# A Characterization of Projective Special Unitary GroupU3(7) by nse

Authors: Shitian Liu
Journal: Algebra
 (2013)

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## Abstract

LetG a group and ω(G) be the set of element orders of G. Let k∈ω(G) and let sk be the number of elements of order k in G. Let nse(G)={sk∣k∈ω(G)}. In Khatami et al. and Liu's works, L3(2) and L3(4) are uniquely determined by nse(G). In this paper, we prove that if G is a group such that nse(G) = nse(U3(7)), then G≅U3(7).


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## Body

## 1. Introduction

A finite groupG is called a simple Kn-group if G is a simple group with |π(G)|=n. In 1987, J. G. Thompson posed a very interesting problem related to algebraic number fields as follows (see [1]).Thompson’s Problem.Let T(G)={(n,sn)∣n∈ω(G) and sn∈nse(G)}, where sn is the number of elements with order n. Suppose that T(G)=T(H). If G is a finite solvable group, is it true that H is also necessarily solvable?It is easy to see that ifG and H are of the same order type, then
(1)nse(G)=nse(H),|G|=|H|.The proof is as follows: letG be a group and M some simple Ki-group, where i=3,4; then G≅M if and only if |G|=|M| and nse(G)=nse(M) (see [2, 3]). And also the group A12 is characterizable by order and nse (see [4]). Recently, all sporadic simple groups are characterizable by nse and order (see [5]).Comparing the sizes of elements of same order but disregarding the actual orders of elements inT(G) of Thompson’s problem, whether it can characterize finite simple groups? Up to now, some groups especial for L2(q), where q∈{7,8,9,11,13}, can be characterized by only the set nse(G) (see [6, 7]). The author has proved that the group L3(4) is characterizable by nse (see [8]). In this paper, it is shown that the group U3(7) also can be characterized by nse.Here, we introduce some notations which will be used. Leta·b denote the product of integer a by integer b. If n is an integer, then we denote by π(n) the set of all prime divisors of n. Let G be a group. The set of element orders of G is denoted by ω(G). Let k∈ω(G) and let sk be the number of elements of order k in G. Let nse(G)={sk∣k∈ω(G)}. Let π(G) denote the set of prime p such that G contains an element of order p. Ln(q) denotes the projective special linear group of degree n over finite fields of order q. Un(q) denotes the projective special unitary group of degree n over finite fields of order q. The other notations are standard (see [9]).


## 2. Some Lemmas

Lemma 1 (see [10]).
Let G be a finite group and m a positive integer dividing|G|. If Lm(G)={g∈G∣gm=1}, then m∣|Lm(G)|.Lemma 2 (see [11]).
Let G be a finite group and letp∈π(G) be odd. Suppose that P is a Sylow p-subgroup of G and n=psm with (p,m)=1. If P is not cyclic and s>1, then the number of elements of order n is always a multiple of ps.Lemma 3 (see [7]).
Let G be a group containing more than two elements. If the maximal number s of elements of the same order in G is finite, then G is finite and|G|≤s(s2-1).Lemma 4 (see [12, Theorem 9.3.1]).
Let G be a finite solvable group and|G|=mn, where m=p1α1⋯prαr, (m,n)=1. Let π={p1,…,pr} and let hm be the number of Hall π-subgroups of G. Then, hm=q1β1⋯qsβs satisfies the following conditions for all i∈{1,2,…,s}: (1)
qiβi≡1 (modpj) for some pj;(2)
the order of some chief factor of G is divided byqiβi.To proveG≅U3(7), we need the structure of simple K4-groups.Lemma 5 (see [13]).
LetG 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:
(a)
L2(r), where r is a prime and r2-1=2a·3b·vc with a≥1, b≥1, c≥1, and v is a prime greater than 3,(b)
L2(2m), where 2m-1=u, 2m+1=3tb with m≥2, u, t are primes, t>3, and b≥1,(c)
L2(3m), where 3m+1=4t, 3m-1=2uc or 3m+1=4tb, 3m-1=2u, with m≥2, u, t are odd primes, b≥1, and c≥1,(4)
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), 2D4(2), or2F4(2).Lemma 6.
LetG be a simple K4-group and 43∥G∥27·3·73·43. Then, G≅U3(7).Proof.
From Lemma5(1-2), order consideration rules out these cases.
So, we consider Lemma5(3). We will deal with this with the following cases.
Case  1. Consider  G≅L2(r), where r∈{3,7,43}. (i)
Letr=3; then |π(r2-1)|=1, which contradicts |π(q2-1)|=3.(ii)
Letr=7; then |π(r2-1)|=2, which contradicts |π(q2-1)|=3.(iii)
Letr=43; then |π(r2-1)|=4, which contradicts |π(q2-1)|=3.
Case  2. Consider  G≅L2(2m), where u∈{3,7,43}. (i)
Letu=3; then m=2 and so 5=3tb. But the equation has no solution in ℕ, which is a contradiction.(ii)
Letu=7; then m=3 and 23+1=3tb. Thus, t=3 and b=1. But t>3, a contradiction.(iii)
Letu=43; then the equation 2m-1=43 has no solution in ℕ, which is a contradiction.
Case  3. Consider  G≅L2(3m).
We will consider the case by the following two subcases.
Subcase 3.1.Consider 3m+1=4t  and 3m-1=2uc.
We can suppose thatt∈{3,7,43}.
Lett=3,7,43; the equation 3m+1=4t has no solution.
Subcase 3.2. Consider
3
m
+
1
=
4
t
b
and
3
m
-
1
=
2
u.
We can suppose thatu∈{3,7,43}.
Letu=3,5,7; then the equation 3m-1=2u has no solution in ℕ, which is a contradiction.
In review of Lemma5(4), G≅U3(7).
This completes the proof of the lemma.


## 3. Proof of Theorem

LetG be a group such that nse(G)=nse(U3(7)), and let sn be the number of elements of order n. By Lemma 3, we have that G is finite. We note that sn=kϕ(n), where k is the number of cyclic subgroups of order n. Also we note that if n>2, then ϕ(n) is even. If m∈ω(G), then by Lemma 1 and the above discussion, we have
(2)ϕ(m)∣sm,m∣∑d∣m‍sd.Theorem 7.
LetG be a group with nse(G)=nse(U3(7))={1, 2107, 92708, 101136, 117468, 117992, 202272, 235984, 471968, 539392, 809088, 943936, 1843968},  where U3(7) is the projective special unitary group of degree 3 over field of order 7. Then, G≅U3(7).Proof.
We prove the theorem by first proving thatπ(G)⫅{2,3,7,43}, second showing that |G|=|U3(7)|, and so G≅U3(7).
By (2), π(G)⫅{2,3,5,7,17,43}. If m>2, then ϕ(m) is even, and then s2=2107, 2∈π(G).
In the following, we prove that17∉π(G). If  17∈π(G), then by (2), s17=539392. If 2.17∈ω(G), then s2.17∉nse(G). Therefore, 2.17∉ω(G). Now we consider Sylow 17-subgroup P17 acts fixed-point-freely on the set of elements of order 2 of G; then |P17|∣s2, which is a contradiction. Hence, we have π(G)⫅{2,3,5,7,43}. Furthermore, by (2) s3=92708,117992,471968,s5=235984, s7=117648, and s43=1843968.If2a∈ω(G), then ϕ(2a)=2a-1∣s2a, and so 0≤a≤9. If a=9, then by (2), s29∉nse(G), and so 0≤a≤8.If3a∈ω(G), then 1≤a≤3.If5a∈ω(G), then a=1.If7a∈ω(G), then 1≤a≤5. If 75∈ω(G), then by (2), s75∉  nse(G). So, 1≤a≤4.If43a∈ω(G), then 1≤a≤2. If 422∈ω(G), then by (2), s432∉nse(G). Thus, a=1.
In the following, we first prove that5∉π(G), then consider the proper set of {2,3,7,43}, and so π(G)={2,3,7,43}. Finally, we get the desired result by [2].
To remove the “bad” prime 5, we should prove that43∈π(G).
Since2∈π(G) and exp(P2)=2, 4, 8, 16, 32, 64, 128, 256, then by Lemma 1, |P2|∣1+s2+⋯+s28 and |P2|∣211 (this case occurs only when s4=92708, s8=s16=101136, s32=s64=s128=809088, and s256=539392).
If3∈π(G), then exp(P3)=3,9,27.(i)
Ifs3=92708, then s32=117648. It follows that s33∉nse(G). So we have |P3|∣1+s3+s32 and so, |P3|∣34.(ii)
Ifs3=117992, then s32=202272 or 809088.
(a)
Lets32=202272; then 33∣1+s3+s32+s33, and so s33=117648. Then, |P3|∣1+s3+s32+s33, and so |P3|∣33.(b)
Lets32=809088; then 33∣1+s3+s32+s33, but s33∉nse(G). Then, |P3|∣32.(iii)
Ifs3=471968, then s32=s33=117648, and so |P3|∣33.
If5∈π(G) and exp(P5)=5, then by Lemma 1, |P5|∣1+s5, and so |P5|=5.
If7∈π(G) and exp(P7)=7,72,73,74, then by Lemma 1, |P7|∣1+s7+s72+s73+s74. By (2), s72  = 101136, 202272, 809088, 1843968. (i)
s72  = 101136, 202272 or 809088, and then 73∣1+s7+s72+s73, but s73∉nse(G). Then, |P7|∣72.(ii)
s72  = 1843968, and then s73=s74=1843968. So, |P7|∣74.
If43∈π(G) and exp(P43)=43, then by Lemma 1, |P43|∣1+s43, and so |P43|=43.
Assume that43∉π(G). We deal with it with the following cases.
Case A.Let 7∈π(G). We know that exp(P7)=7,72,73,74.(i)
Let exp(P7)  = 7. Then by Lemma 1, |P7|∣1+s7, and so |P7|∣76. If |P7|=7, then since n7=s7/ϕ(7), 43∈π(G), which is a contradiction. If |P7|=72, then we can assume that π(G)⫅{2,3,5,7}, 5663616 + 92708k1+101136k2 + 117468k3+117992k4+202272k5+235984k6 + 471968k7+539392k8 + 809088k9+943936k10 + 1843968k11=2m·3n·5p·72, where k1,…,k11, m,n, and p are nonnegative integers and 0≤∑i=111ki≤154. Since 5663616≤|G|=2m·3n·5p·72≤5663616 + 154.1843968, then since m is at most 11, the equation has no solution in ℕ. If |P7|=73, then similarly, (m,n,p)=(7, 3, 1), (8, 3, 1), (9, 3, 1), (10, 3, 1), (11, 3, 1), (9, 2, 1), (10, 2, 1), (11, 2, 1). Let |G|=27·33·5·73. Then by Sylow’s theorem, the number of Sylow 5-subgroups of G is 1, 6, 16, 21, 36, 56, 96, 126, 196, 216, 336, 441, 576, 686, 756, 896, 1176, 2016, 2646, 3136, 3456, 3528, 4116, 7056, 9261, 10976, 12096, 18816, 24696, 42336, 65856, and 395136, and so the number of elements of order 5 is at most 4, 24, 64, 84, 144, 224, 384, 504, 784, 864, 1344, 1764, 2304, 2744, 3024, 3584, 4704, 8064, 10584, 12544, 13824, 14112, 16464, 28224, 37044, 43904, 48384, 75264, 98784, 169344, 263424, and 1580544, but none of which belongs to nse(G), which is a contradiction. Similarly, we can rule out the other cases “(m,n,p)=(8, 3, 1), (9, 3, 1), (9, 2, 1).” Similarly, for |P7|≥74, we can get a contradiction as the case “|P7|=73."(ii)
Let exp(P7)=72. Then by Lemma 1, |P7|∣1+s7+s72 for s72∈{101136, 202272, 809088, 1843968}, and so |P7|∣72. Since n7=s72/ϕ(72) for s72∈{101136, 202272, 809088}, 43∈π(G), which is a contradiction. If s72=1843968/42=43904, by Sylow’s theorem, n7=7k+1 for some integer k, but the equation has no solution in ℕ, which is a contradiction.(iii)
Let exp(P7)=73. Then by Lemma 1, |P7|∣1+s7+s72+s73 and so |P7|∣74. If |P7|=73, then since n7=s73/ϕ(73)=6272, by Sylow’s theorem, n7=7k+1 for some integer k, but the equation has no solution in ℕ, which is a contradiction. If |P7|=74, then we can assume that π(G)⫅{2,3,5,7}, 5663616+92708k1+101136k2 + 117468k3+117992k4 + 202272k5+235984k6 + 471968k7+539392k8+809088k9 + 943936k10+1843968k11=2m.3n.5p.74, where k1,…,k11, m,n, and p are nonnegative integers and 0≤∑i=111ki≤163. Since 5663616≤|G|=2m·3n·5p·74≤5663616 + 163.1843968, then (m,n,p)=(5, 3, 1), (6, 3, 1), (7, 3, 1), (8, 3, 1), (9, 3, 1), (6, 2, 1), (7, 2, 1), (8, 2, 1), (9, 2, 1), (10, 2, 1), (11, 2, 1), (8, 1, 1), (9, 1, 1), (10, 1, 1), and (10, 1, 1). Let |G|=25·33·5·74. Then by Sylow’s theorem, the number of Sylow 5-subgroups of G is 1, 6, 16, 21, 36, 56, 96, 126, 196, 216, 336, 441, 686, 756, 1176, 2401, 2016, 2646, 4116, 7056, 9261, 10976, 14406, 18816, 24696, 28416, 42336, 86436, 148176, and 230496. It follows that the number of elements of order 5 is 4, 24, 64, 84, 144, 224, 384, 504, 784, 864, 1344, 1764, 2744, 3024, 4704, 9604, 8064, 10584, 16464, 28224, 37044, 43904, 57624, 75264, 98784, 113664, 169344, 345744, 592704, and 921984, but none of which belongs to nse(G). Similarly, we can rule the other cases.(iv)
Let exp(P7)=74. Then by Lemma 1, |P7|∣1+s7+s72+s73+s74, and so |P7|∣74. Since n7=s74/ϕ(74)=896, by Sylow’s theorem, n7=7k+1 for some integer k, but the equation has no solution in ℕ, which is a contradiction.
Case B. Let 5∈π(G). We know that exp(P5)   = 5. Then by Lemma 1, |P5|∣1+s5, and so |P5|  = 5. Since n5=s5/ϕ(5), then 43∈π(G), which is a contradiction.
Case C.Let 3∈π(G). We know that exp(P3)=3,9,27. (i)
Let exp(P3)=3. Then by Lemma 1, |P3|∣1+s3 and so |P3|∣32. Since n3=s3/ϕ(3), 43∈π(G), which is a contradiction. Therefore, |P3|=32. From cases A and B, and by assumption “7∉π(G)", we can assume that π(G)={2,3}. Then, 5663616+92708k1 + 101136k2+117468k3+117992k4+202272k5+235984k6 + 471968k7+539392k8+809088k9+943936k10+1843968k11=2m·9, where k1,…,k11 and m are nonnegative integers and 0≤∑i=111ki≤7. Since 5663616≤|G|=2m·9≤5663616+7.1843968, then since m is at most 11, the equation has no solution in ℕ.(ii)
Let exp(P3)=9. Then by Lemma 1, |P3|∣1+s3+s32 and so |P3|∣34. If |P3|=9, then 43∈π(G), which is a contradiction. If |P3|=27, 81, then similarly as the case “exp(P3)=3 and |P3|=9," the equation has no solution in ℕ since m is at most 11.(iii)
Let exp(P3)=27. Then by Lemma 1, |P3|∣1+s3+s32+s33, and so |P3|=27. Since n3=s33/ϕ(33), then 43∈π(G), which is a contradiction.
Therefore,43∈π(G). We prove that 5∉π(G).
Suppose to the contrary that5∈π(G). If 5.43∈ω(G), sets P and Q are Sylow 43-subgroups of G; then P and Q are conjugate in G, and so CG(P) and CG(Q) are also conjugate in G. Therefore, we have s5.43=ϕ(5.43)·n43·k, where k is the number of cyclic subgroups of order 5 in CG(P43). Since n43=s43/ϕ(43)=1843968/42,7375872∣s5.43 and so s5.43=7375872t for some integer t, the equation has no solution in ℕ since s5.43∈nse(G). Therefore, 5.43∉ω(G); it follows that the Sylow 5-subgroup of G acts fixed-point-freely on the set of elements of order 43, and so |P5|∣s43, which is a contradiction. Thus, 5∉π(G).
Therefore, we have that{2,43}⫅π(G)⫅{2,3,7,43}.
Case A (π(G)={2,43}). Therefore, 5663616+92708k1+101136k2+117468k3+117992k4+202272k5 + 235984k6+471968k7+539392k8+809088k9+943936k10+1843968k11=2m·9, where k1,…,k11 and m are nonnegative integers and 0∑i=111ki≤20. Since 5663616≤|G|=2m·43≤5663616+20.1843968, then since m is at most 11, the equation has no solution in ℕ.
Case B (π(G)={2,3,43}). As exp(P43)=43, by Lemma 1, |P43|∣1+s43 and so |P43|=43. Since n43=s43/ϕ(43), 7∈π(G), which is a contradiction.
Case C (π(G)={2,7,43}). Since exp(P43)=43, then by Lemma 1, |P43|∣1+s43, and so |P43|=43. Since n43=s43/ϕ(43), 3∈π(G), which is a contradiction.
Case D   (π(G)={2,3,7,43}). In the following, we first show that |G|=27·3·73·43, |G|=27·3·74·43, |G|=28·3·73·43, |G|=28·3·74·43, |G|=29·3·73·43, or |G|=29·3·74·43. and second prove that G≅U3(7).
Step  1  (|G|=27·3·73·43 or |G|=28·3·73·43). We have known that |P43|=43.
If3.43∈ω(G), sets P and Q are Sylow 43-subgroups of G; then P and Q are conjugate in G, and so CG(P) and CG(Q) are also conjugate in G. Therefore, we have s3.43=ϕ(3.43)·n43·k, where k is the number of cyclic subgroups of order 3 in CG(P43). Since n43=s43/ϕ(43)=1843968/42,3687936∣s3.43, so s3.43=3687936t for some integer t. Since s3.43∈nse(G), the equation has no solution. Therefore, 3.43∉ω(G); it follows that the Sylow 3-subgroup P3 of G acts fixed-point-freely on the set of elements of order 43, |P3|∣s43, and so |P3|=3. Similarly, 3.7,2.43∉ω(G) and |P7|∣73, |P2|∣28.
Therefore, we can assume that|G|=2m·3·7p·43. Since 5663616=27·3·73·43≤|G|=2m·3·7p·43, then |G|=27·3·73·43, or |G|=28·3·73·43.
Step  2   (G≅U3(7)). First prove that there is no group such that |G|=27·3·73·43 or |G|=28·3·73·43 and nse(G)=nse(U3(7)). Then by [2], G≅U3(7).
Let|G|=28·3·73·43 and nse(G)=nse(U3(7)).
IfG is soluble, set H is a Hall {3,7,43}-subgroup of G, and all the Hall {3,7,43}-subgroups of G are conjugate, so the number of Hall {3,7,43}-subgroups of G is |G:NG(H)|∣28. Let n43(H)=3a1+a2+⋯+ar·7b1+b2+⋯+bs by Lemma 4, 3ai≡1 (mod 43), 7bj≡1 (mod 43), i=1,…,r, j=1,2,…,s, where a1,…,ar, b1,…,bs are nonnegative integers and a1+⋯+ar≤1,  b1+⋯+bs≤4. Hence, n43(H)=1. So 42≤s43(G)≤10752 and 42∣s43(G), but s43(G)∉nse(G), which is a contradiction. So G is insoluble.
Therefore,G has a normal series 1⊲K⊲L⊲G such that L/K is isomorphic to a simple Ki-group with i=3, 4 as 9 and 1849 do not divide the order of G.
IfL/K is isomorphic to a simple K3-group, from [14], L/K≅L2(7). Then, |G/L|∣25·73·43. Let A/K:=CG/K(L/K). Then, A/K∩L/K=1. It is easy to see that G/K/A/K≲Aut(L/K)=SL(2,7), and so G/A≲SL(2,7). Since A/K,L/K⊲G/K, A/K×L/K≤G/K. Therefore, |L/K|∣|G/A|, and so G/K≅L2(7) or SL(2,7). That is, |A|=25·73·43 or 24·73·43. By Sylow’s theorem, n43(A)=1. Since A⊲G, we have that n7(A)=n7(G), and so s7(G)=42, which contradicts s7(G)∈nse(G).
Hence,G is isomorphic to a simple K4-group; then by Lemma 6, L/K≅U3(7). So G/A≤Aut(U3(7)). Therefore, G/A≅U3(7) or G/A≅2·U3(7).
IfG/A≅U3(7), then order consideration |A|=2. It follows that A is a normal subgroup generated by a 2-central element of G. So, there exists an element of order 2.43, which is a contradiction.
IfG/A≅2·U3(7), then A=1. But nse(2·U3(7))≠nse(U3(7)), which is a contradiction.
Therefore,|G|=27·3·73·43=|U3(7)|, and by assumption, nse(G) = nse(U3(7)); then by [2], G≅U3(7).
This completes the proof of the theorem.



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