On Groups with the Same Set of Conjugacy Class Sizes as Nilpotent Groups

Let $G$ be a finite group. In [1], Baer defined the index of $x$ in $G$, denoted by ${\operatorname{Ind}}_{G}(x)$, as $|G:{\mathrm{C}}_{G}(x)|$, which represents the size of the conjugacy class of $G$ containing $x$. In [11], Itô defined the conjugate type vector of $G$ as $(n_{1},n_{2},\ldots,n_{r})$, where $n_{1}>n_{2}>\ldots>n_{r}=1$ are the indices of all elements in $G$. Since we are not interested in the ordering of these indices, we will denote the set of indices (sizes of conjugacy classes) by ${\mathrm{N}}(G)$, i.e., ${\mathrm{N}}(G)=\{n_{1},n_{2},\ldots,n_{r}\}$.

Many authors have studied the relationship between the structure of finite groups and the sizes of their conjugacy classes. Itô proved that if ${\mathrm{N}}(G)=\{1,n\}$, then $G$ must be the direct product of a $p$-group and an abelian $p^{\prime}$-group [11]. Ishikawa proved that the nilpotent class of such groups is at most $3$ [10]. More results can be found in [4].

It is easy to see that if $G$ is nilpotent, then ${\mathrm{N}}(G)={\mathrm{N}}(P_{1})\times{\mathrm{N}}(P_{2})\times\ldots\times{\mathrm{N}}(P_{k})$, where $P_{1},P_{2},\ldots,P_{k}$ are the Sylow subgroups of $G$. A natural question is whether the converse holds:

In [6], Cossey proved that every finite set of $p$-powers containing $1$ can be the set of conjugacy class sizes of some $p$-group. Therefore, the above question can be restated as follows: If ${\mathrm{N}}(G)=\Omega_{1}\times\Omega_{2}\times\cdots\times\Omega_{r}$, where $\Omega_{i}$ is a finite set of $p_{i}$-powers containing $1$ and $p_{1},p_{2},\ldots,p_{r}$ are distinct primes, is $G$ nilpotent? The answer is positive in some special cases. For example, if ${\mathrm{N}}(G)=\{1,p_{1}^{m_{1}}\}\times\{1,p_{2}^{m_{2}}\}\times\cdots\times\{1,p_{k}^{m_{k}}\}$, where $p_{1}^{m_{1}},p_{2}^{m_{2}},\ldots,p_{k}^{m_{k}}$ are powers of distinct primes, then $G$ is nilpotent [5]. More generally, if ${\mathrm{N}}(G)=\{1,n_{1}\}\times\{1,n_{2}\}\times\cdots\times\{1,n_{r}\}$, where $n_{1},n_{2},\ldots,n_{r}$ are pairwise coprime integers, then $G$ is nilpotent [9]. A more general question is as follows:

However, the answer to Question 1 is not always true, as some counterexamples are provided in [3]. In that paper, A. R. Camina posed a number of questions about the structure of groups with the same set of conjugacy class sizes as nilpotent groups. One of them is as follows:

Using GAP[7], we find that Question 3 does not have a positive answer in general. The smallest counterexamples are two groups of order $486=3^{5}\times 2$, with the set of conjugacy class sizes $\{1,3,27\}\times\{1,2\}$. One of them is SmallGroup(486, 36), and the other is SmallGroup(486, 38). Moreover, we constructed the following series of counterexamples.

From this theorem, the following corollary can be derived.

A prime number $q$ such that $2q+1$ is also a prime is called a Sophie Germain prime. The largest known proven Sophie Germain prime is $2618163402417\times 2^{1290000}-1$ [2]. It is conjectured that there are infinitely many Sophie Germain primes, but this has not been proven. So we cannot conclude that there are infinitely many counterexamples to Question 3.

Let $G,A,B,H,N$ be as defined in the main theorem. For convenience, we use $(x_{1},x_{2},\ldots,x_{p})$ to represent the element $k_{1}^{x_{1}}k_{2}^{x_{2}}\ldots k_{p}^{x_{p}}N$ of $H$, $x_{1},\ldots,x_{p}\in\mathbb{N}$. Under this notation, we have $(x,x,\ldots,x)=1$, $\forall x\in\mathbb{N}$. We can always set $x_{1}=0$, in which case $x_{2},\ldots,x_{p}$ are determined. Let $h=(0,x_{2},\ldots,x_{p})\in H$, $a\in A$, $b\in B$ and $h,a,b\neq 1$. It is clear that $|G|=p^{p}q$.

(1) $|{\mathrm{C}}_{H}(a)|=p$ and ${\operatorname{Ind}}_{G}(a)=p^{p-2}q$.

By Lemma 1, it suffices to consider the case $a=\alpha$, i.e., when $(0,x_{2},\ldots,x_{p-1},x_{p})^{a}=(x_{p},0,x_{1},\ldots,x_{p-1})$. If $h\in{\mathrm{C}}_{H}(a)$, we have $0-x_{p}\equiv x_{2}-0\equiv\ldots\equiv x_{p}-x_{p-1}\pmod{p}.$ If $x_{p}=1$, then $h=(0,p-1,p-2,\ldots,1)$. In fact, ${\mathrm{C}}_{H}(a)=\langle(0,p-1,p-2,\ldots,1)\rangle$. Therefore $|{\mathrm{C}}_{H}(a)|=p$.

Let $h_{1}a_{1}b_{1}\in{\mathrm{C}}_{G}(a)$, where $h_{1}\in H$, $a_{1}\in A$ and $b_{1}\in B$. We have $h_{1}a_{1}b_{1}=(h_{1}a_{1}b_{1})^{a}=h_{1}^{a}a_{1}b_{1}^{a}=h_{1}^{a}(a_{1}a^{-1}a^{b_{1}^{-1}})b_{1}$. It follows that $h_{1}\in C_{H}(a)$ and $b_{1}=1$. Hence ${\mathrm{C}}_{G}(a)={\mathrm{C}}_{H}(a)A$. Thus, $|{\mathrm{C}}_{G}(a)|=p^{2}$ and so ${\operatorname{Ind}}_{G}(a)=p^{p-2}q$.

(2) $|{\mathrm{C}}_{H}(b)|=p^{2}$ and ${\operatorname{Ind}}_{G}(b)=p^{p-2}$.

It is easy to verify that ${\mathrm{C}}_{G}(b)=\langle k_{m_{1}}\ldots k_{m_{q}}N,k_{n_{1}}\ldots k_{n_{q}}N\rangle$. Therefore, ${\mathrm{C}}_{H}(b)=p^{2}$. Moreover, ${\mathrm{C}}_{H}(a)\cap{\mathrm{C}}_{H}(b)=1$.

If $h_{1}a_{1}b_{1}\in{\mathrm{C}}_{G}(b)$, then $h_{1}a_{1}b_{1}=(h_{1}a_{1}b_{1})^{b}=h_{1}^{b}a_{1}^{b}b_{1}$. It follows that $h_{1}\in{\mathrm{C}}_{H}(b)$ and $a_{1}=1$. Therefore ${\mathrm{C}}_{G}(b)={\mathrm{C}}_{H}(b)B$. We have $|{\mathrm{C}}_{G}(b)|=p^{2}q$ and so ${\operatorname{Ind}}_{G}(b)=p^{p-2}$.

(3) ${\operatorname{Ind}}_{G}(ab)=p^{p-2}$.

By Sylow’s theorems, $AB$ has $p$ Sylow $q$-subgroups. Since that $p(q-1)+p=pq=|AB|$, every element in $AB-A$ has order $q$. Hence $ab$ must be contained in some conjugate of $B$. Thus, ${\operatorname{Ind}}_{G}(ab)={\operatorname{Ind}}_{G}(b)=p^{p-2}$.

(4) $\{{\operatorname{Ind}}_{G}(h)\mid h\in H\}=\{p,q,pq\}$.

It is clear that $H\leq{\mathrm{C}}_{G}(h)$. If $h\in{\mathrm{C}}_{G}(a)$, then ${\mathrm{C}}_{G}(h)=HA$ and ${\operatorname{Ind}}_{G}(h)=q$. If $h\in{\mathrm{C}}_{G}(b)$ or ${\mathrm{C}}_{G}(ab)$, ${\mathrm{C}}_{G}(h)=HB$ or $H\langle ab\rangle$ and ${\operatorname{Ind}}_{G}(h)=p$. The number of such $h$ in all the cases above is at most $|{\mathrm{C}}_{H}(a)|+p|{\mathrm{C}}_{H}(b)|=p^{3}+p$. Here $p$ must be greater than or equal to $5$, so $p^{3}+p<p^{p-1}=|H|$. Hence there exists $h\in H$ such that ${\mathrm{C}}_{G}(H)=H$. For such $h$, ${\operatorname{Ind}}_{G}(h)=pq$.

(5) ${\operatorname{Ind}}_{G}(ha)=p^{p-2}q$.

Let $h_{1}a_{1}b_{1}\in{\mathrm{C}}_{G}(ha)$, where $h_{1}\in H$, $a_{1}\in A$ and $b_{1}\in B$. We have $ha=(ha)^{h_{1}a_{1}b_{1}}=(hh_{1}^{-1}h_{1}^{a^{-1}})^{a_{1}b_{1}}a^{b_{1}}$. Hence $b_{1}=1$ and so ${\mathrm{C}}_{G}(ha)={\mathrm{C}}_{HA}(ha)$. We have $(ha)^{p}=(hh^{a^{-1}}h^{a^{-2}}\ldots h^{a^{1-p}})a^{p}=hh^{a^{-1}}h^{a^{-2}}\ldots h^{a^{1-p}}$. If $h=(x_{1},x_{2},\ldots,x_{p})$, then $hh^{a^{-1}}h^{a^{-2}}\ldots h^{a^{1-p}}=(x_{1}+\ldots+x_{p},\ldots,x_{1}+\ldots+x_{p})=1$. Hence $ha$ is an element of order $p$. Since $\langle ha\rangle\cap H=1$, We have $HA=H\rtimes\langle ha\rangle$. Therefore ${\mathrm{C}}_{G}(ha)={\mathrm{C}}_{H}(ha)\langle ha\rangle={\mathrm{C}}_{H}(a)\langle ha\rangle$. Thus $|{\mathrm{C}}_{G}(ha)|=p^{2}$ and so ${\operatorname{Ind}}_{G}(ha)=p^{p-2}q$.

(6) ${\operatorname{Ind}}_{G}(hab)={\operatorname{Ind}}_{G}(hb)=p^{p-2}$.

By (3), $ab$ and $b$ are conjugate, so $hab$ must be conjugate to $h^{\prime}b$ where $h^{\prime}$ is some element in $H$. Thus, we only need to consider ${\operatorname{Ind}}_{G}(hb)$. Let $h_{1}a_{1}b_{1}\in{\mathrm{C}}_{G}(hb)$ where $h_{1}\in H,a_{1}\in A,b_{1}\in B$. We have $hb=(hb)^{h_{1}a_{1}b_{1}}=(hh_{1}^{-1}h_{1}^{b^{-1}}b)^{a_{1}b_{1}}=(hh_{1}^{-1}h_{1}^{b^{-1}})^{a_{1}b_{1}}(a_{1}^{-1}a_{1}^{b^{-1}})^{b_{1}}b$. Hence $a_{1}=1$ and ${\mathrm{C}}_{G}(hb)={\mathrm{C}}_{HB}(hb)$. It follows that ${\operatorname{Ind}}_{G}(hb)={\operatorname{Ind}}_{HB}(hb)\times p$. By Lemma 2, ${\operatorname{Ind}}_{HB}(hb)=|H:{\mathrm{C}}_{H}(b)|=p^{p-3}$. Therefore, ${\operatorname{Ind}}_{G}(hb)=p^{p-2}$.

From (1)–(6), all nontrivial elements of $G$ have been considered, so ${\mathrm{N}}(G)=\{1,p,p^{p-2}\}\times\{1,q\}$ and ${\mathrm{Z}}(G)=1$. The theorem is proved.