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Recognition of the symplectic simple group PSp4(p) by the order and degree prime-power graph

  • Let G be a finite group, cd(G) the set of all irreducible character degrees of G, and ρ(G) the set of all prime divisors of integers in cd(G). For a prime p and a positive integer n, let np denote the p-part of n. The degree prime-power graph of G is a graph whose vertex set is V(G)={pep(G)pρ(G)}, where pep(G)=max{npncd(G)}, and there is an edge between distinct numbers x,yV(G) if xy divides some integer in cd(G). The authors have previously shown that some non-abelian simple groups can be uniquely determined by their orders and degree prime-power graphs. In this paper, the authors build on this work and demonstrate that the symplectic simple group PSp4(p) can be uniquely identified by its order and degree prime-power graph.

    Citation: Chao Qin, Yu Li, Zhongbi Wang, Guiyun Chen. Recognition of the symplectic simple group PSp4(p) by the order and degree prime-power graph[J]. AIMS Mathematics, 2024, 9(2): 2808-2823. doi: 10.3934/math.2024139

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  • Let G be a finite group, cd(G) the set of all irreducible character degrees of G, and ρ(G) the set of all prime divisors of integers in cd(G). For a prime p and a positive integer n, let np denote the p-part of n. The degree prime-power graph of G is a graph whose vertex set is V(G)={pep(G)pρ(G)}, where pep(G)=max{npncd(G)}, and there is an edge between distinct numbers x,yV(G) if xy divides some integer in cd(G). The authors have previously shown that some non-abelian simple groups can be uniquely determined by their orders and degree prime-power graphs. In this paper, the authors build on this work and demonstrate that the symplectic simple group PSp4(p) can be uniquely identified by its order and degree prime-power graph.



    Throughout the following, all groups are assumed to be finite, and all characters are complex characters. Let G be a group, then Irr(G) denotes the set of all irreducible characters of G, and cd(G) denotes the set of all irreducible character degrees of G. For a positive integer n, we use π(n) to denote the set of all prime divisors of n. For convenience, we write π(G)=π(|G|), and ρ(G)=χIrr(G)π(χ(1)). For a prime p, the p-part of n, denoted by np, is the maximum power of p such that npn. CFSG is the abbreviation for the Classification Theorem of Finite Simple Groups.

    The set cd(G) is very important for studying G. Many results have been obtained about the relationship between the set cd(G) and the structure of G. This paper's primary focus centers on finite non-abelian simple groups. In 2000, Huppert proposed a conjecture: "If S is a non-abelian simple group such that cd(G)=cd(M), then G M×A, where A is an abelian group" (see [1]). Currently, this conjecture has not been fully proven. If Huppert's conjecture proves valid, it would follow naturally that all finite non-abelian simple groups could be uniquely determined by their orders and sets of irreducible character degrees. Interestingly, provided |G|=|M| and fewer numbers in cd(G) require consideration, one can still deduce the same conclusion more efficiently. Researchers have shown that certain simple groups can be uniquely determined by their orders and at most three distinct irreducible character degrees (see [2,3,4,5,6] etc.).

    Other researchers have studied the structure of a finite group with the character degree graph Δ(G). The vertex set of Δ(G) is ρ(G). In Δ(G), two vertices are joined by an edge if the product of them divides some number in cd(G). The character degree graph has been widely studied, for example, in [7]. Clearly, Δ(G) contains much less information of cd(G). An interesting fact is that some simple groups (not all) can be uniquely determined by their orders and degree graphs.

    In [8], Khosravi et al. proved that A5, A6, A7, A8, L3(3), L3(4), L2(64), L2(q) (where q is an odd prime or a square of an odd prime, and q5), and L2(2α) (where α is a positive integer such that 2α1 or 2α+1 is a prime) can be uniquely determined by their orders and character degree graphs. However, in [9] Heydari and Ahanjideh gave a counter example of this characterization, that is, M12 and A4×M11 have the same order and character degree graph. To overcome this shortcoming, Chao Qin and Guiyun Chen defined a new graph by using the information of cd(G) in [10], which is called the degree prime-power graph Γ(G).

    Definition. For every pρ(G), let pep(G)=max{χ(1)pχIrr(G)} and V(G)={pep(G)pρ(G)}. Define the degree prime-power graph Γ(G) of G as follows: V(G) is the vertex set, and there is an edge between distinct numbers x,yV(G) if xy divides some integer in cd(G). Denote the edge between distinct numbers x,yV(G) by xy, and the set of all edges of Γ(G) by E(G).

    Chen, Qin, Wang and et al. have proven that the group M12 can be uniquely determined by its order and degree prime-power graph. They have also shown that all sporadic simple groups, PSL2(p), PSL2(p1) (where p is a Fermat prime), PSL2(p2), and PSL2(p3) can be uniquely determined by their orders and degree prime-power graphs (see [10,11,12,13]). In this paper, we continue this investigation for the symplectic simple group PSp4(p). The following theorem is our main result.

    Main Theorem. Let p be an odd prime, then GPSp4(p) if and only if |G|=|PSp4(p)| and Γ(G)= Γ(PSp4(p)).

    In this section, we present the degree prime-power graph of PSp4(p) when p is an odd prime, along with an important proposition.

    By Atlas [14] and Magma, we obtain the order and the degree prime-power graph of PSp4(p) when p=3,5,7 (see Table 1).

    Table 1.  The degree prime-power graph of PSp4(p) when p=3,5,7.
    Groups PSp4(3) PSp4(5) PSp4(7)
    Order 26345 26325413 28325474
    Γ()

     | Show Table
    DownLoad: CSV

    If p>7, then the order of PSp4(p) is

    |PSp4(p)|=12p4(p21)(p41)=p4(p1)2(p+1)2(p2+12).

    The character degrees of PSp4(p) were determined by Shahabi and Mohtadifar in [15], that is

    cd(PSp4(p))={1,12p(p2+1),12p(p+1)2,p4,12p(p1)2,p41,(p1)2(p2+1),(p1)2(p+1)2,(p+1)(p2+1),(p1)(p2+1),p(p+1)(p2+1),p(p1)(p2+1),12(p41),p(p2+1),12(p2+1),12p2(p2+1),12(p1)2(p2+1),(p1)2(p2+1),12(p+1)2(p2+1),12(p+ϵ)(p2+1),12p(p+ϵ)(p2+1)}

    where ϵ=(1)(p1)/2. We see that PSp4(p) has irreducible characters of the following degrees:

    p4,(p1)2(p+1)2,(p1)2(p2+1),(p+1)22(p2+1). (2.1)

    Note that

    (p1,p+1)=2,(p1,p2+12)=(p+1,p2+12)=1,

    and we have |PSp4(p)|2=((p1)2(p+1)2)2, moreover, 2e2(PSp4(p))=|PSp4(p)|2 by (2.1). Similarly, for every rρ(PSp4(p))=π(PSp4(p)), one has

    rer(PSp4(p))=|PSp4(p)|r.

    That means the vertex set of Γ(PSp4(p)) is

    V(PSp4(p))={|PSp4(p)|r|rπ(PSp4(p))}.

    Observing cd(PSp4(p)), the edges of Γ(PSp4(p)) can be described as

    (1) p4 is an isolated vertex,

    (2) let r1, r2π(PSp4(p)){2,p}. From (2.1) there is an edge

    rer1(PSp4(p))1rer2(PSp4(p))2,i.e.,|PSp4(p)|r1|PSp4(p)|r2,

    (3) for every rπ((p1)2(p+1)2){2}, there is an edge |PSp4(p)|2|PSp4(p)|r,

    (4) for every rπ((p+1)22), there is not an edge |PSp4(p)|2|PSp4(p)|r.

    Apart from the above cases, the following holds.

    Proposition 2.1. If p7 and rπ(PSp4(p)){p,3}, then there is always an edge |PSp4(p)|3|PSp4(p)|r in Γ(PSp4(p)).

    Proof. By the above statements about Γ(PSp4(p)), we only need to prove

    3π((p1)2(p+1)2).

    This is trivial, because we can obtain 3|(p1)(p+1) by 3p.

    Example. The orders of PSp4(11) and PSp4(13) respectively are

    |PSp4(11)|=114(111)2(11+1)2(112+12)=114(263252)61,|PSp4(13)|=134(131)2(13+1)2(132+12)=134(263272)517.

    Using the description above, we obtain the degree prime-power graphs of PSp4(11) and PSp4(13) (see Figure 1).

    Figure 1.  The degree prime-power graphs of PSp4(11) and PSp4(13).

    In this section, we study a subnormal chain of non-solvable groups through some lemmas. These lemmas will play an important role in the proof of the main theorem.

    Lemma 3.1. [10,Lemma 2.5] Let G be a non-solvable group. If T/S is a non-abelian chief factor of G, then there is a normal series 1H<KG such that K/HT/S and G/KOut(T/S).

    Lemma 3.2. [10,Corollary 2.1] Let G be a non-solvable group. Then there is a subnormal series

    G=G0G1>G2>G2k2G2k1>G2k1,(k1) (3.1)

    such that G2k is solvable, G2i1/G2i is a non-abelian chief factor of G2i2, and

    G2i2/G2i1Out(G2i1/G2i)

    for each 1ik.

    For convenience, we denote the subnormal series in Lemma 3.2 by sn2k, or snH2k if the subnormal subgroups in the series are written as Hi, and denote the j-th term in sn2k by sn2k(j), where 0j2k. Generally, sn2k is not unique for G.

    Lemma 3.3. Let G be a finite non-solvable group, and M be a non-abelian composition factor of G. If, for any sn2k and any 1ik, the quotient group sn2k(2i2)/sn2k(2i1) does not have a non-abelian composition factor which is isomorphic to M, then there exists a subnormal series

    snH2t:G=H0H1>H2>H2t2H2t1>H2t1,(t1) (3.2)

    such that H2t1/H2tMm for some integer m.

    Proof. If M is the unique (up to an isomorphism) non-abelian composition factor of G. By Lemma 3.2, the conclusion is obvious. Now suppose that the non-abelian composition factor of G is not unique (up to an isomorphism).

    Assume M1 is a non-abelian composition factor of G and M1 is not isomorphic to M. Suppose T/K is a non-abelian chief factor of G and T/KMm11. By Lemma 3.1, there is a normal series

    G=H0H1>H21

    such that H1/H2T/KMm11 and H0/H1Out(H1/H2). By this assumption, M is not a non-abelian composition factor of H0/H1. Then, M is a non-abelian composition factor of H2.

    If M is the unique (up to an isomorphism) non-abelian composition factor of H2, then the conclusion holds by Lemma 3.2. If M is not unique (up to an isomorphism) non-abelian composition factor of H2, then we assume that M2 is a non-abelian composition factor of H2 and M2 is not isomorphic to M. Similar to the above reason, there is a normal series

    H2H3>H41

    such that H3/H4Mm22 is a non-abelian chief factor of H2, and H2/H3Out(H3/H4). By this assumption, M is a non-abelian composition factor of H4.

    Repeating the above process, we obtain a subnormal series

    G=H0H1>H2>H2r2H2r1>H2r (3.3)

    such that

    (1) M is the unique (up to an isomorphism) non-abelian composition factor of H2r;

    (2) H2i1/H2iMmii is a non-abelian chief factor of H2i2, and H2i2/H2i1Out(H2i1/H2i) where Mi is not isomorphic to M and i=1, , r.

    By Lemma 3.2, there is a subnormal series

    H2rH2r+1>H2r+2>H2t2H2t1>H2t1 (3.4)

    such that H2t is solvable, H2i1/H2iMmi is a non-abelian chief factor of H2i2, and H2i2/H2i1Out(H2i1/H2i) for each r+1it. Therefore, the proof is complete by (3.3) and (3.4).

    Lemma 3.4. Let G be a group, and p be an odd prime. Suppose that

    (1) |G|p10, and

    (2) G has a non-abelian composition factor M satisfying p||M|.

    Then, there exists a subnormal series sn2k such that

    sn2k(2k1)/sn2k(2k)Mm,

    where m is an integer. Specifically, if sn2k(2k)=1, then there exists a subnormal subgroup of G which is isomorphic to M.

    Proof. By Lemma 3.3, we only need to prove that for any snH2t and any 1it the quotient group snH2t(2i2)/snH2t(2i1) does not have a non-abelian composition factor which is isomorphic to M.

    Otherwise, suppose there is a snH2t such that snH2t(2i2)/snH2t(2i1) has a non-abelian composition factor which is isomorphic to M for some 1it. Note that snH2t(2i1)/snH2t(2i) is a non-abelian chief factor of snH2t(2i2), and it is a direct product of isomorphic non-abelian simple groups. Let

    snH2t(2i1)/snH2t(2i)=(M1)m,

    where M1 is a non-abelian simple group. Hence, we have

    Out(snH2t(2i1)/snH2t(2i))=Out(M1)Sm.

    For convenience, denote snH2t(2i2)/snH2t(2i1) as K, then KOut(M1)Sm. Without loss of generality, assume KOut(M1)Sm. Then

    K/K(Out(M1))mK(Out(M1))m/(Out(M1))m(Out(M1)Sm)/(Out(M1))mSm.

    Since M1 is a non-abelian simple group, we know (Out(M1))m is solvable. So, K(Out(M1))m is solvable. Because K=snH2t(2i2)/snH2t(2i1) has a non-abelian composition factor which is isomorphic to M, we have

    MK/K(Out(M1))mSm, then |M|||Sm|.

    Hence, p||Sm| and pm. Therefore

    p10|G||snH2t(2i1)/snH2t(2i)|=|(M1)m|=|M1|m|M1|p60p.

    This is a contradiction, because p1060p is impossible.

    For the main theorem, the necessity is obvious. It is enough to prove the sufficiency. To achieve this, we need the following series of lemmas.

    Lemma 4.1. Let G be a finite group and K be a subnormal subgroup of G. If V(G)={|G|p|pπ(G)}, then the following statements hold.

    (1) G is non-solvable.

    (2) Every minimal subnormal subgroup of G is a non-abelian simple group.

    (3) 1 is the unique solvable subnormal subgroup of G. Furthermore, for any sn2k, we have sn2k(2k)=1.

    (4) If K1, then V(K)={|K|p|pπ(K)}.

    (5) If there are p,rπ(K) such that |G|p|G|r is an edge of Γ(G), then |K|p|K|r is an edge of Γ(K).

    Proof. Statements (1) and (2) are Lemma 2.2 in [10]. Statement (3) is a corollary of (2).

    For (4) and (5), without loss of generality we suppose K is a normal subgroup of G. Suppose that χIrr(G) such that χ(1)p=|G|p where pπ(K). Let θIrr(K) be a constituent of χK. Using Corollary 11.29 in [16], we have χ(1)θ(1)||G/K|. Furthermore,

    χ(1)pθ(1)p||G|p|K|p.

    Then, θ(1)p=|K|p by χ(1)p=|G|p. Hence, |K|pV(K). This means statement (4) holds.

    Let p,rπ(K) such that |G|p|G|r is an edge of Γ(G), then there exists χIrr(G) satisfying (|G|p|G|r)|χ(1). If ϕIrr(K) is a constituent of χK, then χ(1)θ(1)||G/K| by Corollary 11.29 in [16]. Hence, (|K|p|K|r)|ϕ(1), i.e., |K|p|K|r is an edge of Γ(K). Statement (5) holds.

    Lemma 4.2. The following statements about the exceptional group of Lie type 2B2(22n+1) (n1) hold.

    (1) 2B2(22n+1) is the only simple group whose order is prime to 3.

    (2) Let p5 be a prime. If

    p||2B2(22n+1)|and|2B2(22n+1)|||PSp4(p)|,

    then n=1 and p=13.

    Proof. (1) Obviously, for any alternating group An (n5), its order is divisible by 3. According to Atlas [14], any sporadic simple group's order is divisible by 3 as well as the order of 2F4(2).

    If M is a simple group of Lie type over GF(q) with M2F4(2) and 2B2(22n+1) (n1), then q(q21)||M|. Notice that 3|q(q21), and we have 3||M|.

    The order of 2B2(22n+1) is

    |2B2(22n+1)|=22n+1(24n+2+1)(22n+11).

    Since

    \begin{aligned} 2^{2n+1}-1& = \left(3-1\right)^{2n+1}-1 \equiv -2 &\mod 3, \\ 2^{4n+2}-1& = \left(3-1\right)^{4n+2}+1 \equiv\ 2 &\mod 3, \end{aligned}\; \;

    the order of { }^2 B_2\left(2^{2 n+1}\right) is not divisible by 3 . This completes the proof of (1) by CFSG.

    (2) Suppose

    q = 2^{2 n+1}, \quad a = q+\sqrt{2q}+1, \quad b = q-\sqrt{2q}+1, \quad\epsilon_i = 1 \; \text{or }\; -1, \; ( i = 1, \; 2 ).

    It is not hard to prove that q^2+1 = a b , \left(q^2+1, q-1\right) = 1 , and (a, b) = 1 . Hence,

    |{ }^2 B_2\left(2^{2 n+1}\right)| = |{ }^2 B_2\left(q\right)| = q^2(q-1)(q^2+1) = q^2(q-1)ab.

    By p\, \Big|\, |{ }^2 B_2\left(2^{2 n+1}\right)| , we have

    p\, \big|\, (q-1) , \; \; p\, \big|\, a \; \; \text{or}\; \; p\, \big|\, b.

    From |{ }^2 B_2\left(2^{2 n+1}\right)|\, \Big|\, |PSp_4(p)| , it follows that q^2\, \Big|\, (p+1)^2(p-1)^2 , i.e., q\, \big|\, (p+1)(p-1) . Since \bigl(p+1, p-1\bigr) = 2 , we get

    q\, \big|\, 2(p+1)\; \; \text{or} \; \; q\, \big|\, 2(p-1).

    (i) If p\, \big|\, (q-1) and q\, \big|\, 2(p+\epsilon_1) , then there are positive integers k and m such that

    q-1 = pk \; \; \text{and }\; \; 2(p+\epsilon_1) = qm.

    It follows that

    2(p+\epsilon_1) = (pk+1)m, \; \; \text{i.e., }\; \; (mk-2)p = -m+2\epsilon_1.

    Hence p\, \big|\, -m+2\epsilon_1 .

    If -m+2\epsilon_1 = 0 , then m = 2 and \epsilon_1 = 1 , moreover, q = p+1 , i.e., p = q-1 . Therefore,

    |PSp_4(p)| = (q-1)^4(q-2)^2q^2\frac{1}{2}\left(q^2-2q+2\right).

    By |{ }^2 B_2\left(2^{2 n+1}\right)|\, \Big|\, |PSp_4(p)| , we obtain

    \begin{equation} q^2+1\, \Big|\, (q-1)^3(q-2)^2\frac{1}{2}\left(q^2-2q+2\right). \end{equation} (4.1)

    A short calculation reveals that

    q^2+1 = (q-2)(q+2)+5, \; \text{ and }\; q^2-2q+2 = (q^2+1)-2(q-1).

    Then, \left(q^2+1, q-2\right)\Big|5 and \left(q^2+1, q^2-2q+2 \right) = \left(q^2+1, q-1\right) = 1 . Consequently, by (4.1) we know that q^2+1\, \Big|\, (q-2)^2 , and q^2+1 is a power of 5 . This leads to \pi\left(|{ }^2 B_2\left(2^{2 n+1}\right)|\right) = \{2, 5, p\} . This is a contradiction, because |{ }^2 B_2\left(2^{2 n+1}\right)| is not a K_3 -simple group, where the so-called K_3 -simple group is a non-abelian simple group with only three prime factors of its order.

    Now we assume -m+2\epsilon_1\neq0 . By p\, \big|\, -m+2\epsilon_1 , it follows that m\geqslant p+2\epsilon_1\geqslant3 . Then, 2(p+\epsilon_1) = mq\geqslant3(pk+1) , i.e., 0 < (3k-2)p\leqslant 2\epsilon_1-3 < 0 . This is a contradiction.

    (ii) If p\, \big|\, q+\epsilon_2\sqrt{2q}+1 and q\, \big|\, 2(p+\epsilon_1) , then there exist positive integers k and m such that

    \begin{equation} q+\epsilon_2\sqrt{2q}+1 = pk \; \; \text{and }\; \; 2(p+\epsilon_1) = qm. \end{equation} (4.2)

    Then, 2(q+\epsilon_2\sqrt{2q}+1) = 2pk = (qm-2\epsilon_1)k . Substituting q = 2^{2n+1} into this equation yields

    \begin{equation} 2^{n+1}\left(2^{n}+\epsilon_2-2^{n-1}mk\right) = -1-\epsilon_1k. \end{equation} (4.3)

    If 1+\epsilon_1k = 0 , then k = 1 , \epsilon_1 = -1 and 2^{n}+\epsilon_2-2^{n-1}m = 0 , i.e., \epsilon_2 = 2^{n-1}m-2^{n} . It follows that n = 1 and m = \epsilon_2+2 = 1 or 3 . Therefore, q = 2^{2n+1} = 8 and 2(p-1) = 8m = 8 or 24 , by (4.2). This means that p = 5 or 13 .

    If 1+\epsilon_1k\neq0 , then 2^{n+1}\, \big|\, 1+\epsilon_1k by (4.3). We conclude that k\geqslant 2^{n+1}-\epsilon_1 . Therefore,

    2(q+\epsilon_2\sqrt{2q}+1) = (qm-2\epsilon_1)k\geqslant (qm-2\epsilon_1)(2^{n+1}-\epsilon_1).

    By q = 2^{2n+1} , the above inequality can be rewritten as

    2^{2n+1}+\epsilon_22^{n+1}\geqslant 2^{3n+1}m-\epsilon_1(2^{n+1}+2^{2n}m).

    Dividing both sides of this inequality by 2^{n+1} , we get

    2^{n-1}\big[(2^{n+1}-\epsilon_1)m-2\big]\leqslant \epsilon_1+\epsilon_2\leqslant 2

    which implies that (2^{n+1}-\epsilon_1)m\leqslant 2+\frac{2}{2^{n-1}} \leqslant 4 . This means 2^{n+1}-\epsilon_1\leqslant 4 , moreover,

    2^{n+1}\leqslant 4+\epsilon_1\leqslant 5, \; \; \text{i.e.,} \; \; n = 1.

    Therefore, q = 2^{2n+1} = 8 . By p\, \big|\, q+\epsilon_2\sqrt{2q}+1 , we obtain that p\, \big|\, q^2+1 . Then, p = 5 or 13 .

    Because |{ }^2 B_2\left(8\right)| = 8^2(8^2+1)(8-1) = 2^6\cdot(5\cdot13)\cdot7 , |PSp_4(5)| = 2^6 \cdot 3^2 \cdot 5^4 \cdot 13 , |PSp_4(13)| = 13^4 \cdot \left(2^6\cdot 3^2\cdot 7^2\right)\cdot 5\cdot 17 , and |{ }^2 B_2\left(2^{2 n+1}\right)|\, \Big|\, |PSp_4(p)| , we obtain that p can only be 13 .

    Lemma 4.3. If M is a simple group of Lie type over GF(q) , where q = p^f , then |M|_p is an isolated vertex in \Gamma(M) .

    Proof. If M is a simple group of Lie type over some GF(q) with M \neq {}^2F_4(2)^{\prime} , the result can be obtained directly by Lemma 2.4 in [17]. If M = {}^2F_4(2)^{\prime} , then |M|_2 = 2^{11} is an isolated vertex in \Gamma(S) by the character table of {}^2F_4(2)^{\prime} in [14].

    Lemma 4.4. Let G be a group, and p\geqslant 7 be an odd prime. If

    |G| = |PSp_4(p)| = \frac{1}{2} p^{4}\left(p^2-1\right)\left(p^4-1\right), \mathit{\; \; {and\; \; }\;} \Gamma(G) = \Gamma(PSp_4(p)),

    then the following statements hold.

    (1) Let M be a simple group of Lie type over GF(q) where q = r^f . If p\, \big|\, |M| and M is a subnormal subgroup of G , then r = p or M\cong ^2B_2(8) .

    (2) If T is a non-abelian composition factor of G , then p\, \nmid\, |\operatorname{Out}(T) .

    (3) For any sn_{2k} and any 1\leqslant i\leqslant k , the order of \operatorname{Out}\left(sn_{2k}(2i-1)/ sn_{2k}(2i)\right) is not divisible by p , and p\, \nmid\, |sn_{2k}(2i-2)/ sn_{2k}(2i-1)| .

    Proof. (1) By the assumption, the vertex set of \Gamma(G) is V(G) = \{|G|_t\, |\, t\in\pi(G)\} . Then, V(M) = \{|M|_t\, |\, t\in\pi(M)\} by Lemma 4.1(4).

    First, we suppose that r\neq p and M is not isomorphic to { }^2 B_2\left(2^{2 n+1}\right) ( n\geqslant 1 ).

    If r = 3 , then there exists an edge |G|_2\sim |G |_3 by Proposition 2.1. Because M is not isomorphic to { }^2 B_2\left(2^{2 n+1}\right) , we know 3\, \big|\, |M| and |M|_3\sim|M|_2 is an edge of \Gamma(M) by Lemma 4.1(5). This is a contradiction, because |M|_r is an isolated vertex in \Gamma(M) by Lemma 4.3.

    If r\neq 3 , then there exists an edge |G|_r\sim |G |_3 by Proposition 2.1. We can obtain a contradiction using the same method as before.

    Second, assume r\neq p and M is isomorphic to { }^2 B_2\left(2^{2 n+1}\right) ( n\geqslant 1 ). By Lemma 4.2(2), we see that M is isomorphic to { }^2 B_2\left(8\right) and p = 13 . This completes the proof of (1).

    (2) Assume T is a non-abelian composition factor of G , and p\, \big|\, |\operatorname{Out}(T) . Since p is an odd prime, T is not isomorphic to an alternating group, one of 26 sporadic simple groups or the Tits simple group ^2F_4(2)^\prime , because the outer automorphism groups of these simple groups are 1 , C_2 , or C_2\times C_2 . By CFSG, T is a simple group of Lie type over some GF(q) except ^2F_4(2)^\prime , where q = r^f , and r is a prime. Therefore, |\operatorname{Out(T)}| = dfg where d , f , and g are the orders of diagonal, field, and graph automorphisms of T , respectively. By Table 5 in [14], we know that g\, \big|\, 6 , so p\, \nmid\, g . Therefore, if p\big| |Out(T)| , then p\, \big|\, d or p\, \big|\, f .

    If p\, \big|\, d , then T is isomorphic to A_n(q) or {}^2A_n(q) (i.e., PSL_{n+1}(q) or PSU_{n+1}(q) ) by Table 5 in [14]. Assume T\cong A_n(q) , then d = (n+1, q-1) and

    |T| = |A_n(q)| = \frac{1}{(n+1, q-1)}q^{n(n+1)/2}\prod\limits_{i = 1}^n(q^{i+1}-1).

    Hence, n+1\geqslant p\geqslant 7 , p\, \big|\, q-1 , and p\, \big|\, |T| . By Lemma 3.4 and Lemma 4.1(3), there exists a subnormal subgroup K\trianglelefteq \trianglelefteq G such that K\cong T \cong A_n(q) . It follows that q = p^f by (1). It contradicts to p\, \big|\, q-1 . Using the same method, one can demonstrate that T\ncong{}^2A_n(q) .

    Now, we assume p\, \big|\, f . It follows that q = r^{kp}\geqslant 2^p\geqslant 2^7 . If T is not isomorphic to A_1(q) , then one can obtain

    |T|\geqslant q^3(q+1)

    by observing the order of a simple group of Lie type over GF(q) . Hence,

    \frac{1}{2}p^4(p^2-1)(p^4-1) = |G|\geqslant |T| \geqslant q^3(q+1)\geqslant 2^{3p}(2^p+1).

    This implies p\leqslant 5 by Maple. This contradicts p\geqslant 7 .

    If T is isomorphic to A_1(q) , then |T| = \frac{1}{(2, q-1)}q(q^2-1) . When p\geqslant 7 , one can obtain the following inequality by Maple:

    \begin{equation} \frac{1}{2}p^4(p^2-1)(p^4-1) < \frac{1}{2}\cdot3^{p}(3^{2p}-1). \end{equation} (4.4)

    However, since (2, q-1)\leqslant 2 and q = r^{kp} , we have

    \begin{equation} \frac{1}{2}p^4(p^2-1)(p^4-1) = |G|\geqslant |T| = \frac{1}{(2, q-1)}q(q^2-1) \geqslant \frac{1}{2}\cdot r^{kp}(r^{2kp}-1). \end{equation} (4.5)

    By inequalities (4.4) and (4.5), one can see that r^k = 2 , i.e., q = 2^p . Then, |T| = \frac{1}{(2, 2^p-1)}2^p(2^{2p} -1) = 2^p(2^{2p}-1) . Using Maple to solve the inequality |G|\geqslant |T| , we conclude that p\leqslant 11 , i.e., p = 7 or 11 . If p = 7 , then |T| = 2^7\cdot3\cdot43\cdot127 and |G| = 2^8 \cdot 3^2 \cdot 5^4 \cdot 7^4 . This is a contradiction to |T|\, \big|\, |G| . If p = 11 , then |T| = 2^{11}\cdot3\cdot23\cdot89\cdot683 and |G| = 11^4 \cdot\left(2^6 \cdot 3^2 \cdot 5^2\right) \cdot 61 . This is a contradiction. Therefore, f is not divisible by p when p\geqslant7 . This completes the proof of (2).

    (3) Suppose that there exists an sn_{2k} and i such that

    p\, \Big|\, |\operatorname{Out}\left(sn_{2k}(2i-1)/ sn_{2k}(2i)\right)|.

    Notice that {sn} _{2k}(2i-1)/ {sn}_{2k}(2i) is a non-abelian chief factor of {sn} _{2k}(2i-2) , and we have

    \operatorname{Out}\left(sn_{2k}(2i-1)/ sn_{2k}(2i)\right) = \operatorname{Out}(T)\wr S_m

    where T is a non-abelian composition factor of G , and {sn} _{2k}(2i-1)/ {sn}_{2k}(2i) is a direct product of m copies of T . By the proof of Lemma 3.4, we know p\, \nmid\, |S_m| . Hence, p\, \big|\, |\operatorname{Out}(T)| . However, this is impossible by (2). Therefore, statement (3) is proved.

    Now we can present a proof of the main theorem based on the above lemmas.

    Proof of Main Theorem. Obviously, it is enough to prove the sufficiency. We first prove the sufficiency when p = 5 . If p = 5 , then |G| = |PSp_4(5)| = 2^6 \cdot 3^2 \cdot 5^4 \cdot 13 and \Gamma(G) = \Gamma(PSp_4(5)) . Hence,

    V(G) = \{|G_r\, |\, r\in\pi(G)\} = \{2^6 , 3^2 , 5^4 , 13\}, \; \; \; \; E(G) = \{2^6\sim 3^2\}.

    By Lemma 4.1, G is non-solvable and every minimal subnormal subgroup of G is a non-abelian simple group. Let N be a minimal normal subgroup of G . Then, N is a direct product of isomorphic non-abelian simple groups. It is enough to prove N\cong PSp_4(5) . By |N|\, \big|\, |G| , we know that N is isomorphic to one of A_5 , A_5\times A_5 , A_6 , PSL_3(3) , PSL_2(25) , PSU_3(4) , PSp_4(5) by [14]. If N is isomorphic to one of A_5 , A_5\times A_5 , A_6 , PSL_3(3) , PSU_3(4) , then |N|_2\sim|N|_3 is not an edge of \Gamma(N) by checking their character table in [14]. However, there exists an edge |N|_2\sim|N|_3 in \Gamma(N) by Lemma 4.1(5). This is a contradiction. If N\cong PSL_2(25) , then

    G\Big/ \big(C_G(N)\times N \big) \lesssim \operatorname{Out}(N).

    By |N| = |PSL_2(25)| = 2^3\cdot3\cdot5^2\cdot13 and \operatorname{Out}(PSL_2(25))\cong C_2\times C_2 , we have |C_G(N)| = 2^\alpha \cdot3\cdot5^2 , where \alpha = 1 , 2 , or 3 . Notice that C_G(N)\vartriangleleft G , and we know there exists a minimal subnormal subgroup of G , denoted by T , such that T\leqslant C_G(N) . Hence, T is isomorphic to A_5 . By Lemma 4.1(5), there exists an edge |T|_2\sim|T|_3 in \Gamma(T) , a contradiction. Therefore, N\cong PSp_4(5) , moreover, G\cong PSp_4(5) .

    When p \geqslant 7 , we will discuss all non-abelian composition factors of G whose order can be divisible by p , covering all possible cases. Our proof for this purpose will consist of 5 steps. Now, assume that T is a non-abelian composition factor of G satisfying p\, \big|\, |T| .

    Step 1. Give information about the degree-power graph \Gamma(T) .

    By Lemma 4.1(3) and Lemma 3.4, there exists a subnormal subgroup of G which is isomorphic to T . Let r\in\pi(T)\setminus\{p, 3\} . Then there is an edge |T|_3\sim|T|_r in \Gamma(T) , by Lemma 4.1(5) and Proposition 2.1.

    Step 2. Prove T is not isomorphic to an alternating group A_n .

    On the contrary, T is isomorphic to an alternating group A_n . By p\, \big|\, |T| , we know p\leqslant n , moreover, \frac{p!}{2}\, \big|\, |T| . Notice that |T|\, \big|\, |G| , and we see that

    p!\, \Big|\, p^4(p^2-1)(p^4-1).

    By \left(\frac{p+1}{e} \right)^p < p! ( e is the natural constant), the following inequality holds:

    \left(\frac{p+1}{e} \right)^p < p^4(p^2-1)(p^4-1).

    Using Maple, it implies p\leqslant13 . If p = 13 , then p! = 2^{10}\cdot3^5\cdot 5^2\cdot7\cdot11\cdot13 , p^4(p^2-1)(p^4-1) = 2^7\cdot3^2\cdot5\cdot7^2\cdot13^4\cdot17 . This contradicts p!\, \big|\, p^4(p^2-1)(p^4-1) . If p = 11 , then p! = 2^{8}\cdot3^4\cdot 5^2\cdot7\cdot11 , p^4(p^2-1)(p^4-1) = 2^7\cdot3^2\cdot5^2\cdot11^4\cdot61 , a contradiction. If p = 7 , then |G| = 2^8 \cdot 3^2 \cdot 5^4 \cdot 7^4 . Therefore, T only is isomorphic to A_7 or A_8 by |T|\, \big|\, |G| . Observing the character tables of A_7 and A_8 in [14], there is not an edge |T|_2\sim |T|_3 of \Gamma(T) . However, |T|_2\sim |T|_3 should be an edge of \Gamma(T) by {Step 1.} This is a contradiction.

    Step 3. Prove T is not isomorphic to any sporadic simple group.

    We assume that T is isomorphic to one of following simple groups:

    \begin{gather*} M_{11}, \; M_{12}, \; M_{22}, \; M_{23}, \; M_{24}, \; J_1, \; J_2, \; J_3, \; Co_1, \; Co_2 , \; Co_3, \; Fi_{22}, \; Fi_{23}, \; Fi_{24}^{\prime}, \; \\ Suz , \; HS , \; McL, \; He , \; HN, \; Th, \; B , \; M , \; Ru. \end{gather*}

    Observing the degree prime-power graphs of the above 23 sporadic simple groups in [11], there is not an edge |T|_2\sim|T|_3 in \Gamma(T) . This is a contradiction, by Step 1. Next, we analyze the remaining three sporadic simple groups: O^{\prime}N , Ly , J_4 .

    If T is isomorphic to O^{\prime}N , then \pi(T) = \{2, 3, 5, 7, 11, 19, 31\} . By Step 1, for every r\in\pi(T)\setminus\{p, 3\} , there is an edge |T|_3\sim|T|_r in \Gamma(T) . This indicates that there are at least 5 edges at vertex |T|_3 . However, by the character table of O^{\prime}N , we know there are only 3 edges at vertex |T|_3 , i.e., |T|_3\sim|T|_2 , |T|_3\sim|T|_5 and |T|_3\sim|T|_{31} . This is a contradiction. Using the same method, one can obtain a contradiction when T is isomorphic to Ly .

    If T\cong J_4 , then |T| = 2^{21} \cdot 3^3 \cdot 5 \cdot 7 \cdot 11^3 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 43 . Since p\, \Big|\, |T| , we obtain p = 7 , 11 , 23 , 29 , 31 , 37 or 43 . Through calculations using the Magma, we have |PSp_4(p)|_2\Big| 2^{13} when p\in \{7, 11, 23, 29, 31, 37, 43\} . This contradicts |T|\, \Big|\, |PSp_4(p)| .

    Therefore, T is not isomorphic to any sporadic simple group.

    Step 4. Prove T is isomorphic to one of A_1(p^3)( = PSL_2(p^3)) , A_2(p)( = PSL_3(p)) , {}^2A_2(p)( = PSU_3(p)) , and B_2(p)( = PSp_4(p)) .

    By Steps 2 and 3, T is a simple group of Lie type over some GF(q) where q = r^f . Hence, by Lemma 4.4(1), we know r = p or T\cong {}^2B_2(8) .

    Assume T\ncong {}^2B_2(8) , then r = p\geqslant7 . Since |G|_p = p^4 , we can obtain that T is isomorphic to one of

    \begin{aligned} A_1(p^i)( = PSL_2(p^i)), \; \; i = 1, 2, 3, 4; & \; \; A_2(p)( = PSL_3(p)), \\ B_2(p)( = PSp_4(p)), \; \; & {}^2A_2(p)( = PSU_3(p)). \end{aligned}

    Notice that

    \begin{aligned} &|A_1(p)| = \frac{1}{2}p(p^2-1), \; \; |A_1(p^2)| = \frac{1}{2}p^2(p^2-1)(p^2+1), \\ &|A_1(p^3)| = \frac{1}{2}p^3(p^6-1) = \frac{1}{2}p^3(p-1)(p^2+p+1)(p^3+1), \\ &|A_1(p^4)| = \frac{1}{2}p^4(p^4-1)(p^4+1), \; \; |A_2(p)| = \frac{p^3(p^2-1)}{(3, p-1)}(p^3-1), \\ &|{}^2A_2(p)| = \frac{p^3(p^2-1)}{(3, p-1)}(p^3+1). \\ \end{aligned}

    By [18], the degrees of A_1(p) and A_1(p^2) are

    \begin{aligned} \operatorname{cd}(A_1(p))& = \{1, p, \frac{p+\varepsilon}{2}, p-1, p+1\}, \quad ( \varepsilon = (-1)^{(p-1)/2}), \\ \operatorname{cd}(A_1(p^2))& = \{1, p^2, \frac{p^2+1}{2}, p^2-1, p^2+1\} \end{aligned}

    respectively.

    If T\cong A_1(p^2) , then we can obtain that |T|_3\sim |T|_s is not an edge of \Gamma(T) for every s\in\pi(\frac{p^2+1}{2}) , because 3\, \big|\, (p^2-1) and (p^2-1, p^2+1) = 2 . By Step 1, this is a contradiction.

    If T\cong A_1(p) and \frac{p-1}{2} is an even number, then \epsilon = -1 and \operatorname{cd}(T) = \{1, p, \frac{p+1}{2}, p-1, p+1\} where \frac{p+1}{2} = \frac{p-1}{2}+1 is an odd number. Hence, for every s\in\pi(\frac{p+1}{2}) and t\in\pi(p-1) there is not an edge |T|_s\sim|T|_t in \Gamma(T) . Since s , t\in \pi\left((p-1)^2(p+1)^2\right) and PSp_4(p) has an irreducible character of degree (p-1)^2(p+1)^2 , then |G|_s\sim |G|_t is an edge of \Gamma(G) . By Step 1, T is isomorphic to a subnormal subgroup of G . Hence, |T|_s\sim|T|_t is an edge of \Gamma(T) by Lemma 4.1(5). So, it is a contradiction. Using the same method, there is a contradiction when \frac{p-1}{2} is an odd number.

    If T\cong A_1(p^4) , then |T| > |G| obviously. This is a contradiction.

    Assume T\cong{}^2B_2(8) , then p = 13 by Lemma 4.2. Hence, |T| = 2^6\cdot5\cdot7\cdot13 and |G| = |PSp_4(13)| = 13^4 \cdot \left(2^6\cdot 3^2\cdot 7^2\right)\cdot 5\cdot 17 . Observing |T|_2 = |G|_2 , we can conclude that G has only one non-abelian composition factor. By Lemma 3.2 and 4.1(2), there is a subnormal series

    sn^G_2: G = G_0 \geqslant G_1 > G_2 = 1

    such that G_1/G_2\cong T and G_0/G_1 \, \lesssim \, \operatorname{Out}\left(T\right) . This will lead to a contradiction. Since the outer automorphism group is C_3 , we obtain |G| = |G_1/G_2|\cdot|G_0/G_1|\leqslant |T|\cdot |C_3| = 2^6\cdot5\cdot7\cdot13\cdot 3 . That is impossible.

    Based on the above discussion, T is isomorphic to one of A_1(p^3)( = PSL_2(p^3)) , A_2(p)( = PSL_3(p)) , {}^2A_2(p)( = PSU_3(p)) , and B_2(p)( = PSp_4(p)) .

    Step 5. Prove G is isomorphic to PSp_4(p) .

    Take a subnormal series sn_k^{G}

    sn_{2k}^{G}: G = G_0 \geqslant G_1 > G_2 \geqslant \cdots > G_{2 k-2}\geqslant G_{2 k-1} > G_{2 k} \geqslant 1, \quad(k \geqslant 1).

    By Lemmas 4.1(3) and 4.4(3), we have G_{2k} = 1 and p\nmid |G_{2i-2}/G_{2i-1}| where i = 1 , 2 , \cdots , k . It follows that

    \begin{equation} p^4\, \Bigg|\, \prod\limits_{i = 1}^{k} |G_{2i-1}/G_{2i}|. \end{equation} (4.6)

    Notice that every G_{2i-1}/G_{2i} is a product of isomorphic non-abelian simple groups, and we assume that the direct product factor of G_{2i-1}/G_{2i} is T_i for i = 1, 2, \cdots, k . Then there exists a T_{j_1} such that its order is divisible by p . By Step 4, T_{j_1} is isomorphic to one of PSL_2(p^3) , PSL_3(p) , PSU_3(p) , and PSp_4(p) . We claim T_{j_1} is isomorphic to PSp_4(p) . Otherwise, T_{j_1} is isomorphic to PSL_2(p^3) , PSL_3(p) , or PSU_3(p) . Then |T_{j_1}|_p = p^3 , moreover G_{2j_1-1}/G_{2j_1}\cong T_{j_1} by |G|_p = p^4 . Hence, there is a T_{j_2} such that its order is divisible by p , and |T_{j_2}|_p can only be p^1 . This contradicts the result in {Step 4}. Therefore, G has a non-abelian composition factor that is isomorphic to PSp_4(p) . This means G\cong PSp_4(p) . This completes the proof of Main Theorem.

    We have primarily investigated whether the symplectic simple group PSp_4(p) can be uniquely determined by its order and degree prime-power graph. In the main theorem, we provide a positive answer to this question. In Lemma 3.3, a sufficient condition for adjusting the positions of specific non-abelian composition factors in the subnormal series sn_{2k} is provided. This condition plays a crucial role in the proof of the main theorem. We believe that this condition is valuable for studying the characterization of more non-abelian simple groups. Based on the results of this paper and [10,11,12,13], we conjecture that most non-abelian simple groups can be uniquely determined by their orders and degree prime-power graphs.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported by the Doctoral Through Train Scientific Research Project of Chongqing (No. sl202100000324), Guangxi Natural Science Foundation (No. 2022GXNSFBA035572), Scientific and Technological Research Program of Chongqing Municipal Education Commission (No. KJQN202300846) and National Natural Science Foundation of China (Nos. 12301021 and 12071376).

    The authors declare no conflicts of interest.



    [1] B. Huppert, Some simple groups which are determined by the set of their character degrees I, Illinois J. Math., 44 (2000), 828–842. http://dx.doi.org/10.1215/ijm/1255984694 doi: 10.1215/ijm/1255984694
    [2] H. Behravesh, M. Ghaffarzadeh, M. Ghasemi, S. Hekmatara, Recognition of Janko groups and some simple K_4-groups by the order and one irreducible character degree or character degree graph, Int. J. Group Theory, 10 (2021), 1–10. http://dx.doi.org/10.22108/IJGT.2019.113029.1502 doi: 10.22108/IJGT.2019.113029.1502
    [3] S. Heydari, N. Ahanjideh, Characterization of some simple K_4-groups by some irreducible complex character degrees, Int. J. Group Theory, 5 (2016), 61–74. http://dx.doi.org/10.22108/IJGT.2016.8233 doi: 10.22108/IJGT.2016.8233
    [4] B. Khosravi, B. Khosravi, B. Khosravi, Z. Momen, A new characterization for the simple group PSL(2, p^2) by order and some character degrees, Czech. Math. J., 65 (2015), 271–280. http://dx.doi.org/10.1007/s10587-015-0173-6 doi: 10.1007/s10587-015-0173-6
    [5] H. Xu, G. Chen, Y. Yan, A new characterization of simple K_3-groups by their orders and large degrees of their irreducible characters, Commun. Algebra, 42 (2014), 5374–5380. http://dx.doi.org/10.1080/00927872.2013.842242 doi: 10.1080/00927872.2013.842242
    [6] H. Xu, Y. Yan, G. Chen, A new characterization of Mathieu-groups by the order and one irreducible character degree, J. Inequal. Appl., 2013 (2013), 209. http://dx.doi.org/10.1186/1029-242X-2013-209 doi: 10.1186/1029-242X-2013-209
    [7] M. Lewis, An overview of graphs associated with character degrees and conjugacy class sizes in finite groups, Rocky Mountain J. Math., 38 (2008), 175–211. http://dx.doi.org/10.1216/RMJ-2008-38-1-175 doi: 10.1216/RMJ-2008-38-1-175
    [8] B. Khosravi, B. Khosravi, B. Khosravi, Z. Momen, Recognition of some simple groups by character degree graph and order, Math. Reports, 18 (2016), 51–61.
    [9] S. Heydari, N. Ahanjideh, Some simple groups which are determined by their character degree graphs, Sib. Elektron. Math. Re., 13 (2016), 1290–1299. http://dx.doi.org/10.17377/semi.2016.13.101 doi: 10.17377/semi.2016.13.101
    [10] C. Qin, Y. Yan, K. Shum, G. Chen, Mathieu groups and its degree prime-power graphs, Commun. Algebra, 47 (2019), 4173–4180. http://dx.doi.org/10.1080/00927872.2019.1579342 doi: 10.1080/00927872.2019.1579342
    [11] C. Qin, Research on the sporadic simple groups and their degree prime-power graphs of irreducible character degrees, Ph.D thesis, Southwest University, 2019.
    [12] Z. Wang, C. Qin, H. Lv, Y. Yan, G. Chen, A new characterization of L_2(p^2), Open Math., 18 (2020), 907–915. http://dx.doi.org/10.1515/math-2020-0048 doi: 10.1515/math-2020-0048
    [13] Z. Wang, G. Chen, A new characterization of L_2(p^3), Commun. Algebra, 50 (2022), 4000–4008. http://dx.doi.org/10.1080/00927872.2022.2057509 doi: 10.1080/00927872.2022.2057509
    [14] J. Conway, R. Curtis, S. Norton, R. Parker, R. Wilson, Atlas of finite groups, New York: Oxford University Press, 1985.
    [15] M. Shahabi, H. Mohtadifar, The characters of finite projective symplectic group PSp(4, q), In: Groups St Andrews 2001 in Oxford, Cambridge: Cambridge University Press, 2003,496–527. http://dx.doi.org/10.1017/CBO9780511542787.018
    [16] I. Isaacs, Character theory of finite groups, New York: Academic Press, 1976.
    [17] H. Tong-Viet, The simple Ree groups {}^2F_4(q^2) are determined by the set of their character degrees, J. Algebra, 339 (2011), 357–369, http://dx.doi.org/10.1016/j.jalgebra.2011.04.034 doi: 10.1016/j.jalgebra.2011.04.034
    [18] D. White, Character degrees of extensions of PSL_2(q) and SL_2(q), J. Group Theory, 16 (2013), 1–33. http://dx.doi.org/10.1515/jgt-2012-0026 doi: 10.1515/jgt-2012-0026
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