Note on the Number of Finite Groups of a Given Order

TL;DR

This paper investigates the number of non-isomorphic finite groups of a given order, providing new proofs for cases where there are 2 such groups and solving the case where there are 3.

Contribution

It offers a novel proof for the case of exactly two finite groups of a given order and solves the problem for exactly three groups, advancing understanding of group classification.

Findings

New proof for G(n) = 2 case

Solution to G(n) = 3 case

Enhanced understanding of finite group enumeration

Abstract

Let $n$ be a positive integer and $G(n)$ denote the number of non-isomorphic finite groups of order $n$. It is well-known that $G(n) = 1$ if and only if $(n,\phi(n)) = 1$, where $\phi(n)$ and $(a, b)$ denote the Euler's totient function and the greatest common divisor of $a$ and $b$, respectively. The aim of this paper is to first present a new proof for the case of $G(n) = 2$ and then give a solution to the equation of $G(n) = 3$.

Equations

$$G(n)=\sum _S\left(\prod _{j=1}^r\frac{p_{S(j)}^{c_{S(j)}}-1}{p_{S(j)}-1}\right),$$

$$G(n)=\sum _{d|n}\prod _{p|d}\frac{f(p,\frac{n}{d}-1)}{p-1},$$

$$F_{p,q}=⟨a,b|a^p=b^q=1,b^{-1}ab=a^u⟩,$$

$$Z_{\frac{n}{p_i^2{p}_j^2}}\times Z_{p_i}\times Z_{p_i}\times Z_{p_j^2},Z_{\frac{n}{p_i^2{p}_j^2}}\times Z_{p_i^2{p}_j^2},Z_{\frac{n}{p_i^2{p}_j^2}}\times Z_{p_i^2}\times Z_{p_j}\times Z_{p_j},Z_{\frac{n}{p_i^2{p}_j^2}}\times Z_{p_i}\times Z_{p_i}\times Z_{p_j}\times Z_{p_j}$$

$$Z_n,$$

$$Z_{\frac{n}{p_1^2{p}_i{p}_j}}\times Z_{p_1^2}\times F_{p_j,p_i},$$

$$Z_{\frac{n}{p_1^2{p}_i{p}_j}}\times Z_{p_1}\times Z_{p_1}\times Z_{p_i{p}_j}$$

$$Z_{\frac{n}{p_1^2{p}_i{p}_j}}\times Z_{p_1}\times Z_{p_1}\times F_{p_j,p_i},$$

$$Aut(Z_{p_{1}s})$$

$$Aut(Z_{p_1}\times Z_s)$$

$${\phi }_1:P_l$$

$$x$$

$${\phi }_2:P_l$$

$$x$$

$${\phi }_3:P_l$$

$$x$$

Full text

Keywords: Finite group, cyclic number, abelian number, square-free order group.

2010 AMS Subject Classification Number: $20D99$.

1 Introduction

Throughout this paper all groups are assumed to be finite. The function $G(n)$ and $\phi(n)$ denote the number of non-isomorphic finite groups of order $n$ and the number of positive integers $i$ such that $1\leq i\leq n$ and $(i,n)=1$, respectively. For any other notation not defined here, we refer the reader to [2, 11]. Our calculations are done with the aid of GAP [13].

The problem of finding exact value of $G(n)$ in general seems to be very difficult, but there are some very good estimations for this function in literature. M. R. Murty and V. K. Murty [9], used this well-known fact that groups of square-free order are supersolvable to prove that for a square free number $n$, $G(n)\leq\phi(n)$.

Dickson determined those positive integers $n$ for which every group of order $n$ is abelian. As a consequence of his result it can be proved that $G(n)=1$ if and only if $(n,\phi(n))=1$. We encourage the interested readers to consult papers [4, 8] for a simpler proof of Dickson’s result. This result is our motivation to study the equation $G(n)=k$, for some small values of $k$. In an exact phrase,

Question 1.1

For which positive integers $n$ does $G(n)=2$? For which $n$ does $G(n)=3$?

This paper will be concerned with Question 1.1. For the sake of completeness, we mention here some results which are crucial throughout this paper.

Given a group $G$ of square-free order greater than 1, a theorem of H$\ddot{\rm o}$lder [6] tells us that $G$ is metacyclic. If $n=p_{1}p_{2}\ldots p_{s}$, where $p_{1}>p_{2}>\ldots>p_{s}$ are primes then

[TABLE]

where the summation is taken over all the subsets $S=\{S(1),5(2),\ldots,S(r)\}$ of $\{2,3,...,n\}$ and $c_{S(j)}$ is the number of differences $p_{i}-1,i\not\in S$, which are divisible by $p$. If we define $f(n,m)=\prod_{q|m}(n,q-1)$, then one can see that

[TABLE]

where $p$ and $q$ are primes, see [1] for details.

Suppose $G$ is a finite group and $H$ is a normal subgroup such that $p\nmid|H|$ and $\frac{|G|}{|H|}=p^{\alpha}$, where $\alpha$ is a positive integer. Then $H$ is called a normal $p-$complement for $G$. In [5, p. 252], the following result is proved:

Theorem 1.2

(Burnside’s Transfer Theorem). Let $G$ be a finite group and $P$ be a Sylow $p-$subgroup of $G$. If $P\leq Z(N_{G}(P))$ then $G$ has a normal $p-$complement.

Suppose $n=\prod_{i=1}^{r}p_{i}^{\alpha_{i}}$ and define $\Phi(n)=\prod_{i=1}^{r}(p^{\alpha_{i}}-1)$. Redei [12, Satz 10], characterized the positive integers $n$ with this property that all abelian groups of order $n$ are abelian. Such a positive integer is called an abelian number. He proved that:

Theorem 1.3

$n=\prod_{i=1}^{r}p_{i}^{\alpha_{i}}$* is an abelian number if and only if $(n,\Phi(n))=1$ , for each $i$, $1\leq i\leq r$.*

In the same manner, we call a positive integer $n$ a nilpotent number if every group of order $n$ is nilpotent. Pakianathan and Shankar [10], proved that a positive integer $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\cdots p_{3}^{\alpha_{t}}$, $p_{i}$’s are distinct primes, is a nilpotent number if and only if $p_{i}^{k}\not\equiv 1\ (modp_{j})$, for all integers $i,$ $j$ and $k$ with $1\leq k\leq\alpha_{i}$.

In this paper, our interest is the study of the function $G(n)$. The main result of this paper is as follows:

Theorem 1.4

Suppose $n=\prod_{i=1}^{r}p_{i}^{\alpha_{i}}$.

$G(n)=2$* if and only if one of the following are hold:*

1. (a)

$r=1$* and $\alpha_{1}=2$;* 2. (b)

$r\geq 2$, $n$ is square-free and there is a unique pair $(i,j)$ such that $p_{i}|p_{j}-1$; 3. (c)

$n=p_{1}^{2}p_{2}\ldots p_{r}$, $r\geq 2$, and for each $i,j$ such that $1\leq i,j\leq k$, we have $p_{i}\nmid p_{j}-1$; and for each $k$, $2\leq k\leq r$, $p_{k}\nmid p_{1}^{2}-1$. 2. 2.

$G(n)=3$* if and only if one of the following are hold:*

1. (a)

$r\geq 3$, $n$ is square-free and there is a unique triple $(i,j,l)$ such that $p_{i}|p_{j}-1$ and $p_{j}$ $|$ $p_{l}-1$; 2. (b)

$n=p_{1}^{2}p_{2}\ldots p_{r}$, $r\geq 2$, and for each $i,j$ such that $1\leq i,j\leq r$, we have $p_{i}\nmid p_{j}-1$; and there exists a unique positive integer $l$, $2\leq l\leq r$ such that $p_{l}|p_{1}^{2}-1$.

2 Proof of Main Result

The aim of this section is to prove our main results. At first, we note that there are exactly five groups of order $p^{3}$, where $p$ is prime. It is well-known that $Z_{8}$, $Z_{2}\times Z_{4}$, $Z_{2}\times Z_{2}\times Z_{2}$, $D_{8}$ and $Q_{8}$ are all groups of order eight and if $p$ is an odd prime then $Z_{p^{3}}$, $Z_{p}\times Z_{p^{2}}$ and $Z_{p}\times Z_{p}\times Z_{p}$ are only abelian groups of order $p^{3}$ and there are exactly two non-abelian groups of this order as follows:

$G_{1}=\langle x,y\ |\ x^{p}=y^{p^{2}}=1,x^{-1}yx=y^{p+1}\rangle$, 2. 2.

$G_{2}=\langle x,y,z\ |\ x^{p}=y^{p}=z^{p}=1,yz=zy,zx=xz,x^{-1}yx=yz\rangle$.

This proves that $G(p^{3})=5$. The number of groups of order $p^{2}q$, $p$ and $q$ are primes, are recorded in Table 1. This table is first appeared in the famous book of Burnside [2].

Table 1. The Number of Groups of Order $p^{2}q$, $p$ and $q$ are primes.

[TABLE]

Suppose $p$ is prime and $q|p-1$. The Frobenius group $F_{p,q}$ can be presented as follows:

[TABLE]

where $u$ is an element of order $q$ in the unit group of the ring $Z_{p}$, [7, p. 290].

2.1 The Case that $G(n)=2$.

Suppose $G(n)=2$, where $n=\prod_{i=1}^{r}p_{i}^{r_{i}}$. If there exists $i$ such that $r_{i}\geq 3$ then there are five non-isomorphic finite groups of order $p_{i}^{3}$. This shows that we have at least five groups $Z_{p_{i}^{r_{i}-3}p_{1}^{r_{1}}\ldots p_{i-1}^{r_{i-1}}p_{i+1}^{r_{i+1}}\ldots p_{k}^{r_{k}}}\times A$, where $A$ is a group of order $p_{i}^{3}$. Also, if there are $i$ and $j$, $i\neq j$ and $r_{i},r_{j}\geq 2$ then

[TABLE]

are four non-isomorphic finite groups of order $n$ which is not possible. Thus, $n$ is square-free or $n$ is a cube-free number such that there exists exactly one prime number $p$ such that $p^{2}|n$. If $n$ is square-free number then by Equation (1), there exists only one pair $(i,j)$ such that $p_{i}|p_{j}-1$, as desired. So, we can assume that $n$ is a cube-free number such that there exists exactly one prime number $p$ such that $p^{2}|n$. Since $Z_{n}$ and $Z_{p}\times Z_{\frac{n}{p}}$ are two non-isomorphic abelian group of order $n$, every group of order $n$ has to be abelian. By Theorem 1.3, the condition $1(c)$ in Theorem 1.4 is satisfied.

Note that if there are only two finite groups of order $n$ then $(n,\phi(n))$ is primes, but the converse is not generally correct. To see this, it is enough to check that $G(75)=3$ and $(75,\phi(75))=(75,40)=5$.

2.2 The Case that $G(n)=3$.

In this case the finite groups with $G(n)=3$ will be characterized. By the proof of Theorem 1.4(1), such a number $n$ is square-free or it is a cube-free integer that there exists a unique prime number $p$ such that $p^{2}|n$. Our main proof will consider two different cases as follows:

$n=p_{1}p_{2}\cdots p_{r}$, $r\geq 3$, is square-free. By Equation (1), there are exactly two pairs $(i,j)$ and $(i^{\prime},j^{\prime})$ such that $p_{i}|p_{j}-1$ and $p_{i^{\prime}}|p_{j^{\prime}}-1$.

1. (a)

$p_{i}=p_{i^{\prime}}$* and $p_{j}\neq p_{j^{\prime}}$*. In this case, by Equation (1), there are exactly $p_{i}+2\geq 4$ finite groups of order $n$, which is a contradiction. 2. (b)

$p_{i}\neq p_{i^{\prime}}$* and $p_{j}=p_{j^{\prime}}$*. Under this conditions there are four groups $Z_{n}$, $Z_{\frac{n}{p_{i}p_{j}}}$ $\times$ $F_{p_{j},p_{i}}$, $Z_{\frac{n}{p_{i}^{\prime}p_{j}^{\prime}}}$ $\times$ $F_{p_{j}^{\prime},p_{i}^{\prime}}$ and $Z_{\frac{n}{p_{i}p_{j}p_{i}^{\prime}}}$ $\times$ $F_{p_{j},p_{i}p_{i}^{\prime}}$ are four non-isomorphic groups of order $n$, which is impossible. 3. (c)

$p_{i}=p_{j^{\prime}}$. By Equation (1), there are exactly three finite groups of order $n$, as desired. 2. 2.

$n=p_{1}^{2}p_{2}\ldots p_{r}$, where $p_{i}$’s are distinct prime integers and $r\geq 2$. Suppose there are $i$ and $j$, $2\leq i,j\leq r$, such that $p_{i}|p_{j}-1$. Then there are four non-isomorphic groups as follows:

[TABLE]

which is not possible. If there exists $i$, $2\leq i\leq r$, such that $p_{1}|p_{i}-1$ or $p_{i}|p_{1}-1$ then by Table 1, there are at least four groups of order $n$ leads again to a contradiction. We now assume that the prime integers $p_{1},\cdots,$ $p_{r}$ are satisfied the condition $2(b)$ in Theorem 1.4. Choose $P_{l}$ to be the Sylow $p_{{}_{l}}-$subgroup of $G$ and $N_{l}=N_{G}(P_{l})$. Since $(|\frac{N_{l}}{P_{l}}|,|Aut(P_{l})|)=1$, $P_{l}\leq Z(N_{l})$ and by Burnside’e transfer theorem $G$ has a normal subgroup $N$ of order $p_{1}^{2}p_{2}\ldots p_{l-1}p_{l+1}\ldots p_{k}$ = $p_{1}s$. By Part (1) of this theorem, $N\cong Z_{p_{1}s}$ or $Z_{p_{1}}\times Z_{s}$. Since

[TABLE]

and $p_{l}\nmid|Aut(Z_{p_{1}s})|$, there exists a unique homomorphism $\varphi_{1}$ from $P_{l}$ into $Aut(Z_{p_{1}s})$. This homomorphism has to be trivial. On the other hand, $|GL(2,p_{1})|=(p_{1}^{2}-p_{1})(p_{1}^{2}-1)$, $p_{l}\mid|Aut(Z_{p_{1}}\times Z_{s})|$ and all Sylow $p_{l}$ subgroups of $GL(2,p_{1})$ are cyclic. This shows that all subgroups of order $p_{l}$ are conjugate in $GL(2,p_{1})$. If $\varphi$ and $\psi$ are two non-trivial homomorphism from $P_{l}$ into $Aut(Z_{P_{1}}\times Z_{s})$ then $P_{l}\times_{\psi}(Z_{p_{1}}\times Z_{s})\cong P_{l}\times_{\varphi}(Z_{p_{1}}\times Z_{s})$. Therefore, there are three groups of order $n$ which can be presented as semi-direct product by the following homomorphism:

[TABLE]

Here, $x$ and $y$ are two elements of order $p_{l}$ in the groups $P_{l}$ and $Aut(Z_{p_{1}}\times Z_{s})$, respectively.

This completes the proof.

3 Concluding Remarks

In this paper the equation $G(n)=2,3$ is considered into account. In the case that $G(n)=2$, we present a new proof for an old result of M. R. Murty and V. K. Murty. Then we continue the pioneering work of these authors to solve the equation $G(n)=3$. We check also the case for $G(n)=4$, but the problem for this case is so difficult. So, the complete solution of $G(n)=4$ is a problem for future.

Acknowledgement. The research of the authors are partially supported by the University of Kashan under grant no 364988/101.