A sufficient conditon for solvability of finite groups

TL;DR

This paper proves that finite groups lacking elements of certain orders (2, 3, 4, 5) are necessarily solvable, establishing specific conditions based on element order sets.

Contribution

The paper introduces new sufficient conditions for the solvability of finite groups based on the absence of elements with specific orders.

Findings

If no elements of order 2 exist, the group is solvable.

If no elements of orders 3 or 4 exist, the group is solvable.

If no elements of orders 3 or 5 exist, the group is solvable.

Abstract

The following theorem is proved: Let $G$ be a finite group and $\pi_e(G)$ be the set of element orders in $G$. If $\pi_e(G) \cap \{2\}=\emptyset$; or $\pi_e(G) \cap \{3, 4\}=\emptyset$; or $\pi_e(G) \cap \{3,5\}=\emptyset$, then $G$ is solvable. Moreover, using the intersection with $\pi_e(G)$ being empty set to judge $G$ is solvable or not, only the above three cases.

Equations

$$(2^m-1,2^n-1)=2^d-1.$$

Full text

Dept. of Math., Chongqing Univ. of Arts and Science, Chongqing 402160

00footnotetext: The author gratefully acknowledges the support by National Natural Science Foundation of China (Grant No. 11171364, 11271301, 11671063).

AMS Subject Classification: 20D10; 20D60 Key words and phrases: finite group, solvability, set of element orders

1 Introduction

Let $G$ be a finite group. We have two basic sets: $|G|$ and $\pi_{e}(G)$. There are many famous works about $|G|$ in the history of group theory. The set $\pi_{e}(G)$ was studied first by author in [1]. The main results in [1] are:

Lemma 1.1

Let $G$ be a finite group. If $\pi_{e}(G)=\{1,2,3,5\}$, then $G\cong A_{5}$.

One can easily get the following conclusion from [2]:

Lemma 1.2

Let $G$ be a finite group with $\pi_{e}(G)=\{2,$ the factors of $(2^{n}-1)$ and $(2^{n}+1),n\geq 2\}$. Then $G\cong L_{2}(2^{n})$.

For the simple group $Sz(2^{2m+1})$, we have the following result([3]):

Lemma 1.3

Let $G$ be a finite group. If $\pi_{e}(G)=\{2,4,$ the factors of $(2^{2m+1}-1),(2^{2m+1}-2^{m+1}+1)$ and $(2^{2m+1}+2^{m+1}+1),m\geq 1\}$, then $G\cong Sz(2^{2m+1})$.

We will use the above three lemmas in the following discussion.

Compare with the study of $|G|$, we can also ask similar question about $\pi_{e}(G)$:

In [4], one can find the definition of $h$ function: for a set $S$ of element orders of a finite group, $h(S)$ is defined to be the number of non-isomorphic groups $G$ with $\pi_{e}(G)=S$. There are many study about the $h$ function with $h(S)=1$, which means that the group $G$ can be characterized by $\pi_{e}(G)$. The recent study can be find in [5].

Same as CLT-group ([6]), we study COE-group in [7].

As the solvability can be decided by the order of a group ([8]), we give condition about solvabilty by $\pi_{e}(G)$. This can be seen as a supplement of the following theorem in [9]:

Theorem 1.4

Let $G$ be a finite group and $\pi_{e}(G)$ be the set of element orders. Let $|\pi(G)|$ be the number of prime in $\pi_{e}(G)$ and $|\chi(G)|$ be the number of composite number in $\pi_{e}(G)$. Then $|\pi(G)|\leq|\chi(G)|+3$, and $G$ is simple if $|\pi(G)|=|\chi(G)|+3$.

Definition 1.5

Let $G$ be a finite group and $\pi_{e}(G)$ be the set of element orders of $G$. A set $T$ is called IES if $T\cap\pi_{e}(G)=\emptyset$ implies $G$ is solvable.

If an IES-set $T$ with $|T|=1$, then by Feit-Thompson theorem, we get $T=\{2\}$. For the other set $T$, there are examples of non-solvable groups. So we get

For an IES-set, $|T|=1$ if and only if $T=\{2\}$.

We can consider the IES-set $T$ with $|T|=2$. If $T=\{2,*\}$, where $*$ is an integer $>2$, we can get $G$ is solvable and this is trivial. We need to consider the case that $T=\{3,**\}$, where $**>3$.

J. G. Thompson classified all the minimal simple groups in [10]:

Lemma 1.6

The minimal simple groups are:

(1) $L_{2}(p)$, $p>3$, $5\nmid(p^{2}-1)$, where $p$ is a prime;

(2) $L_{2}(2^{p})$, where $p$ is a prime;

(3) $L_{2}(3^{p})$, where $p$ is a odd prime;

(4) $L_{3}(3)$;

(5) $Sz(2^{2m+1})$, where $2m+1$ is a prime.

Notice that in the above lemma, the simple groups in (1)-(4) all have an element of order $3$. For the simple group $Sz(2^{2m+1})$, since $\pi_{e}(G)=\{2,4,$ the factors of $(2^{2m+1}-1),(2^{2m+1}-2^{m+1}+1)$ and $(2^{2m+1}+2^{m+1}+1),m\geq 1\}$, we get that $3\not\in\pi_{e}(Sz(2^{2m+1})))$, but $4,5\in\pi_{e}(Sz(2^{2m+1}))$($5\mid(2^{2m+1}-2^{m+1}+1)(2^{2m+1}+2^{m+1}+1)$ for $5\mid(2^{4m+2}+1)$). Hence if $|T|=2$, then $T=\{3,4\}$ and $\{3,5\}$ are IES-sets. We claim that there are no other IES-set $T$ with $|T|=2$.

(a) Let $T=\{3,x\}$, where $x>5$. Notice $\pi_{e}(Sz(2^{3}))=\{1,2,4,5,7,13\}$ and $\pi_{e}(Sz(2^{5}))=\{1,2,4,5,25,31,41\}$. We get that the common element orders in the above two minimal simple groups are $\{1,2,4,5\}$. So we can get a conterexample for any $x>5$.

(b) Let $T=\{4,y\}$, where $y>4$. Notice $\pi_{e}(A_{5})=\{1,2,3,5\}$. To exclude this case, we know $y=5$. Since $\pi_{e}(L_{2}(2^{3}))=\{1,2,3,7,9\}$, thus such $y$ such that $T$ is an IES-set does not exist.

(c) Let $T=\{5,z\}$, where $z>5$. Since $\pi_{e}(L_{2}(2^{3}))=\{1,2,3,7,9\}$ and $\pi_{e}(L_{2}(2^{5}))=\{1,2,3,11,31,33\}$, we get that such $T$ does not exists by the same reason.

For the other $T$ with $|T|=2$, we can find a counterexample from $\pi_{e}(A_{5})=\{1,2,3,5\}$.

Next we consider the case of $|T|=3$. Let $T=\{n_{1},n_{2},n_{3}\}$. From the above discussion, $n_{1},n_{2},n_{3}$ are odd. We can assume that $n_{1}<n_{2}<n_{3}$.

Suppose $n_{1}=3$. If $n_{2}$ or $n_{3}\in\{4,5\}$, it is a trivial consequence of the above discussion.

So we can assume that $n_{2}>5$ and $m_{3}>6$. Since $\pi_{e}(Sz(2^{3}))=\{1,2,4,5,7,13\}$, $\pi_{e}(Sz(2^{5}))=\{1,2,4,5,25,31,41\}$ and $\pi_{e}(Sz(2^{7}))=\{1,2,4,5,29,113,127,145\}$, we can get an counterexample for any $n_{2},n_{3}$. Hence such $T$ exists.

Suppose $n_{1}=4$. We can assume that $n_{2}>4$ and $n_{3}>5$. Since $\pi_{e}(L_{2}(2^{3}))=\{1,2,3,7,9\}$, $\pi_{e}(L_{2}(2^{5}))=\{1,2,3,11,31,33\}$ and $\pi_{e}(L_{2}(2^{7}))=\{1,2,3,43,127,129\}$, we can get an counterexample for any $n_{2},n_{3}$. Hence such IES-set does not exist. We can also get the same conclusion for $n_{1}=5$.

For $n_{1}>5$, $\pi_{e}(A_{5})=\{1,2,3,5\}$ will give us an counterexample.

Therefore, the nontrivial IES-set $T$ with $|T|=3$ does not exist.

Finally, we consider the case of IES-set $T$ with $|T|>3$. We claim that no such IES-set exists. To do this, we give the following lemma.

Lemma 1.7

Let $m,n$ be two integer with $d=(m,n)$. Then $(2^{m}-1,2^{n}-1)=2^{d}-1$.

Proof: Suppose that $m=nq+r$, where $0\leq r<n$. Then 2^{m}-1=2^{nq+r}-1=2^{r}(2^{nq}-1)+2^{r}-1$$\equiv$$2^{r}-1(\mod{2^{n}-1}). So $(2^{m}-1,2^{n}-1)=(2^{r}-1,2^{n}-1)$. In this way, we can get

[TABLE]

$\Box$

This lemma is a direct consequence of Theorem 1 of Section 7.4 in [11].

Corollary 1.8

Let $p,q$ be two different primes. Then $\pi_{e}(L_{2}(2^{p}))\cap\pi_{e}(L_{2}(2^{q}))=\{1,2,3\}$.

Proof: It is well-known that $|L_{2}(2^{p})|=2^{p}(2^{2p}-1)$, $|L_{2}(2^{q})|=2^{q}(2^{2q}-1)$, and $(2p,2q)=2$. By Lemma 1.7, there is no common element order in $L_{2}(2^{p})$ and $L_{2}(2^{q})$ except $1,2,3$. $\Box$

Corollary 1.9

Let $p,q$ are two different odd primes. Then $\pi_{e}(Sz(2^{p}))\cap\pi_{e}(Sz(2^{q}))=\{1,2,4,5\}$.

Proof: We first prove that $(2^{2p}+1,2^{2q}+1)=5$ for any different odd prime $p,q$.

Clearly, $(2^{2p}+1,2^{2q}+1)\mid(2^{4p}-1,2^{4q}-1)$. By Lemma 1.7, $(2^{4p}-1,2^{4q}-1)=2^{4}-1=3\times 5$. Since $3\nmid 2^{2p}+1$, $(2^{2p}+1,2^{2q}+1)=5$.

In the same way, we can get that $(2^{2p}+1,2^{q}-1)=(2^{2q}+1,2^{p}-1)=1$.

It is easy to get that $\pi_{e}(Sz(2^{p}))\cap\pi_{e}(Sz(2^{q}))=\{1,2,4,5\}$. $\Box$

Now we continue the discussion of IES-set $T$ with $|T|=s>3$. Let $T=\{n_{1},n_{2},\cdots,n_{s}\}$, where $n_{1}<n_{2}<\cdots<n_{s}$ and $s>3$.

As above, the case that $n_{1}=2$, or $\{n_{1},n_{2}\}=\{3,4\}$, or $\{n_{1},n_{2}\}=\{3,5\}$ is trivial. If $n_{1}>5$, $A_{5}$ will be a counterexample. So we need to consider the following cases:

(d) Let $T=\{3,n_{2},\cdots,n_{s}\}$, where $n_{2}>5$. Notice that $\pi_{e}(Sz(2^{p}))\cap\pi_{e}(Sz(2^{q}))=\{1,2,4,5\}$. We get that there exists $k\in\pi_{e}(Sz(2^{p}))-\{1,2,4,5\}$ and $k\not\in\pi_{e}(Sz(2^{q}))$. Hence for any $s=|T|$, there are enough large primes $p$ to provide counterexample. Thus no such $T$ exists.

(e) Let $T=\{4,n_{2},\cdots,n_{s}\}$, where $n_{2}>4$. Notice $\pi_{e}(L_{2}(2^{p}))\cap\pi_{e}(L_{2}(2^{q}))=\{1,2,3\}$. We get that there exists $k\in\pi_{e}(L_{2}(2^{p}))-\{1,2,3\}$ and $k\not\in\pi_{e}(L_{2}(2^{q})),q\neq p$. Hence for any $s=|T|$, there are enough large primes $p$ to provide counterexample. Thus no such $T$ exists.

Hence we get:

Theorem 1.10

Let $G$ be a finite group and $\pi_{e}(G)$ be the set of element orders of $G$. If $2\not\in\pi_{e}(G)$, $\pi_{e}(G)\cap\{3,4\}=\emptyset$, or $\pi_{e}(G)\cap\{3,5\}=\emptyset$, then $G$ is solvable. Furthermore, $T$ is an IES-set if and only if $T=\{2\}$, $\{3,4\}$ or $\{3,5\}$.

Acknowledgements The author would like to thank Prof. Ming Luo for his help in number theory.