#### NSE characterization of the Chevalley group \(\varvec{G}_{\varvec{2}} {\varvec{(4)}}\)

Arabian Journal of Mathematics
NSE characterization of the Chevalley group G2(4)
Maryam Jahandideh Khangheshlaghi 0 1
Mohammad Reza Darafsheh 0 1
Mathematics Subject Classification 0 1
0 M. R. Darafsheh School of Mathematics , Statistics and Computer Science , College of Science, University of Tehran , Tehran , Iran
1 M. J. Khangheshlaghi (
Let G be a group and ω(G) = {o(g)|g ∈ G} be the set of element orders of G. Let k ∈ ω(G) and sk = |{g ∈ G|o(g) = k}|. Let nse(G) = {sk |k ∈ ω(G)}. In this paper, we prove that if G is a group and G2(4) is the Chevalley group such that nse(G) = nse(G2(4)), then G ∼= G2(4). Let G be a finite group and ω(G) be the set of element orders of G. If k ∈ ω(G), then sk is the number of elements of order k in G. Let nse(G) = {sk |k ∈ ω(G)}. If n is a positive integer, the set of all prime divisors of n is denoted by π(n). The number of the Sylow p-subgroups Pp of G is denoted by n p or n p(G). We set π(G) = π(|G|). To see notations concerning finite simple group, we refer the reader to [1]. A finite group G is called a simple Kn-group, if G is a simple group and |π(G)| = n. In 1987, J. G. Thompson posed the following problem related to algebraic number fields [2]: Let T (G) = {(k, sk )|k ∈ ω(G), sk ∈ nse(G)}. Suppose that T (G) = T (H ) for some finite group H. If G is a finite solvable group, is it true that H is also necessarily solvable? If the groups G and H have the same order type, then |G| = |H | and nse(G) = nse(H ). Due to this problem, many results were obtained as follows: Result 1 [3,4] Let G be a group and S a simple Ki -group, where i = 3, 4. Then, G ∼= S if and only if |G| = |S| and nse(G) = nse(S). Result 2 [5,6] The two groups A12 and A13 are characterizable by order and nse. Result 3 [7] All sporadic simple groups are characterizable by nse and order.
1 Introduction
Result 4 [8] L2(2m ), with 2m + 1 prime, is 2m − 1 prime, is characterized by nse and order.
Result 5 [4,9] L2(q), where q ∈ {7, 8, 9, 11, 13}, can be characterized by only the nse.
Result 6 [10] L3(4) is characterized by nse.
Result 7 [11] L5(
2
) is characterized by nse.
Result 8 [12] U3(5) is characterized by nse.
Up to now, some groups can be characterized by only the set nse(G). The aim of this paper is to prove that
the Chevalley group is characterizable by nse.
Main Theorem. Let G be a group such that nse(G) = nse(G2(4)) where G2(4) is the Chevalley group of
type G2 over G F (4). Then G ∼= G2(4).
2 Preliminaries
In this section, we will give some lemmas which will be used in the proof of the main theorem.
Lemma 2.1 [13] Let G be a finite group and n be a positive integer dividing |G|. If Ln (G) = {g ∈ G|gn = 1},
then |n||Ln(G)|.
Lemma 2.2 [14] Let G be a finite group and p ∈ π(G) be odd. Suppose that P is a Sylow p-subgroup of G
and n = ps m with ( p, m) = 1. If P is not cyclic, the number of elements of order n is always a multiple of ps .
Lemma 2.3 [4] Let G be a group containing more than two elements. If the maximum number s of elements
of the same order in G is finite, then G is finite and |G| ≤ s(s2 − 1).
Lemma 2.4 [15] Let G be a finite solvable group and |G| = mn, where m = pα1
1 · · · prαr , (m, n) = 1. Let
π = { p1, . . . , pr } and hm be the number of Hall π -subgroups of G. Then hm = q1β1 · · · qsβs , qi ,s should be
prime numbers satisfying the following conditions for all i ∈ {1, 2, . . . , s}:
1. qiβi ≡ 1(mod p j ) for some p j .
2. The order of some chief factor of G is divided by qiβi . We need the structure of simple Kn-group with
n = 4, 5.
Lemma 2.5 [16] Let G be a simple K4-group. Then G is isomorphic to one of the following groups:
1. A7, A8, A9 or A10.
2. M11, M12 or J2.
3. One of the following:
(i) L2(r ), where r is a prime and r 2 − 1 = 2a . 3b.vc with a ≥ 1, b ≥ 1, c ≥ 1, and v is a prime greater than
3.
(iii) L2(3m ), where 3m −+ 11 == u4,t2,m3m − 1 = 2uc or 3m + 1 = 4t b, 3m − 1 = 2u, with m ≥ 2, u, t are odd
(ii) L2(2m ), where 2m + 1 = 3t b with m ≥ 2, u, t are primes, t > 3, b ≥ 1.
primes, b ≥ 1, c ≥ 1.
(iv) One of the following 28 simple groups:
L2(16), L2(25), L2(49), L2(81), L3(4), L3(5), L3(7), L3(8), L3(17), L4(
3
), S4(4), S4(5), S4(7), S4(9),
S6(
2
), O8+(
2
), G2(
3
), U3(4), U3(5), U3(7), U3(8), U3(9), U4(
3
), U5(
2
), Sz(8), Sz(32),2 D4(
2
) or 2 F4(
2
).
Lemma 2.6 [6] Each simple K5-group is isomorphic to one of the following simple groups:
1. L2(q) with |π(q2 − 1)| = 4.
2. L3(q) with |π((q2 − 1)(q3 − 1))| = 4.
3. U3(q) with q satisfies |π((q2 − 1)(q3 + 1))| = 4.
4. O5(q) with |π(q4 − 1)| = 4.
5. Sz(22m+1) with |π((22m+1 − 1)(24m+1 + 1))| = 4.
6. R(q), where q is an odd power of 3, and |π(q2 − 1)| = 3 and |π(q2 − q + 1)| = 1.
7. The following 30 simple groups:
A11, A12, M22, J3, H S, H e, M cL , L4(4), L4(5), L4(7), L5(
2
), L5(
3
), L6(
2
), O7(
3
), O9(
2
), P S P8(
2
),
U4(4), U4(5), U4(7), U4(9), U5(
3
), U6(
2
), O8+(
3
), O8−(
2
),3 D4(
3
), G2(4), G2(5), G2(7) or G2(9).
TLheemnm,Ga 2is.7onLeeotfGthbeefoallsoiwmipnlge gKronu-gprso:uGp2w(4it)h, Snz=(8)4,,L52(a2n5d),1L32d(i1v3id),esL|2G(2|7a)n,2dF|4G(2|)d,iUvi3d(e4s)2.12.33.52.7.13.
Proof We prove this lemma in the following two cases:
Case 1. G is a simple K4-group. Order consideration rules out of the cases (
1
) and (
2
) of Lemma 2.5. Hence,
we use Lemma 2.5 (
3
). In this case, it is easily seen that G is one of the following groups:
Sz(8), L2(25), L2(13), L2(27),2 F4(
2
), U3(4).
Case 2. G is a simple K5-group. By Lemma 2.6, we only have G = G2(4).
3 Proof of the main theorem
In this section, we prove that if G is a group and nse(G) = nse(G2(4)), then G ∼= G2(4). However, we have
some observations and lemmas before that.
Observation 3.1 If sn is the number of elements of order n in G, then sn = kϕ(n), such that k is the number
of cyclic subgroups of order n in G. .
Observation 3.2 If n > 2, then ϕ(n) is even.
Lemma 3.3 If m ∈ ω(G), then ϕ(m)|sm and m| d|m sd .
Proof It follows from Lemma 2.1.
Lemma 3.4 Let G be a group such that nse(G) = nse(G2(4)), then G is a finite group.
Proof It follows from Lemma 2.3.
Theorem 3.5 Let G be a group with nse(G) = nse(G2(4)) = {(
1, 69615, 1401920, 982800, 3354624,
22276800, 11980800, 15724800, 50319360, 38707200, 67092480, 23961600
)}, where G2(4) is the Chevalley
group of type G2 over G F(4). Then G ∼= G2(4).
Proof of the main theorem. It follows from Lemma 3.4, G is a finite group. According to Lemma 2.1,
π(G) ⊆ {2, 3, 5, 7, 13, 17, 337, 982801, 23961601}. The first claim is π(G) = {2, 3, 5, 7, 13}.
According to the Observation 3.2, s2 = 69615 and 2 ∈ π(G). Since s337 is not equal to none of nse
values, 337 ∈/ π(G). Now, we show that 982801 ∈/ π(G). If 982801 ∈ π(G), s982801 = 982800. If
2.982801 ∈ π(G), ϕ(2.982801)|s2.982801, and by Lemma, 2.1 s2.982801 = 982800 or 15724800, and on
the other hand, 2.982801|1 + s2 + s982801 + s2.982801(982800, 15724800), a contradiction. Now, we consider
bSeycloawuse92if80P198-2s8u01bggr=ou{pa P∈98P29808128o0f1G|gaan=d g} == 1{,gw∈heGre|og(∈g) =, th2e}.n PG98h2a80s1aancetslefimxeendt opfoionrtdferre2el.y98o2n801,.
Hence, |P982801||| | = s2, a contradiction. As a result, 982801 ∈/ π(G). Similarly, we can prove that the
prime numbers 24897601 and 23961601 do not belong to π(G). Hence we have π(G) ⊆ {2, 3, 5, 7, 13, 17}.
We have:
1. If 2a ∈ ω(G), then 2a−1|s2a and 0 ≤ a ≤ 15.
2. If 3a ∈ ω(G), then 2.3a−1|s3a and 0 ≤ a ≤ 4.
3. If 5a ∈ ω(G), then 4.5a−1|s5a and 0 ≤ a ≤ 3.
4. If 7a ∈ ω(G), then 6.7a−1|s7a and 0 ≤ a ≤ 2.
5. If 13a ∈ ω(G), then 12.13a−1|s13a and 0 ≤ a ≤ 2.
6. If 17a ∈ ω(G), then 16.17a−1|s17a and 0 ≤ a ≤ 1.
We show that the prime number 17 does not belong to π(G). To prove this, we distinguish several cases.
Case 1. If π(G) = {2, 17}, then s17 = 11980800. If P17 is a Sylow 17-subgroup of G, according to Lemma
2.1, |P17||1 + s17 and |P17| = 17. Thus
s17
n17 = ϕ(17) = 748800 = [G : NG (P17)] ⇒ 3, 5, 13 ∈ π(G),
which is impossible.
Case 2. If π(G) = {2, 3, 17}, the same as case 1, we have 5 ∈ π(G). But it is impossible.
0 <
10
i=1
Case 3. If π(G) = {2, 3, 5, 17} or π(G) = {2, 3, 5, 7, 17}, , similar to case 1, we have 13 ∈ π(G), which is
not possible.
Case 4. If π(G) = {2, 3, 5, 7, 13, 17}, |G| = 2m1.3m2.5m3.7m4.13m5.17 and 251596800 + 1401920k1 +
982800k2 + 3354624k3 + 22276800k4 + 11980800k5 + 15724800k6 + 50319360k7 + 38707200k8 +
67092480k9 + 23961600k10 = 2m1.3m2.5m3.7m4.13m5.17 where k1, k2, k10, m1, . . . , m6 are nonnegative
integers and
ki ≤ 5748 ⇒ 251596800 < 2m1.3m2.5m3.7m4.13m5.17 < 5748.67092480.
10
i=1
This inequality does not hold. Hence, 17 does not belong to π(G) and π(G) ⊆ {2, 3, 5, 7, 13}. Now, we prove
π(G) = {2, 3, 5, 7, 13}. To do this, we distinguish several cases.
Case 1. π(G) = {2}.
In this case, ω(G) ⊆ {1, 2, 22, 215} and 251596800 + 1401920k1 + 982800k2 + 3354624k3 + 22276800k4 +
11980800k5 + 15724800k6 + 50319360k7 + i13=0817k0i7≤2040kb8ut+th6is70e9q2u4a8ti0okn9h+as 2n3o9s6o1l6u0ti0okn10in =inte2gmerssu.ch that
k1, k2, k10, m are nonnegative integers. 0 <
Case 2. π(G) = {2, 3}.
If 2a.3b ∈ ω(G) according to Lemma 2.1, 0 ≤ a ≤ 15, 0 ≤ b ≤ 4. We have ex p(P3) = 3, 9, 27 or 81. Suppose
that exp (P3) = 3. In this case, it follows: |P3||1+s3(1401920) ⇒ |P3||1401921 ⇒ |P3||27. If |P3| = 3, n3 =
ϕs(33) = 14012920 = 700960 ⇒ [G : NG(P3)] = 700960, then 5 ∈ π(G), a contradiction. If |P3| = 9, |G| =
2m.32, such that 0 ≤ m ≤ 15. In addition, calculations show: 251596800 + 1401920k1 + 982800k2 +
3354624k3 + 22276800k4 + 11980800k5 + 15724800k6 + 50319360k7 + 38707200k8 + 67092480k9 +
23961600k10 = 2m.32, such that k1, k2, k10, m are nonnegative integers. Then
0 <
ki ≤ 36 ⇒ 251596800 ⇒ 2m.32 ≤ 36.67092480 ⇒ m > 15,
which is a contradiction.
If |P3| = 27, |G| = 2m.32, such that 0 ≤ m ≤ 15. Similarly, as |P3| = 9, we also get a contradiction.
Now, we suppose that exp (P3) = 9, according to Lemma 2.1 |P3||1 + s3 + s9. There are several cases for
s9. Namely, 982800, 3354624, 22276800, 11980800, 15724800, 50319360, 15724800, 38707200, 67092480,
and 23961600. For all cases, we have n3 = ϕs(99) = [G : NG(P3)], then 5 ∈ π(G), a contradiction.
s27
If exp(P3) = 27, s27 = 9828200, 3354624, 22276800, 11980800 and 15724800 also n3 = ϕ(27) = [G :
NG(P3)], then 5 or 13 or 7 ∈ π(G), a contradiction.
If exp(P3) = 81, s81 = 15724800, 50319360 and 38707200. Similar as in the previous case, we obtain a
contradiction.
Case 3. π(G) = {2, 5}.
If 2a.5b belongs to π(G), 0 ≤ a ≤ 15, 0 ≤ b ≤ 3. In this case, exp(P5) = 5, 25 or 125. If exp(P5) =
5, |P5||1 + s5(3354624), then |P5||125.
If |P5| = 5, n5 = ϕs(55) = [G : NG(P5)] ⇒ 2, 7 and 13 ∈ π(G), a contradiction.
If |P5| = 25, |G| = 2a.52 and also 251596800 + 1401920k1 + 982800k2 + 3354624k3 + 22276800k4 +
11980800k5 + 15724800k6 + 50319360k7 + 38707200k8 + 67092480k9 + 23961600k10 = 2m.52, such
that k1, k2, . . . , k10, m are nonnegative integers. Then, 0 < i1=01 ki ≤ 36 ⇒ 251596800 ≤ 2m.52 ≤
36.67092480, then m > 15, which is a contradiction. If |P5| = 125, similarly ,we have a contradiction.
If exp(P5) = 25, then |P5||1 + s5 + s25(982800, 22276800, 11980800, 15724800, 38707200, 67092480 or
23961600), then |P5||25.|P5| can be 5 or 25, similar to exp(P5) = 5 we get a contradiction. Now if exp(P5) =
125, s125 = 982800, 22276800, 11980800 or 15724800.
In this case, n5 = ϕs(112255) = [G : NG(P5)], then 3 ∈ π(G), a contradiction. Case 4. π(G) = {2, 7}.
In this case, s7 = 11980800 and if 72 ∈ ω(G), s72 = 38707200, |P7||1 + s7 + s49. Hence |P7||49.
s7 s49
If |P7| = 7, n7 = ϕ(7) = 1996800 ⇒ 5, 3 and 13 ∈ π(G), a contradiction. If |P7| = 49, n7 = ϕ(49) =
921600, then 3 and 5 ∈ π(G), a contradiction. So 72 does not belong to ω(G) and |P7||1 + s7. Since |P7| = 7,
similar as above, we have a contradiction.
Case 5. π(G) = {2, 13}.
We have s13 = 38707200 and s132 = 11980800. According to Lemma 2.1, |P13||1 + s13 + s132 and |P13||132.
s13 s132
If | P13| = 13, n13 = ϕ(13) = 3225600, then 5 ∈ π(G), a contradiction. If | P13| = 132, n13 = ϕ(132) =
3225600, then 5 ∈ π(G), a contradiction. Hence 132 ∈/ ω(G) and |G| = 2a .13, | P13| = 13. Again we get a
contradiction similar as above.
Case 6. π(G) = {2, 3, 5}.
In this case, |G| = 2a .3b.5c, where 0 ≤ a ≤ 15, 0 ≤ b ≤ 4, 0 ≤ c ≤ 3. 251596800+1401920k1 +982800k2 +
3354624k3 + 22276800k4 + 11980800k5 + 15724800k6 + 50319360k7 + 38707200k8 +i1=0617k0i92≤48300k89 ⇒+
23961600k10 = 2a .3b.5c, such that k1, k2, . . . , k10, a, b, c are nonnegative integers. 0 <
251596800 ≤ 2a .3b.5c ≤ 308.67092480, then a > 15, it is a contradiction. In the remaining cases, in the
same way, we obtain a contradiction. Therefore, π(G) = {2, 3, 5, 7, 13}. Now, we prove that 2.7 ∈/ ω(G). If
2.7 belongs to ω(G), let P7 and Q7 be Sylow 7-subgroups of G, P7 and Q7 are conjugate in G and CG ( P7)
and CG (Q7) are conjugate in G. Therefore, s14 = ϕ(14).n7.k, where k is the number of cyclic subgroups
s7
of order 5 in CG ( P7). Hence, n7 = ϕ(7) = 1996800 and s14 = 6.1996800.k ⇒ s14 = 11980800, since 14
does not divide 1 + s2 + s7 + s14, we conclude that 2.7 does not belong to ω(G). It follows that the Sylow
2-group of G acts fixed point freely on the set of elements of order 7 so | P2||s7 and | P2||212. Next, it is
shown by the way of contradiction, 3.13 ∈/ ω(G). If 3.13 ∈ ω(G), let P13 and Q13 be Sylow 13-subgroups
of G, then s39 = ϕ(39).n13.k, where k is the number of cyclic subgroups of order 3 in CG ( P13). Therefore,
s13
n13 = ϕ(13) = 3225600 and s39 = 24.3225600.k. Due to the values of nse and 77414400|s39, it is impossible
and it follows that the Sylow 3-group of G acts fixed point freely on the set of elements of order 13, so | P3||s13;
therefore, | P3||33. If 5.13 ∈ ω(G), s65 does not belong to nse and similarly as before case | P5||s13; therefore,
| P3||33. If 5.13 ∈ ω(G), s65 does not belong to nse and similarly as before case | P5||s13 and | P5||52.
Therefore, |G| = 2m .3n.5k .7.13 and also sk ∈nse(G) sk = 251596800 = 212.33.52.7.13 ≤ |G| =
2m .3n.5k .7.13 ≤ 212.33.52.7.13; hence, |G| = 212.33.52.7.13. Therefore, we proved |G| = |G2(4)|. Now, if
we supposed that G G2(4), we first prove that G is not solvable. If G is solvable, then, by Lemma 2.4, for
every prime p, p p j ≡ 1(modq) for some q. Thus, we have 7 ≡ 1(mod13), a contradiction.
Therefore, G is nonsolvable. Hence, G has a normal series 1 K L G, such that KL is isomorphic to a
simple Ki -group, where i = 3, 4, 5.
If KL is isomorphic to one of the simple K3-group, by [16], KL is isomorphic to A5, A6, L2(7), L2(8), U3(
3
)
or U4(
2
). Suppose that KL ∼= A5, hence, n5( KL ) = n5( A5) = 6. Therefore, n5(G) = 6t , such that 5 does not
divide t ; then, t = 2s54 .6. There are several values for s5, but for all cases 5|t and it is a contradiction. Similarly,
for the groups A6, L2(7), L2(8), U3(
3
) and U4(
2
), we also can rule out these cases.
If KL is isomorphic to a simple Kn -group with n = 4, 5, then by Lemma 2.7, L/K is isomorphic to
G2(4), Sz(8), L2(25), L2(13), L2(27),2 F4(
2
), U3(4).
If KL ∼= L2(13); then from [3], 14 = n13( KL ) = n13(L2(13)), and so n13(G) = 14t , such that 13
does not divide t for some integer t . Then, s13 = 12.14t = 12.3225600, and then, t = 230400. Therefore
210.52.32||K ||210.32.52; hence, |K | = 210.32.52 and NK ( P13) = 1. Therefore, K ∩ NK ( P13) = K ∩CK ( P13),
which means that K × P13 is a Frobenius group and so | P13||| Aut (K )|, a contradiction. Similarly, for the other
cases, we can get a contradiction. The last case KL ∼= G2(4). In this case, G2(4) ∼= KL ≤ GK and |G| = |G2(4)|,
then K = 1 and G ∼= G2(4). It is a contradiction to the primary hypothesis. In general, we recognized all steps
and obtained that G ∼= G2(4). This complete the proof.
Acknowledgements The first author gratefully acknowledges the financial support provided by Mahshahr Branch, Islamic Azad
University, Iran to perform research project entitled: “A characterization of simple groups by the number of the same element
orders (nse) and the order of group”.
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