A Characterization of Projective Special Unitary Group U₃(7) by nse

Abstract

Let 𝐺 a group and 𝜔(𝐺) be the set of element orders of 𝐺. Let 𝑘∈𝜔(𝐺) and let 𝑠_𝑘 be the number of elements of order 𝑘 in 𝐺. Let nse(𝐺)={𝑠_𝑘∣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).

1. Introduction

A finite group 𝐺 is called a simple 𝐾_𝑛-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,𝑠_𝑛)∣𝑛∈𝜔(𝐺) and 𝑠_𝑛∈nse(𝐺)}, where 𝑠_𝑛 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 if 𝐺 and 𝐻 are of the same order type, then nse(𝐺)=nse(𝐻),|𝐺|=|𝐻|. (1)

The proof is as follows: let 𝐺 be a group and 𝑀 some simple 𝐾_𝑖-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 in 𝑇(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. Let 𝑎⋅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 𝑠_𝑘 be the number of elements of order 𝑘 in 𝐺. Let nse(𝐺)={𝑠_𝑘∣𝑘∈𝜔(𝐺)}. Let 𝜋(𝐺) denote the set of prime 𝑝 such that 𝐺 contains an element of order 𝑝. 𝐿_𝑛(𝑞) denotes the projective special linear group of degree 𝑛 over finite fields of order 𝑞. 𝑈_𝑛(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 𝐿_𝑚(𝐺)={𝑔∈𝐺∣𝑔^𝑚=1}, then 𝑚∣|𝐿_𝑚(𝐺)|.

Lemma 2 (see [11]). Let G be a finite group and let 𝑝∈𝜋(𝐺) be odd. Suppose that P is a Sylow p-subgroup of G and 𝑛=𝑝^𝑠𝑚 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 𝑝^𝑠.

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=𝑝^𝛼1₁⋯𝑝^𝛼𝑟_𝑟, (𝑚,𝑛)=1. Let 𝜋={𝑝₁,…,𝑝_𝑟} and let ℎ_𝑚 be the number of Hall 𝜋-subgroups of G. Then, ℎ_𝑚=𝑞^𝛽1₁⋯𝑞^𝛽𝑠_𝑠 satisfies the following conditions for all 𝑖∈{1,2,…,𝑠}: (1)𝑞^𝛽𝑖_𝑖≡1 (mod𝑝_𝑗) for some 𝑝_𝑗;(2)the order of some chief factor of G is divided by 𝑞^𝛽𝑖_𝑖.

To prove 𝐺≅𝑈₃(7), we need the structure of simple 𝐾₄-groups.

Lemma 5 (see [13]). 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: (a)𝐿₂(𝑟), where 𝑟 is a prime and 𝑟²−1=2^𝑎⋅3^𝑏⋅𝑣^𝑐 with 𝑎≥1, 𝑏≥1, 𝑐≥1, and 𝑣 is a prime greater than 3, (b)𝐿₂(2^𝑚), where 2^𝑚−1=𝑢, 2^𝑚+1=3𝑡^𝑏 with 𝑚≥2, 𝑢, 𝑡 are primes, 𝑡>3, and 𝑏≥1, (c)𝐿₂(3^𝑚), where 3^𝑚+1=4𝑡, 3^𝑚−1=2𝑢^𝑐 or 3^𝑚+1=4𝑡^𝑏, 3^𝑚−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), 𝑂⁺₈(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. Let 𝐺 be a simple 𝐾₄-group and 43‖𝐺‖2⁷⋅3⋅7³⋅43. Then, 𝐺≅𝑈₃(7).

Proof. From Lemma 5(1-2), order consideration rules out these cases.
So, we consider Lemma 5(3). We will deal with this with the following cases.
Case  1. Consider  𝐺≅𝐿₂(𝑟), where 𝑟∈{3,7,43}. (i)Let 𝑟=3; then |𝜋(𝑟²−1)|=1, which contradicts |𝜋(𝑞²−1)|=3. (ii)Let 𝑟=7; then |𝜋(𝑟²−1)|=2, which contradicts |𝜋(𝑞²−1)|=3. (iii)Let 𝑟=43; then |𝜋(𝑟²−1)|=4, which contradicts |𝜋(𝑞²−1)|=3.
Case  2. Consider  𝐺≅𝐿₂(2^𝑚), where 𝑢∈{3,7,43}. (i)Let 𝑢=3; then 𝑚=2 and so 5=3𝑡^𝑏. But the equation has no solution in ℕ, which is a contradiction. (ii)Let 𝑢=7; then 𝑚=3 and 2³+1=3𝑡^𝑏. Thus, 𝑡=3 and 𝑏=1. But 𝑡>3, a contradiction. (iii)Let 𝑢=43; then the equation 2^𝑚−1=43 has no solution in ℕ, which is a contradiction.
Case  3. Consider  𝐺≅𝐿₂(3^𝑚).
We will consider the case by the following two subcases.
Subcase 3.1. Consider 3^𝑚+1=4𝑡  and 3^𝑚−1=2𝑢^𝑐.
We can suppose that 𝑡∈{3,7,43}.
Let 𝑡=3,7,43; the equation 3^𝑚+1=4𝑡 has no solution.
Subcase 3.2. Consider 3^𝑚+1=4𝑡^𝑏and 3^𝑚−1=2𝑢.
We can suppose that 𝑢∈{3,7,43}.
Let 𝑢=3,5,7; then the equation 3^𝑚−1=2𝑢 has no solution in ℕ, which is a contradiction.
In review of Lemma 5(4), 𝐺≅𝑈₃(7).
This completes the proof of the lemma.

3. Proof of Theorem

Let 𝐺 be a group such that nse(𝐺)=nse(𝑈₃(7)), and let 𝑠_𝑛 be the number of elements of order 𝑛. By Lemma 3, we have that 𝐺 is finite. We note that 𝑠_𝑛=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 𝜙(𝑚)∣𝑠_𝑚,𝑚∣_𝑑∣𝑚𝑠_𝑑. (2)

Theorem 7. Let 𝐺 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 that 𝜋(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 that 17∉𝜋(𝐺). If  17∈𝜋(𝐺), then by (2), 𝑠₁₇=539392. If 2.17∈𝜔(𝐺), then 𝑠_2.17∉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. If 2^𝑎∈𝜔(𝐺), then 𝜙(2^𝑎)=2^𝑎−1∣𝑠₂^𝑎, and so 0≤𝑎≤9. If 𝑎=9, then by (2), 𝑠₂⁹∉nse(𝐺), and so 0≤𝑎≤8. If 3^𝑎∈𝜔(𝐺), then 1≤𝑎≤3. If 5^𝑎∈𝜔(𝐺), then 𝑎=1. If 7^𝑎∈𝜔(𝐺), then 1≤𝑎≤5. If 7⁵∈𝜔(𝐺), then by (2), 𝑠₇⁵∉  nse(𝐺). So, 1≤𝑎≤4. If 43^𝑎∈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 that 5∉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 that 43∈𝜋(𝐺).
Since 2∈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).
If 3∈𝜋(𝐺), then exp(𝑃₃)=3,9,27.(i)If 𝑠₃=92708, then 𝑠₃²=117648. It follows that 𝑠₃³∉nse(𝐺). So we have |𝑃₃|∣1+𝑠₃+𝑠₃² and so, |𝑃₃|∣3⁴.(ii)If 𝑠₃=117992, then 𝑠₃²=202272 or 809088.(a)Let 𝑠₃²=202272; then 3³∣1+𝑠₃+𝑠₃²+𝑠₃³, and so 𝑠₃³=117648. Then, |𝑃₃|∣1+𝑠₃+𝑠₃²+𝑠₃³, and so |𝑃₃|∣3³. (b)Let 𝑠₃²=809088; then 3³∣1+𝑠₃+𝑠₃²+𝑠₃³, but 𝑠₃³∉nse(𝐺). Then, |𝑃₃|∣3². (iii)If 𝑠₃=471968, then 𝑠₃²=𝑠₃³=117648, and so |𝑃₃|∣3³.
If 5∈𝜋(𝐺) and exp(𝑃₅)=5, then by Lemma 1, |𝑃₅|∣1+𝑠₅, and so |𝑃₅|=5.
If 7∈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.
If 43∈𝜋(𝐺) and exp(𝑃₄₃)=43, then by Lemma 1, |𝑃₄₃|∣1+𝑠₄₃, and so |𝑃₄₃|=43.
Assume that 43∉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=2^𝑚⋅3^𝑛⋅5^𝑝⋅7², where 𝑘₁,…,𝑘₁₁, 𝑚,𝑛, and 𝑝 are nonnegative integers and 0≤∑¹¹_𝑖=1𝑘_𝑖≤154. Since 5663616≤|𝐺|=2^𝑚⋅3^𝑛⋅5^𝑝⋅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=2^𝑚.3^𝑛.5^𝑝.7⁴, where 𝑘₁,…,𝑘₁₁, 𝑚,𝑛, and 𝑝 are nonnegative integers and 0≤∑¹¹_𝑖=1𝑘_𝑖≤163. Since 5663616≤|𝐺|=2^𝑚⋅3^𝑛⋅5^𝑝⋅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=2^𝑚⋅9, where 𝑘₁,…,𝑘₁₁ and 𝑚 are nonnegative integers and 0≤∑¹¹_𝑖=1𝑘_𝑖≤7. Since 5663616≤|𝐺|=2^𝑚⋅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 that 5∈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 𝐶_𝐺(𝑃) and 𝐶_𝐺(𝑄) are also conjugate in 𝐺. Therefore, we have 𝑠_5.43=𝜙(5.43)⋅𝑛₄₃⋅𝑘, where 𝑘 is the number of cyclic subgroups of order 5 in 𝐶_𝐺(𝑃₄₃). Since 𝑛₄₃=𝑠₄₃/𝜙(43)=1843968/42,7375872∣𝑠_5.43 and so 𝑠_5.43=7375872𝑡 for some integer 𝑡, the equation has no solution in ℕ since 𝑠_5.43∈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=2^𝑚⋅9, where 𝑘₁,…,𝑘₁₁ and 𝑚 are nonnegative integers and 0∑¹¹_𝑖=1𝑘_𝑖≤20. Since 5663616≤|𝐺|=2^𝑚⋅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.
If 3.43∈𝜔(𝐺), sets 𝑃 and 𝑄 are Sylow 43-subgroups of 𝐺; then 𝑃 and 𝑄 are conjugate in 𝐺, and so 𝐶_𝐺(𝑃) and 𝐶_𝐺(𝑄) are also conjugate in 𝐺. Therefore, we have 𝑠_3.43=𝜙(3.43)⋅𝑛₄₃⋅𝑘, where 𝑘 is the number of cyclic subgroups of order 3 in 𝐶_𝐺(𝑃₄₃). Since 𝑛₄₃=𝑠₄₃/𝜙(43)=1843968/42,3687936∣𝑠_3.43, so 𝑠_3.43=3687936𝑡 for some integer 𝑡. Since 𝑠_3.43∈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|=2^𝑚⋅3⋅7^𝑝⋅43. Since 5663616=2⁷⋅3⋅7³⋅43≤|𝐺|=2^𝑚⋅3⋅7^𝑝⋅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)).
If 𝐺 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∶𝑁_𝐺(𝐻)|∣2⁸. Let 𝑛₄₃(𝐻)=3^{𝑎1+𝑎2+⋯+𝑎𝑟}⋅7^{𝑏1+𝑏2+⋯+𝑏𝑠} by Lemma 4, 3^𝑎𝑖≡1 (mod 43), 7^𝑏𝑗≡1 (mod 43), 𝑖=1,…,𝑟, 𝑗=1,2,…,𝑠, where 𝑎₁,…,𝑎_𝑟, 𝑏₁,…,𝑏_𝑠 are nonnegative integers and 𝑎₁+⋯+𝑎_𝑟≤1,  𝑏₁+⋯+𝑏_𝑠≤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 𝐾_𝑖-group with 𝑖=3, 4 as 9 and 1849 do not divide the order of 𝐺.
If 𝐿/𝐾 is isomorphic to a simple 𝐾₃-group, from [14], 𝐿/𝐾≅𝐿₂(7). Then, |𝐺/𝐿|∣2⁵⋅7³⋅43. Let 𝐴/𝐾∶=𝐶_𝐺/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).
If 𝐺/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.
If 𝐺/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.

Acknowledgments

The author is supported by the Department of Education of Sichuan Province (Grants nos. 12ZB085, 12ZB291, and 13ZA0119). The author is very grateful for the helpful suggestions of the referee.