Recognition of Some Symmetric Groups by the Set of the Order of Their Elements

A. R. Moghaddamfar, M. R. Pournaki^{*}

Abstract For a finite group G, let π_{e}(G) be the set of order of elements in G and denote S_{n} the symmetric group on n letters. We will show that if π_{e}(G) = π_{e}(H), where H is S_{p} or S_{p+1} and p is a prime with 50 < p < 100, then G ≅ H.

Keywords: Order of elements, Prime graph, Symmetric group. 2000 Mathematics Subject Classification: Primary 20D05; Secondary 20D06.

1 Introduction

For a finite group G, we denote by π_{e}(G) the set of order of elements of G. If for a given finite group H, π_{e}(G) = π_{e}(H) implies that G ≅ H, then we say H is a characterizable group. There are scattered results in the literature showing that certain groups are characterizable. It is known that the symmetric groups S_{n}, n = 2, 3, 4, 5, 6, 8 are not characterizable, but the symmetric groups S_{n}, n = 7, 9, 11, 12, 13, 14 are characterizable (see [1], [3], [6] and [8]). In [4], the authors showed that the symmetric groups S_{p} and S_{p+1}, where p ∈ {23, 29, 31, 41, 43, 47}, are characterizable. In connection with these results, the following has been conjectured in [4]:

Conjecture For all primes p ≥ 7, the symmetric groups S_{p} and S_{p+1} (with the exception of S_{8}) are characterizable.

In this paper we consider some symmetric groups and prove that they are characterizable. More precisely, the symmetric groups S_{p} and S_{p+1}, where 50 < p < 100 is a prime, will be added to the list of groups with the above property.

Main Theorem The symmetric groups S_{p} and S_{p+1}, where p is a prime and 50 < p < 100, are characterizable.

Throughout this paper, all groups are assumed to be finite and all simple groups are non-abelian. We denote by π(G) the set of all prime divisors of the order of G. We use A_{m} and S_{m} to refer to the alternating and symmetric group of degree m, respectively. Further notations are standard and can be found, for example, in [2]. We will freely use the classification of all finite simple groups.

*This research was in part supported by a grant from IPM.

2 Preliminary Lemmas

The prime graph Γ(G) of a group G is the graph whose vertex set is π(G) and two vertices p, q ∈ π(G) are adjacent if and only if G contains an element of order pq. Denote by t(G) the number of connected components of Γ(G), and by π_{i} = π_{i}(G); i = 1, 2, …, t(G); the connected components of Γ(G). If |G| is even, we will always assume that 2 is a vertex of π_{1}. We will use the following unpublished result of K. W. Gruenberg and O. H. Kegel (see [9, Theorem A]).

Lemma 2.1 If G is a group whose prime graph has more than one component, then G has one of the following structures:

(a) Frobenius or 2-Frobenius,

(b) simple,

(c) an extension of a π_{1}-group by a simple group,

(d) simple by π_{1},

(e) π_{1} by simple by π_{1}.

Note that in the above lemma, the group G is called 2-Frobenius if there exists a normal series 1 ⊴ H ⊴ K ⊴ G of G such that H is the Frobenius kernel of K and K/H is the Frobenius kernel of G/H.

Remark The Lemma 2.1 implies that, G has a normal series 1 ⊴ N ◁ G_{1} ⊴ G such that N is a nilpotent π_{1}-group, G_{1}/N is a simple group, and G/G_{1} is a solvable π_{1}-group.

Lemma 2.2 Let G be a non-solvable group and 15 ∈ π_{e}(G). Then G is neither Frobenius nor 2-Frobenius.

Proof. Since G is non-solvable, G cannot be 2-Frobenius. If G is a Frobenius group with kernel K and complement H, then H is non-solvable. Now by the structure theorem of non-solvable Frobenius complements (see [7, Theorem 18.6]), we know that H has a normal subgroup H_{0} of index ≤ 2 such that H_{0} ≅ SL(2, 5) × Z, where every Sylow subgroup of Z is cyclic and π(Z) ∩ {2, 3, 5} = ∅. Since π_{e}(SL(2, 5)) = {1, 2, 3, 4, 5, 6, 10} and 15 ∈ π_{e}(G), we have 15 ∈ π_{e}(H) \ π_{e}(H_{0}). Therefore, there exists 1 ≠ x ∈ H \ H_{0} such that o(x) = 15. Since |H : H_{0}| = 2, it follows that x^{2} ∈ H_{0} which implies that o(x^{2}) = 15 ∈ π_{e}(H_{0}). This is a contradiction and the proof is complete. □

We will utilize the following result concerning the set of order of elements of the symmetric groups (see [10]).

Lemma 2.3 The group S_{n} (resp. A_{n}) has an element of order m = p_{1}^{α_{1}}p_{2}^{α_{2}} … p_{s}^{α_{s}}, where p_{1}, p_{2}, ..., p_{s} are distinct primes and α_{1}, α_{2}, ..., α_{s} are natural numbers, if and only if p_{1}^{α_{1}} + p_{2}^{α_{2}} + … + p_{s}^{α_{s}} ≤ n (resp. p_{1}^{α_{1}} + p_{2}^{α_{2}} + … + p_{s}^{α_{s}} ≤ n for m odd, and p_{1}^{α_{1}} + p_{2}^{α_{2}} + … + p_{s}^{α_{s}} ≤ n − 2 for m even).

The following result of G. Higman will be used (see [5]):

Lemma 2.4 Let G be a solvable group all of whose elements are of prime power order. Then |π(G)| ≤ 2.

We can now prove the following lemma which will be useful.

Lemma 2.5 Let G be a group and π_{e}(G) = π_{e}(S_{p}), where p is a prime number such that p > 50. Then G is non-solvable. Moreover, G is neither Frobenius nor 2-Frobenius.

Proof. We know that |S_{p}| = p! = p_{1}^{α_{1}}p_{2}^{α_{2}} … p_{k}^{α_{k}} where p_{i} is the i-th prime, p_{k} = p and for all i such that p_{i} > (p − 1)/2, α_{i} = 1 . If G were solvable, there would exist more than 6 primes between (p − 1)/2 and p (see [1]). So we can choose a {p_{k}, p_{k−1}, p_{k−2}}-Hall subgroup H of G. Now by Lemma 2.3, S_{p} does not contain any element of order p_{k}p_{k−1}, p_{k}p_{k−2} and p_{k−1}p_{k−2}. Thus all elements of H will be of prime power order. This contradicts Lemma 2.4 and so G must be non-solvable. The rest follows from Lemma 2.2. □

Now the following result is in order.

Lemma 2.6 Let G be a simple group. If {p} ⊂ π(G) ⊆ π(p!), where p is a prime such that 50 < p < 100, then G is isomorphic to one of the following groups:

(a) if p = 53 : A_{n}, n = 53, 54, 55, 56, 57, 58; A_{1}(23^{2}), B_{2}(23), ^{2}A_{3}(23); A_{1}(53),
(b) if p = 59 : A_{n}, n = 59, 60; A_{1}(59),
(c) if p = 61 : A_{n}, n = 61, 62, 63, 64, 65, 66; A_{4}(3^{2}), A_{1}(3^{5}), B_{5}(3), C_{5}(3);
A_{3}(11), A_{1}(11^{2}), A_{2}(11^{2}), B_{2}(11), B_{3}(11), C_{3}(11), D_{4}(11),
2A_{3}(11); A_{2}(13), A_{3}(13); A_{2}(47), A_{3}(47); A_{1}(61),

(d) if p = 67 : A_{n}, n = 67, 68, 69, 70; Ly; A_{2}(29); A_{2}(37), A_{1}(37^{3}), G_{2}(37);
A_{1}(67),
(e) if p = 71 : A_{n}, n = 71, 72; F_{1}; A_{4}(5), A_{5}(5); A_{1}(71),
(f) if p = 73 : A_{n}, n = 73, 74, 75, 76, 77, 78; A_{2}(2^{3}), A_{1}(2^{9}), G_{2}(2^{3}), C_{4}(2^{3}),

E_{6}(2), B_{3}(2^{2}), C_{3}(2^{3}); A_{1}(3^{6}), A_{5}(3^{2}), B_{6}(3), B_{2}(3^{3}), C_{6}(3),

D_{4}(3^{2}), G_{2}(3^{2}), F_{4}(3), ^{2}A_{2}(3^{2}), ^{2}A_{3}(3^{3}), ^{2}D_{6}(3), ^{2}E_{6}(3);
A_{1}(73), A_{1}(73^{2}), B_{2}(73),
(g) if p = 79 : A_{n}, n = 79, 80, 81, 82; A_{2}(23), A_{3}(23), A_{2}(23^{2}), A_{1}(23^{3}),
B_{3}(23), C_{3}(23), D_{4}(23), G_{2}(23); A_{1}(79), A_{2}(79),
(h) if p = 83 : A_{n}, n = 83, 84, 85, 86, 87, 88; A_{1}(83), A_{1}(83^{2}),
(i) if p = 89 : A_{n}, n = 89, 90, 91, 92, 93, 94, 95, 96; A_{1}(89),
(j) if p = 97 : A_{n}, n = 97, 98, 99, 100; A_{2}(61); A_{1}(97).

Proof. Let p = 53. Then {53} ⊂ π(G) ⊆ π(53!). First, suppose G is an alternating group A_{n}. We know that \frac{1}{2}(n!) is divisible by 53 if and only if n ≥ 53. This implies that the only possibilities for n are 53, 54, 55, 56, 57 or 58. Next, suppose G is a sporadic simple group. The orders of sporadic simple groups are listed in [2]. From this, it is easy to see that G cannot be isomorphic to a sporadic simple group. Finally, we suppose that G is a group of Lie type over a finite field of order q = r^{s} (r prime). Since r divides |G|, r must be equal to 2, 3, 5, …, 51 or 53. Now suppose r = 2. Since the order of 2 modulo 53 is 52, and 53|2^{k} − 1 and 2^{k} − 1 | |G| it follows that k is a multiple of 52. From [2, Table 6], we see that no candidate for G exists in this case. Similarly for r ∈ π(53!)\{23, 53} we do not get a group. Suppose r = 23. Since the order of 23 modulo 53 is 4, from [2, Table 6], the only candidates from G are A_{1}(23^{2}), B_{2}(23) and ^{2}A_{3}(23). Finally we assume that r = 53. Similarly, in this case the only possible group is A_{1}(53).

The rest of the proof is repetition of similar processes. □

We will need the following results (see [10]).

Lemma 2.7 If S_{m} contains a Frobenius subgroup N⟨γ⟩ with kernel N of order n and cyclic complement ⟨γ⟩ of order p^{k}, where p is a prime, then for every natural power t = p^{r} of p there is a Frobenius subgroup in S_{mt} with kernel of order n^{t} and cyclic complement of order tp^{k} = p^{k+r}.

Remark If p is odd prime, the Frobenius group stated in Lemma 2.7 lies in A_{mt}. This is also true for p = 2 provided that r ≥ 1 and, the number m together with the permutation γ are both even or odd.

Lemma 2.8 Let N be a normal elementary abelian p-subgroup of G, K ≅ G/N and let G_{1} = NK be the natural semidirect product. Then π_{e}(G_{1}) ⊆ π_{e}(G).

According to the discussion of A. V. Zavarnitsin and V. D. Mazurov concerning the action of an alternating and symmetric group on an elementary abelian p-group in [10, Propositions 2 and 3] we have the following result:

Lemma 2.9 Let p be an odd prime and V be an elementary abelian p-group, on which the group A_{m} (m ≥ 3) acts. Then π_{e}(V A_{m}) ⊈ π_{e}(S_{m}), where V A_{m} is the natural semidirect product.

We will need the following.

Lemma 2.10 (see [6]) Let G be a group, N a normal subgroup of G and G/N a Frobenius group with Frobenius kernel F and cyclic complement C. If (|F |, |N|) = 1 and F is not contained in NC_{G}(N)/N, then p|C| ∈ π_{e}(G) for some prime p of |N|.

Lemma 2.11 (see [10]) Let G be a semidirect product of an elementary abelian p-group V with A_{m}. Suppose that A_{m} contains Frobenius subgroups F_{1}, F_{2}, …, F_{t} with cyclic complements ⟨d_{i}⟩ of orders p_{i}^{s_{i}}, where p_{i}’s are pairwise distinct primes, such that the supports, supp F_{i}, are pairwise disjoint. Moreover, suppose that the order of the kernel of each F_{i} is prime p, and, all a_{i} = |supp F_{i}| are distinct. Then p o(d_{1}) … o(d_{t}) ∈ π_{e}(G).

We now give the following.

Lemma 2.12 Let G be a group and π_{e}(G) ⊆ π_{e}(S_{p}) ( resp. π_{e}(S_{p+1})) where p is a prime such that 50 < p < 100. Moreover, assume that G possesses a normal elementary abelian r-subgroup N such that G/N ≅ A_{p} ( resp. A_{p+1}). Then N = 1.

Proof. Assume that N ≠ 1. It is clear that from Lemma 2.8 that π_{e}(NA_{p}) ⊆ π_{e}(G) ⊆ π_{e}(S_{p}). On the other hand, Lemma 2.9 forces r = 2. Now, we deal with each prime number separately.

(i) p ∈ {53, 59, 61}: Since S_{9} contains a Frobenius subgroup of order 8 × 9 with cyclic complement of order 8, by Lemma 2.7, it follows that S_{36} has a Frobenius subgroup with cyclic complement of order 32. Clearly S_{36} < A_{p}. By Lemma 2.10, we conclude that 64 ∈ π_{e}(G), a contradiction.

(ii) p ∈ {71, 73, 79, 83, 89, 97}: Since S_{17} contains a Frobenius subgroup of order 16 × 17 with cyclic complement of order 16, by Lemma 2.7, S_{68} contains a Frobenius subgroup with cyclic complement of order 64. On the other hand, S_{68} < A_{p} and so by Lemma 2.10 we obtain 128 ∈ π_{e}(G), again a contradiction.

(iii) p=67: As discussed before, S_{36} contains a Frobenius subgroup of order 32 × 9^{4} with cyclic complement of order 32. Moreover, S_{36} < A_{p}. On the other hand, A_{11} contains a Frobenius subgroup of order 5 × 11, and A_{11} < A_{p}. Therefore, A_{p} has a pair of Frobenius subgroups of orders 32 × 9^{4} and 5 × 11 satisfying the hypothesis of Lemma 2.11, and hence 5 × 64 ∈ π_{e}(G). This contradiction completes the proof of the lemma. □

3 Proof of the Main Theorem

Suppose G is a group such that π_{e}(G) = π_{e}(S_{p}), where p is a prime with 50 < p < 100. The components of the prime graph of G are

π_{1}(G) = {2, 3, 5, …, r_{p−1}} and π_{2}(G) = {p},

where r_{p−1} denotes the largest prime not exceeding p − 1. We must prove that G is isomorphic to S_{p}. By Lemma 2.5, G is neither Frobenius nor 2-Frobenius. Next, from Lemma 2.1, G has a normal series 1 ⊴ N ◁ G_{1} ⊴ G such that N is a nilpotent π_{1}-group, \overline{G}_{1} := G_{1}/N is simple, and G/G_{1} is a solvable π_{1}-group. Furthermore, we have

\frac{G}{N} ≅ \frac{N_{G/N}(\overline{G}_{1})}{C_{G/N}(\overline{G}_{1})} ↪ Aut(\overline{G}_{1}).

Thus we may assume that G/N ≤ Aut(\overline{G}_{1}). Note that \{2, p\} ⊆ π(\overline{G}_{1}) ⊆ π(G) = π(p!), π_{1}(\overline{G}_{1}) ⊆ π_{1}(G) = \{2, 3, …, r_{p−1}\} and π_{2}(\overline{G}_{1}) = π_{2}(G) = \{p\}. Hence \overline{G}_{1} is a simple group such that p | |\overline{G}_{1}|. Now we claim that \overline{G}_{1} ≅ A_{p}.

According to the classification of finite simple groups, we know that the possibilities for \overline{G}_{1} are the alternating groups A_{n}, with n ≥ 5, sporadic simple groups, and finite groups of Lie type. We deal with the above cases separately.

First, suppose \overline{G}_{1} is a simple group of Lie type or sporadic. By Lemma 2.6, \overline{G}_{1} can only be isomorphic to one of the listed simple groups of Lie type or sporadic. For each prime p appearing in (a) – (j) of Lemma 2.6, there are two other primes r_{1}, r_{2} ∈ π(G)\π(Aut(\overline{G}_{1})) with r_{1} + r_{2} > p, where \overline{G}_{1} is any simple group on the appropriate list:

\begin{matrix}p & 53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 & 89 & 97 \\ r_{1} & 43 & 47 & 53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 \\ r_{2} & 47 & 53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 & 89 \end{matrix}

Moreover, from G/N ≤ Aut(\overline{G}_{1}), it follows that r_{1}, r_{2} ∈ π(N), and the nilpotency of N forces r_{1}r_{2} ∈ π_{e}(N) ⊂ π_{e}(G) = π_{e}(S_{p}), which is a contradiction, because r_{1} + r_{2} > p.

Next, suppose \overline{G}_{1} is a simple group of alternating type listed in Lemma 2.6. Then in all cases, there exist primes q_{1} and q_{2} such that q_{1} + q_{2} = p + 1:

\begin{matrix}p & 53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 & 89 & 97 \\ q_{1} & 7 & 7 & 3 & 7 & 5 & 3 & 7 & 5 & 7 & 19 \\ q_{2} & 47 & 53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 & 79 \end{matrix}

But then q_{1}q_{2} ∈ π_{e}(A_{n}) \ π_{e}(G) for n > p, by Lemma 2.3. Therefore, \overline{G}_{1} ≅ A̸_{n} for n > p. Hence \overline{G}_{1} ≅ A_{p} and the claim has been proved.

Finally, we show that G ≅ S_{p}. By Lemma 2.12, N = 1. On the other hand, since G_{1} ≤ G ≤ Aut(G_{1}) and |Out(G_{1})| = 2, it follows that G ≅ A_{p} or G ≅ S_{p}. But since π_{e}(A_{p}) ≠ π_{e}(S_{p}), we have G ≅ S_{p}.

Next, suppose G is a group such that π_{e}(G) = π_{e}(S_{p+1}), where p is a prime with 50 < p < 100. By using similar arguments to those in the proof of the previous case, G has a normal series 1 ⊴ N ◁ G_{1} ⊴ G such that N is a nilpotent π_{1}-group, \overline{G}_{1} := G_{1}/N is simple, and G/G_{1} is a solvable π_{1}-group. We may assume that G/N ≤ Aut(\overline{G}_{1}). Now we claim that \overline{G}_{1} ≅ A_{p+1}.

The proof of this claim is similar to those in the proof of the previous case. For example, we proceed for p = 53. Using similar arguments to those in the proof of the previous case, we see that \overline{G}_{1} ≅ A_{53} or A_{54}. If \overline{G}_{1} is isomorphic to A_{53}, then since |Out(A_{53})| = 2 and 7 × 47 ∈ π_{e}(G)\π_{e}(A_{53}), it follows that N ≠ 1 and 7 ∈ π(N) or 47 ∈ π(N). Let N be a 7-group. Then \overline{G}_{1} has a pair of Frobenius subgroups of orders 32 × 9^{2} and 7 × 8 satisfying the hypothesis of Lemma 2.11, and hence 32 × 49 ∈ π_{e}(G). This is a contradiction. Thus N is a 47-group. Yet, in this case, \overline{G}_{1} contains a Frobenius subgroup of order 8 × 9 with cyclic complement of order 8, and by Lemma 2.10 we get 8 × 47 ∈ π_{e}(G), which contradicts the assumption. Therefore, \overline{G}_{1} ≅ A_{54}.

The rest of the proof is similar to the proof in the previous case and we have G ≅ S_{p+1}. □

Acknowledgment: This work was done while the second author was a Postdoctoral Research Associate at the School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics (IPM). He would like to thank the IPM for the financial support. The authors would also like to thank the referee for very helpful suggestions and comments.

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The Authors’ Addresses

A. R. Moghaddamfar, Department of Mathematics, K.N.T. University of Technology, Tehran, Iran.

E-mail: moghadam@iust.ac.ir

M. R. Pournaki, School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics, P.O. Box 19395-5746, Tehran, Iran.

E-mail: pournaki@ipm.ir