arXiv:1704.01487v1 [] 5 Apr 2017

A sufficient conditon for solvability of finite groups
Wujie Shi

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

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

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Introduction

Let G be a finite group. We have two basic sets: |G| and πe (G). There are many famous
works about |G| in the history of group theory. The set πe (G) was studied first by author
in [1]. The main results in [1] are:
Lemma 1.1 Let G be a finite group. If πe (G) = {1, 2, 3, 5}, then G ∼
= A5 .
One can easily get the following conclusion from [2]:
Lemma 1.2 Let G be a finite group with πe (G) = {2, the factors of (2n − 1) and (2n +
1), n ≥ 2}. Then G ∼
= L2 (2n ).
For the simple group Sz(22m+1 ), we have the following result([3]):
Lemma 1.3 Let G be a finite group. If πe (G) = {2, 4, the factors of (22m+1 − 1), (22m+1 −
2m+1 + 1) and (22m+1 + 2m+1 + 1), m ≥ 1}, then G ∼
= Sz(22m+1 ).
0

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

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Solvability of a finite group

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We will use the above three lemmas in the following discussion.
Compare with the study of |G|, we can also ask similar question about πe (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 πe (G) = S.
There are many study about the h function with h(S) = 1, which means that the group
G can be characterized by πe (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 πe (G). This can be seen as a supplement of the following theorem in [9]:
Theorem 1.4 Let G be a finite group and πe (G) be the set of element orders. Let |π(G)|
be the number of prime in πe (G) and |χ(G)| be the number of composite number in πe (G).
Then |π(G)| ≤ |χ(G)| + 3, and G is simple if |π(G)| = |χ(G)| + 3.
Definition 1.5 Let G be a finite group and πe (G) be the set of element orders of G. A
set T is called IES if T ∩ πe (G) = ∅ 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) L2 (p), p > 3, 5 ∤ (p2 − 1), where p is a prime;
(2) L2 (2p ), where p is a prime;
(3) L2 (3p ), where p is a odd prime;
(4) L3 (3);
(5) Sz(22m+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(22m+1 ), since πe (G) = {2, 4, the factors of (22m+1 −
1), (22m+1 − 2m+1 + 1) and (22m+1 + 2m+1 + 1), m ≥ 1}, we get that 3 6∈ πe (Sz(22m+1 ))),

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Solvability of a finite group

but 4, 5 ∈ πe (Sz(22m+1 ))(5 | (22m+1 − 2m+1 + 1)(22m+1 + 2m+1 + 1) for 5 | (24m+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 πe (Sz(23 )) = {1, 2, 4, 5, 7, 13} and πe (Sz(25 )) =
{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 πe (A5 ) = {1, 2, 3, 5}. To exclude this case, we
know y = 5. Since πe (L2 (23 )) = {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 πe (L2 (23 )) = {1, 2, 3, 7, 9} and πe (L2 (25 )) =
{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 πe (A5 ) = {1, 2, 3, 5}.
Next we consider the case of |T | = 3. Let T = {n1 , n2 , n3 }. From the above discussion,
n1 , n2 , n3 are odd. We can assume that n1 < n2 < n3 .
Suppose n1 = 3. If n2 or n3 ∈ {4, 5}, it is a trivial consequence of the above discussion.
So we can assume that n2 > 5 and m3 > 6. Since πe (Sz(23 )) = {1, 2, 4, 5, 7, 13},
πe (Sz(25 )) = {1, 2, 4, 5, 25, 31, 41} and πe (Sz(27 )) = {1, 2, 4, 5, 29, 113, 127, 145}, we can
get an counterexample for any n2 , n3 . Hence such T exists.
Suppose n1 = 4. We can assume that n2 > 4 and n3 > 5. Since πe (L2 (23 )) =
{1, 2, 3, 7, 9}, πe (L2 (25 )) = {1, 2, 3, 11, 31, 33} and πe (L2 (27 )) = {1, 2, 3, 43, 127, 129}, we
can get an counterexample for any n2 , n3 . Hence such IES-set does not exist. We can also
get the same conclusion for n1 = 5.
For n1 > 5, πe (A5 ) = {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 (2m − 1, 2n − 1) = 2d − 1.
Proof: Suppose that m = nq + r, where 0 ≤ r < n. Then 2m − 1 = 2nq+r − 1 =
2r (2nq − 1) + 2r − 1≡2r − 1( mod 2n − 1). So (2m − 1, 2n − 1) = (2r − 1, 2n − 1). In this
way, we can get
(2m − 1, 2n − 1) = 2d − 1.
✷

Solvability of a finite group

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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 πe (L2 (2p )) ∩ πe (L2 (2q )) = {1, 2, 3}.
Proof: It is well-known that |L2 (2p )| = 2p (22p − 1), |L2 (2q )| = 2q (22q − 1), and (2p, 2q) = 2.
By Lemma 1.7, there is no common element order in L2 (2p ) and L2 (2q ) except 1, 2, 3. ✷

Corollary 1.9 Let p, q are two different odd primes. Then πe (Sz(2p )) ∩ πe (Sz(2q )) =
{1, 2, 4, 5}.
Proof: We first prove that (22p + 1, 22q + 1) = 5 for any different odd prime p, q.
Clearly, (22p + 1, 22q + 1) | (24p − 1, 24q − 1). By Lemma 1.7, (24p − 1, 24q − 1) = 24 − 1 =
3 × 5. Since 3 ∤ 22p + 1, (22p + 1, 22q + 1) = 5.
In the same way, we can get that (22p + 1, 2q − 1) = (22q + 1, 2p − 1) = 1.
It is easy to get that πe (Sz(2p )) ∩ πe (Sz(2q )) = {1, 2, 4, 5}.

✷

Now we continue the discussion of IES-set T with |T | = s > 3. Let T = {n1 , n2 , · · · , ns },
where n1 < n2 < · · · < ns and s > 3.
As above, the case that n1 = 2, or {n1 , n2 } = {3, 4}, or {n1 , n2 } = {3, 5} is trivial. If
n1 > 5, A5 will be a counterexample. So we need to consider the following cases:
(d) Let T = {3, n2 , · · · , ns }, where n2 > 5. Notice that πe (Sz(2p )) ∩ πe (Sz(2q )) =
{1, 2, 4, 5}. We get that there exists k ∈ πe (Sz(2p )) − {1, 2, 4, 5} and k 6∈ πe (Sz(2q )).
Hence for any s = |T |, there are enough large primes p to provide counterexample. Thus
no such T exists.
(e) Let T = {4, n2 , · · · , ns }, where n2 > 4. Notice πe (L2 (2p )) ∩ πe (L2 (2q )) = {1, 2, 3}.
We get that there exists k ∈ πe (L2 (2p )) − {1, 2, 3} and k 6∈ πe (L2 (2q )), q 6= 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 πe (G) be the set of element orders of G. If
2 6∈ πe (G), πe (G) ∩ {3, 4} = ∅, or πe (G) ∩ {3, 5} = ∅, 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.

Solvability of a finite group

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