $1\leq i\leq s$. Therefore, $\delta(G)=[G:K]\delta(K)=\frac{(p^{n}-1)}{2}\operatorname{rank}(K)$. This completes the proof of the theorem. ∎

Let $G$ be a nonabelian finite group whose all nonlinear irreducible characters have distinct degrees. Then, due to the above theorem, we can deal with such groups case by case.
Case 1. $G$ is an extraspecial $2$-group.
In this case, $\delta(G)=\delta_{irr}(G)=2^{r}$, in view of Theorem 3.1(6).
Case 2. $G$ is a Frobenius group of order $p^{n}(p^{n}-1)$ for some prime power $p^{n}$ with an abelian Frobenius kernel $K$ of order $p^{n}$ and a cyclic Frobenius complement $H$.
It follows from Theorem 5.1 of [25] that all the irreducible characters of $G$ are obtained from the extensions of the irreducible characters of $H$, and the rest irreducible characters of $G$ are induced from the representatives of $G$-conjugacy classes of non principal irreducible characters of $K$. As $K$ is abelian, the degree of each nonlinear irreducible character is $[G:K]$. By assumption, all non-linear irreducible characters of the group $G$ have distinct degrees. This yields that there is a unique non-linear irreducible character of degree $[G:K]$, which is faithful, due to Lemma 1.4 of [19]. Consequently, $\delta(G)=\delta_{irr}(G)=p^{n}-1$.
Case 3. $G$ is a Frobenius group of order $72$ in which the Frobenius complement is isomorphic to the quaternion group.
From Lemma 1.4 of [19], it follows that $G$ has a unique nonlinear irreducible faithful character of degree 8. This completes the proof of the theorem. ∎