Proof. It suffices to prove the sufficiency. We have $|L_2(p)| = p(p-1)(p+1)/2$ and $\mathrm {cd}(H) = \{1, p, p-1, p+1, (p + \varepsilon )/2\}$ , where $\varepsilon = (-1)^{(p-1)/2}$ and $p \geq 5$ is a prime. Thus, $|G| = p(p-1)(p+1)/2$ and $m^c(G) = m^c(L_2(p)) = p(p-1)/2$ .

Note that p is the largest prime divisor of $|G|$ . Since $p < p(p-1)/2 = m^c(G)$ , it follows from Lemma 2.2 that $G = G'$ and, hence, G is nonsolvable. Therefore, G has a maximal normal subgroup M such that $G/M \cong S$ and $m^c(G) \leq m^c(S)$ , where S is a nonabelian simple group. The proof proceeds in three steps.

Step 1. We prove that S is not a sporadic group. Suppose, to the contrary, that S is a sporadic group.

Assume $p \mid |S|$ . Note that $|S| \mid |G| = \frac{1}{2}p(p-1)(p+1)$ and p is the largest prime divisor of $|S|$ . Thus, for any prime $r \mid |S|$ with $r < p$ , we have $r \mid (p-1)(p+1)$ . However, by Table 1, for every sporadic group except $M_{11}$ and $M_{12}$ , there exists a prime r such that $r \nmid (p-1)(p+1)$ . If $S = M_{11}$ ( $|M_{11}| = 2^4 \cdot 3^2 \cdot 5 \cdot 11$ ) or $M_{12}$ ( $|M_{12}| = 2^6 \cdot 3^3 \cdot 5 \cdot 11$ ), then $p = 11$ , so $|G| = 2^2 \cdot 3 \cdot 5 \cdot 11$ . Hence, $|G| < |S|$ , which is a contradiction.

Table 1

Sporadic groups except $M_{11}$ and $M_{12}$ .

Table 1 Long description

The table is organized into two identical sets of three columns: Group, p, and r.

Left side entries:

* J sub 1: p = 19, r = 11

* J sub 2: p = 7, r = 5

* J sub 3: p = 19, r = 17

* M c L: p = 11, r = 7

* R u: p = 29, r = 13

* O'N: p = 31, r = 19

* C o sub 2: p = 23, r = 7

* H N: p = 19, r = 11

* T h: p = 31, r = 19

* C o sub 1: p = 23, r = 13

* F i super prime sub 24: p = 29, r = 23

* M: p = 71, r = 59

Right side entries:

* M sub 22: p = 11, r = 7

* H S: p = 11, r = 7

* M sub 24: p = 23, r = 7

* H e: p = 17, r = 7

* S u z: p = 13, r = 11

* C o sub 3: p = 23, r = 7

* F i sub 22: p = 13, r = 11

* L y: p = 67, r = 37

* F i sub 23: p = 23, r = 17

* J sub 4: p = 43, r = 37

* B: p = 47, r = 31

* M sub 23: p = 23, r = 7

Now, assume $p \nmid |S|$ . In this case, $|S| \mid \tfrac 12(p-1)(p+1)$ . Then, $p \geq \sqrt {2|S| + 1}> \sqrt {2|S|}$ and, thus, $m^c(G) = \tfrac 12p(p-1)> |S| - \sqrt {2|S|}$ . By Lemma 2.5, $\sqrt [3]{|S|} < b(S)$ . Since ${|S|}/{b(S)}$ is a codegree, it follows that $m^c(S) \leq {|S|}/{b(S)} \leq |S|^{2/3}$ . Consider the function ${f(x) = x - \sqrt {2x} - x^{2/3}}$ for $x> 2^6$ . This function is strictly increasing and satisfies $f(2^6)> 0$ , so $f(|S|)> 0$ for all sporadic groups S. That is, $|S| - \sqrt {2|S|} - |S|^{2/3}> 0$ , which implies $m^c(G)> |S| - \sqrt {2|S|} > |S|^{2/3} \geq m^c(S)$ , again being a contradiction.

Therefore, S is not a sporadic group.

Step 2. When S is an alternating group, we prove that $S = A_5$ and $G = S$ .

Assume $S = A_n$ with $n = 7$ or $n \geq 9$ . Since $|S| \mid |L_2(p)|$ , it follows from Lemma 2.3 that $m^c(S) < m^c(L_2(p))$ . Therefore, $m^c(G) \leq m^c(S) < m^c(L_2(p))$ , which is a contradiction.

When $S = A_5$ , we have $m^c(S) = 12$ . Since $m^c(G) = \tfrac 12p(p-1) \leq m^c(S)$ , it follows that $p = 5$ and $|G| = 2^2 \cdot 3 \cdot 5$ . Comparing orders immediately yields $G = S = A_5$ .

When $S = A_6$ , we have $m^c(S) = 36$ . Since $m^c(G) = \tfrac 12p(p-1) \leq m^c(S)$ , we conclude that $p = 5$ or $p = 7$ . Hence, $|G| = 2^2 \cdot 3 \cdot 5$ or $2^3 \cdot 3 \cdot 7$ , both of which are less than $|S|$ , which is a contradiction.

When $S = A_8$ , we have $m^c(S) = 2^5 \cdot 3^2$ . Since $m^c(G) = \tfrac 12p(p-1) \leq m^c(S)$ , the possible values of p are $5, 7, 11, 13, 17, 19$ or $23$ . Checking each case individually, we find that $|S| \nmid |G|$ , again being a contradiction.

Therefore, $G = S = A_5$ when S is an alternating group.

Step 3. When S is a simple group of Lie type, we prove that $S \cong L_2(p)$ and $G = S$ .

Assume that S is a simple group of Lie type defined over a field of characteristic r. We divide the proof into two cases: $r = p$ and $r \neq p$ .

Case 1. Assume $r = p$ . Then, $|S| = p |S|_{p'}$ and $|S|_{p'} \mid |G|_{p'} = \tfrac 12(p+1)(p-1)$ . If $|S|_{p'} < \tfrac 12(p+1)(p-1)$ , then $|S|_{p'} \leq \tfrac 14(p+1)(p-1)$ . However, since $m^c(G) = \tfrac 12p(p-1) \leq m^c(S) \leq \mathrm {{cod}}(\mathrm {St}) = |S|_{p'} \leq \tfrac 14(p+1)(p-1)$ , where St denotes the Steinberg character, this leads to a contradiction. Therefore, $|S|_{p'} = \tfrac 12(p+1)(p-1)$ . It follows that $|S| = |G| = |L_2(p)|$ . By Lemma 2.6, we conclude that $S \cong L_2(p)$ and, hence, $G = L_2(p)$ .

Case 2. Assume $r \neq p$ . By [Reference Michler12, Reference Willems21], considering the degree of the Steinberg character $\mathrm {St}(1)_r$ of S, we have $\mathrm {St}(1)_r = |S|_r$ and $\mathrm {{cod}}(\mathrm {St})_r = |S|_{r'}$ . Thus, $m^c(G) \leq m^c(S) \leq |S|_{r'}$ . We proceed with the proof by distinguishing whether S is of type $A_1$ or not.

Case 2.1. Assume S is not of type $A_1$ . Since $|S| {\kern-1pt}\mid{\kern-1pt} \tfrac 12p(p{\kern-1pt}-{\kern-1pt}1)(p{\kern-1pt}+{\kern-1pt}1)$ and $(p-1, p+1) = 2$ , it follows that $|S|_r \mid p-1$ or $|S|_r \mid p+1$ .

First, assume $|S|_r \mid p-1$ . If $|S|_r < p-1$ , then $|S|_r \leq \tfrac 12(p-1)$ . By Lemma 2.4,

$$ \begin{align*} \tfrac{1}{2}p(p-1) = m^c(G)\leq m^c(S)\leq |S|_{r'} \leq |S|_r^2 \leq \tfrac{1}{4}(p-1)^2, \end{align*} $$

which implies that $-1 \leq p \leq 3$ , a contradiction. Therefore, $|S|_r = p-1$ . It follows that $|S|_{r'} \mid \tfrac 12p(p+1)$ .

If $|S|_{r'} = \tfrac 12p(p+1)$ , then $|S| = |L_2(p)|$ and, by Lemma 2.6, $S \cong L_2(p)$ , which is of type $A_1$ , contrary to the assumption for this case. If $|S|_{r'} < \tfrac 12p(p+1)$ , then ${|S|_{r'} \leq \tfrac 14p(p+1)}$ . Since $\tfrac 12p(p-1) = m^c(G) \leq |S|_{r'}$ , we deduce $ \tfrac 12p(p-1) \leq \tfrac 14p(p+1) $ and, hence, $0 \leq p \leq 3$ , again a contradiction.

Second, assume $|S|_r \mid p+1$ . If $|S|_r < p+1$ , then $|S|_r \leq \tfrac 12(p+1)$ . By Lemma 2.4,

$$ \begin{align*} \tfrac{1}{2}p(p-1) = m^c(G) \leq m^c(S) \leq |S|_{r'} \leq |S|_r^2 \leq \tfrac{1}{4}(p+1)^2, \end{align*} $$

which implies $2 - \sqrt {5} \leq p \leq 2 + \sqrt {5}$ , a contradiction. Therefore, ${|S|_r = p+1}$ . It follows that ${|S|_{r'} \mid \tfrac 12p(p-1)}$ . Since $\tfrac 12p(p-1) = m^c(G) \leq |S|_{r'}$ , we deduce $|S|_{r'} = \tfrac 12p(p-1)$ . Hence, $|S| = |L_2(p)|$ . By Lemma 2.6, $S \cong L_2(p)$ , which is of type $A_1$ , again contrary to the assumption in this case.

Case 2.2. Assume S is of type $A_1$ . Let $S = L_2(r^n)$ . If $r \neq 2$ , then we have ${m^c(S) = \tfrac 12r^n(r^n - 1)}$ . Since $m^c(G) \leq m^c(S)$ , it follows that $p \leq r^n$ . Hence, $|S| \geq |L_2(p)|$ and, therefore, $|S| = |L_2(p)|$ and $G = S = L_2(p)$ .

If $r = 2$ , then $|S| = 2^n(2^n - 1)(2^n + 1)$ and $m^c(S) = 2^n(2^n - 1)$ . Clearly, $2^n \mid p-1$ or $2^n \mid p+1$ .

When $2^n \mid p-1$ , we claim that $2^n = p-1$ . Otherwise, $2^n \leq \tfrac 12(p-1)$ and

$$ \begin{align*} m^c(S) \leq \tfrac{1}{2}(p-1)(\tfrac{1}{2}(p-1) - 1) < \tfrac{1}{2}p(p-1) = m^c(G), \end{align*} $$

which is a contradiction. Thus, $2^n = p-1$ . Since $|S| \mid \tfrac 12p(p-1)(p+1)$ , it follows that $|S|_{2'} \mid \tfrac 12p(p+1)$ . We claim that $|S|_{2'} = \tfrac 12p(p+1)$ . Otherwise, $|S|_{2'} \leq \tfrac 14p(p+1)$ and

$$ \begin{align*} m^c(G) = \tfrac{1}{2}p(p-1)\leq m^c(S) \leq |S|_{2'} \leq \tfrac{1}{4}p(p+1), \end{align*} $$

which implies $0 \leq p \leq 3$ , a contradiction. Therefore, $|S|_{2'} = \tfrac 12p(p+1)$ . Together with $2^n = p-1$ , we obtain $|G| = |S| = |L_2(p)|$ , which contradicts Lemma 2.6.

When $2^n \mid p+1$ , we claim that $2^n = p+1$ . Otherwise, $2^n \leq \tfrac 12(p+1)$ and

$$ \begin{align*} m^c(S) \leq \tfrac{1}{2}(p+1)(\tfrac{1}{2}(p+1) - 1) < \tfrac{1}{2}p(p-1) = m^c(G), \end{align*} $$

which is a contradiction. Since $2^n = p+1$ ,

$$ \begin{align*} |S| = 2^n(2^n - 1)(2^n + 1) = (p+1)p(p+2)> \tfrac{1}{2}p(p-1)(p+1) = |G|, \end{align*} $$

again a contradiction.

Therefore, $G = S = L_2(p)$ when S is a simple group of Lie type.

Proof. The lemma follows straightforwardly from [Reference Hung and Moretó5, Lemma 5.4(1), Proposition 8.1 and the second paragraph of the proof of Proposition 8.2].

Proof. Assume $S \cong A_n(q)$ . Since $q^{n(n+1)/2}\ \|\ |S| \mid |A_{39}| = 2^{34} \cdot 3^{18} \cdot 5^8 \cdot 7^5 \cdot 11^3 \cdot 13^3 \cdot 17^2 \cdot 19^2 \cdot 23 \cdot 29 \cdot 31 \cdot 37$ , we have $\tfrac 12n(n+1) \leq 34$ , so $n \leq 7$ . Therefore, if S is not a group listed in [Reference Conway, Curtis, Wilson, Norton and Parker3], then S must be one of the following simple groups of Lie type: $A_1(q)$ with $q> 125$ , $A_2(q)$ with $q> 31$ , $A_3(q)$ with $q> 11$ , $A_4(q)$ with $q> 11$ , $A_5(q)$ with $q> 5$ , $A_6(q)$ with $q> 3$ or $A_7(q)$ with $q> 2$ .

If $S = A_1(q)$ with $q> 125$ , given that $|S|$ divides $|A_{39}|$ , it follows that q must be one of $2^t$ for $7 \leq t \leq 34$ , $3^t$ for $5 \leq t \leq 18$ , $5^t$ for $4 \leq t \leq 8$ , $7^t$ for $3 \leq t \leq 5$ , $11^3$ , $13^t$ for $2 \leq t \leq 3$ , $17^2$ or $19^2$ .

For these values of q, checking the orders of $S = A_1(q)$ such that $|S| \mid |A_{39}|$ , we find that S must be one of the following: $A_1(11^3)$ (with order $2^2 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11^3 \cdot 19 \cdot 37$ ), $A_1(13^2)$ (with order $2^3 \cdot 3 \cdot 5 \cdot 7 \cdot 13^2 \cdot 17$ ) or $A_1(17^2)$ (with order $2^5 \cdot 3^2 \cdot 5 \cdot 17^2 \cdot 29$ ).

If $S = A_2(q)$ with $q> 31$ , then $q^3$ divides one of $2^{34}$ , $3^{18}$ , $5^8$ , $7^5$ , $11^3$ or $13^3$ . Hence, q must be one of $2^t$ for $5 \leq t \leq 11$ or $3^t$ for $4 \leq t \leq 6$ . By computing the orders of $S = A_2(q)$ for these cases, we conclude that $|S| \nmid |A_{39}|$ , which is a contradiction.

Using the same approach, we obtain contradictions for $S = A_3(q)$ with $q> 11$ , $A_4(q)$ with $q> 11$ , $A_5(q)$ with $q> 5$ , $A_6(q)$ with $q> 3$ and $A_7(q)$ with $q> 2$ .

Now, assume $S \cong {}^2A_n(q)$ . From $q^{n(n+1)/2}\ \|\ |S| \mid |A_{39}|$ and the fact that the largest exponent of a prime power dividing $|A_{39}|$ is 34, it follows that $n \leq 7$ . Therefore, if S is not listed in [Reference Conway, Curtis, Wilson, Norton and Parker3], then S must be one of ${}^2A_2(q)$ with $q> 32$ , ${}^2A_3(q)$ with $q> 11$ , ${}^2A_4(q)$ with $q> 11$ , ${}^2A_5(q)$ with $q> 5$ or ${}^2A_6(q)$ with $q> 3$ .

If $S = {}^2A_2(q)$ with $q> 32$ , then $q^3$ divides one of $2^{34}$ , $3^{18}$ , $5^8$ , $7^5$ , $11^3$ or $13^3$ . Thus, q is either $2^t$ for $6 \leq t \leq 11$ or $3^t$ for $4 \leq t \leq 6$ . Computing the orders of $S = {}^2A_2(q)$ for these values yields $|S| \nmid |A_{39}|$ , which is a contradiction.

For $S = {}^2A_3(q)$ with $q> 11$ , ${}^2A_4(q)$ with $q> 11$ , ${}^2A_5(q)$ with $q> 5$ or ${}^2A_6(q)$ with $q> 3$ , the same method leads to a contradiction.

Finally, suppose S is any other simple group of Lie type not yet considered. By the same argument as above, all such S satisfying $|S| \mid |A_{39}|$ are contained in the list of [Reference Conway, Curtis, Wilson, Norton and Parker3].

Proof. Assume S is defined over a field of characteristic p. By Lemma 2.4, $m^c(S) < |S|_p^2 \leq |A_n|_p^2$ . When $n> 10$ , we have $m^c(S) < |A_n|_2^2 \leq 2^{2n-2}$ by Lemma 3.1. By Lemma 3.2, $m^c(A_n) \geq \sqrt {n!}/2$ . Therefore, the result holds if $\tfrac 12\sqrt {n!} \geq 2^{2n-2}$ . This inequality is verified using MATLAB and holds for all $n \geq 40$ . It remains to consider the case ${5 \leq n \leq 39}$ .

Through direct computation, we obtain Table 2, which lists all simple groups of Lie type whose orders divide $|A_n|$ , but do not divide $|A_{n-1}|$ for $5 \leq n \leq 39$ . Upon inspection, for each simple group of Lie type $S \neq L_3(4)$ such that $|S| \mid |A_n|$ and $|S| \nmid |A_{n-1}|$ , we find that $m^c(S) < m^c(A_n)$ . Thus, the conclusion follows from Lemma 3.2. For $S = L_3(4)$ , [Reference Conway, Curtis, Wilson, Norton and Parker3] shows that $3^2 \cdot 5 \cdot 7 = m^c(S) < m^c(A_9) = 2^3 \cdot 3 \cdot 5 \cdot 7$ . In conclusion, by Lemma 3.2, $m^c(S) < m^c(A_n)$ for all $n \geq 9$ .

Table 2

Simple groups of Lie type S satisfying $\vert S\vert \mid \vert A_n\vert $ , but $\vert S\vert \nmid \vert A_{n-1}\vert $ .

Table 2 Long description

The table is organized into four columns, effectively presenting two sets of A sub n and S pairings side-by-side.

Row 1: A sub 5 corresponds to L sub 2 (5) (A sub 5); A sub 23 corresponds to L sub 2 (23).

Row 2: A sub 6 corresponds to L sub 2 (9) (A sub 6); A sub 24 corresponds to L sub 5 (3) O super minus sub 10 (2).

Row 3: A sub 7 corresponds to L sub 2 (7) L sub 2 (8); A sub 25 has no corresponding S value.

Row 4: A sub 8 corresponds to L sub 3 (4) L sub 4 (2) (A sub 8); A sub 26 corresponds to L sub 2 (13 super 2).

Row 5: A sub 9 corresponds to U sub 3 (3) U sub 4 (2); A sub 27 corresponds to O super plus sub 8 (3).

Row 6: A sub 10 has no corresponding S value; A sub 28 corresponds to S sub 4 (7) F sub 4 (2).

Row 7: A sub 11 corresponds to L sub 2 (11); A sub 29 corresponds to L sub 2 (29).

Row 8: A sub 12 corresponds to S sub 4 (2); A sub 30 has no corresponding S value.

Row 9: A sub 13 corresponds to L sub 2 (13) L sub 3 (3) L sub 2 (25) L sub 2 (27); A sub 31 corresponds to L sub 2 (31) L sub 2 (32) L sub 3 (5) L sub 2 (125).

Row 10: Additional S values for A sub 13 include Sz (8) U sub 3 (4) L sub 2 (64); Additional S values for A sub 31 include L sub 5 (2) G sub 2 (5) L sub 4 (5) L sub 6 (2).

Row 11: Further S values for A sub 31 include L sub 3 (25) O super plus sub 10 (2) L sub 5 (4) S sub 10 (2).

Row 12: A sub 14 corresponds to L sub 2 (49) U sub 5 (2); A sub 32 corresponds to O super plus sub 12 (2) O super minus sub 12 (2) L sub 6 (4).

Row 13: A sub 15 corresponds to U sub 3 (5) U sub 4 (3) G sub 2 (3) L sub 4 (3) L sub 3 (9); A sub 33 corresponds to L sub 3 (11) L sub 6 (3).

Row 14: A sub 16 corresponds to super 2 F sub 4 (2) prime O super plus sub 8 (2) super 3 D sub 4 (2) G sub 2 (4) S sub 4 (8); A sub 34 corresponds to O super plus sub 8 (4) L sub 2 (17 super 2).

Row 15: A sub 17 corresponds to L sub 2 (17) L sub 2 (16) S sub 4 (4) O super minus sub 8 (2); A sub 35 has no corresponding S value.

Row 16: Additional S values for A sub 17 include L sub 4 (4) U sub 4 (4) L sub 3 (16).

Row 17: A sub 18 corresponds to U sub 6 (2); A sub 36 has no corresponding S value.

Row 18: A sub 19 corresponds to L sub 2 (19) U sub 3 (8); A sub 37 corresponds to L sub 2 (37) U sub 3 (11) R (27) U sub 3 (27) L sub 2 (11 super 3).

Row 19: A sub 20 corresponds to S sub 4 (5) U sub 4 (5) S sub 8 (2); A sub 38 has no corresponding S value.

Row 20: A sub 21 corresponds to L sub 3 (7) S sub 6 (3) O sub 7 (3); A sub 39 has no corresponding S value.

Row 21: A sub 22 corresponds to S sub 6 (4) U sub 4 (8).

Proof. It suffices to prove the sufficiency.

We first show that G is simple. Assume, for a contradiction, that G is not simple and let M be a maximal normal subgroup of G. Then, $M \neq 1$ and $G/M \cong S$ , where S is a simple group. If S is nonabelian, it follows from $|S| \mid |A_n|$ and $|S| < |A_n|$ that $m^c(S) < m^c(A_n)$ by Lemmas 3.2 and 3.4. Since $m^c(G) \leq m^c(S)$ , we obtain ${m^c(G) < m^c(A_n)}$ , which is a contradiction. If S is abelian, then $m^c(S)$ is prime and

$$ \begin{align*} m^c(S) \leq \max\pi(A_n) \leq n \leq \frac{\sqrt{n!}}{2} < m^c(A_n), \end{align*} $$

again a contradiction. Therefore, G must be simple.

Since $|G| = |H|$ and G is simple, it follows from Lemma 2.6 that either $G \cong H$ or $G = L_3(4)$ and $H = A_8$ . If $G = L_3(4)$ and $H = A_8$ , then

$$ \begin{align*} m^c(A_8) = 2^5 \cdot 3^2 < 3^2 \cdot 5 \cdot 7 = m^c(G), \end{align*} $$

which is a contradiction. Hence, $G \cong H = A_n$ and the theorem follows.

Proof. Let $b = b(S)$ be the largest irreducible character degree of S. Note that $b^2 < |S|$ , so $b < |S|/b = m^c(S)$ . Moreover, by [Reference Conway, Curtis, Wilson, Norton and Parker3], $p < b(S)$ for the largest prime divisor p of $|S|$ . Therefore, $p < m^c(S)$ and the lemma follows.

Proof. It is sufficient to prove the sufficiency. First, it follows from Lemmas 4.2 and 2.2 that $G = G'$ and G is nonsolvable. We claim that G is simple. Suppose, for contradiction, that G is not simple. Then, there exists a maximal nontrivial normal subgroup M of G such that $G/M = S$ , where S is a nonabelian simple group. In the subsequent arguments, we shall repeatedly apply the fact that $|S|$ divides $|H|$ and that $|S| < |H|$ .

Case 1. Assume that H is a Mathieu group. By Lemma 4.1 and [Reference Conway, Curtis, Wilson, Norton and Parker3], if $H = M_{11}$ , then S is one of $A_5$ , $A_6$ or $L_2(11)$ . If $H = M_{12}$ , then S is one of

(4.1) $$ \begin{align} A_5,\ A_6,\ L_2(11)\text{ or } M_{11}. \end{align} $$

If $H = M_{22}$ , then S is one of

(4.2) $$ \begin{align} A_5,\ L_2(7),\ A_6,\ L_2(8),\ L_2(11),\ A_7,\ A_8,\ L_3(4)\text{ or } M_{11}. \end{align} $$

If $H = M_{23}$ , since $|M_{12}|$ divides $|M_{23}|$ , S is either one of the groups listed in (4.1), or one of $M_{22}$ or $L_2(23)$ . If $H = M_{24}$ , since $|M_{22}|$ divides $|M_{24}|$ , S is either one of the groups listed in (4.2), or one of $M_{22}$ , $L_2(23)$ or $M_{23}$ .

According to [Reference Conway, Curtis, Wilson, Norton and Parker3], we always have $m^c(S) < m^c(H)$ . Since $m^c(G) \leq m^c(S)$ , it follows that $m^c(G) < m^c(H)$ , which leads to a contradiction.

Case 2. Assume that H is a Janko group. By Lemma 4.1 and [Reference Conway, Curtis, Wilson, Norton and Parker3], if $H = J_1$ , then S is one of $A_5$ , $L_2(7)$ or $L_2(11)$ . If $H = J_2$ , then S is one of

(4.3) $$ \begin{align} A_5,\ L_2(7),\ A_6,\ L_2(8),\ A_7,\ U_3(3),\ A_8\text{ or } L_3(4). \end{align} $$

If $H = J_3$ , then S is one of

$$ \begin{align*} A_5,\ A_6,\ L_2(17),\ L_2(19),\ L_2(16)\text{ or } U_4(2). \end{align*} $$

If $H = J_4$ , then S is one of

$$ \begin{align*} \begin{split} A_5,\ &L_2(7),\ A_6,\ L_2(8),\ L_2(11),\ A_7,\ U_3(3),\ M_{11},\ L_2(23),\ L_2(29),\ L_2(31), \\ & A_8,\ L_3(4),\ L_2(32),\ L_2(43),\ M_{12},\ M_{22},\ L_5(2),\ M_{23}\text{ or } M_{24}. \end{split} \end{align*} $$

According to [Reference Conway, Curtis, Wilson, Norton and Parker3], we always have $m^c(S) < m^c(H)$ . Since $m^c(G) \leq m^c(S)$ , it follows that $m^c(G) < m^c(H)$ , which leads to a contradiction.

Case 3. Assume that H is one of the following sporadic simple groups: $HS$ , $M^cL$ , $\textit {Suz}$ , $Ly$ , $He$ , $Ru$ , $\textit {O'N}$ , $\textit {Th}$ or $\textit {HN}$ . By Lemma 4.1 and [Reference Conway, Curtis, Wilson, Norton and Parker3], if $H = HS$ , then S is one of

(4.4) $$ \begin{align} A_5,\ L_2(7),\ A_6,\ L_2(8),\ L_2(11),\ A_7,\ A_8,\ L_3(4),\ M_{11},\ U_3(5)\text{ or } M_{22}. \end{align} $$

If $H = M^cL$ , then S is one of

(4.5) $$ \begin{align} &A_5,\ L_2(7),\ A_6,\ L_2(8),\ L_2(11),\ A_7,\ U_3(3),\ A_8,\ L_3(4),\ M_{11}, \nonumber\\ &\quad U_4(2),\ M_{12}, \ U_3(5),\ A_9,\ M_{22},\ J_2,\ A_{10},\ U_4(3) \text{ or } A_{11}. \end{align} $$

If $H = \textit {Suz}$ , since $|M_{12}|$ , $|M_{22}|$ and $|J_2|$ divide $| \textit {Suz}|$ , then S is either one of the groups listed in (4.1), (4.2) and (4.3), or one of

(4.6) $$ \begin{align} &A_n\ (9 \leq n \leq 13),\ L_2(13),\ L_3(3),\ L_2(25),\ L_2(27),\ U_4(2),\ Sz(8),\ U_3(4),\ M_{12},\ L_2(64), \nonumber\\ &\quad M_{22},\ J_2,\ S_6(2),\ U_4(3),\ G_2(3),\ L_4(3),\ U_5(2),\ {}^2F_4(2)',\ L_3(9),\ O_8^+(2) \text{ or } G_2(4). \end{align} $$

If $H = Ly$ , since $|M^cL|$ divides $|Ly|$ , then S is either one of the groups listed in (4.5), or one of

$$ \begin{align*} L_2(31),\ L_2(32),\ L_3(5),\ L_2(125),\ M^cL\text{ or } G_2(5). \end{align*} $$

If $H = He$ , since $|J_2|$ divides $|He|$ , then S is either one of the groups listed in (4.3), or one of

(4.7) $$ \begin{align} L_2(17),\ L_2(16),\ L_2(49),\ J_2\text{ or } S_4(4). \end{align} $$

If $H = Ru$ , since $|J_2|$ divides $|Ru|$ , then S is either one of the groups listed in (4.3), or one of

(4.8) $$ \begin{align} & L_2(13),\ L_3(3),\ L_2(25),\ L_2(27),\ L_2(29),\ Sz(8), \nonumber\\ &\quad U_3(4), U_3(5),\ L_2(64),\ J_2,\ G_2(4)\ \text{ or} \ {}^2F_4(2)'. \end{align} $$

If $H = \textit {O'N}$ , since $|M_{22}|$ divides $|\textit {O'N}|$ , then S is either one of the groups listed in (4.2), or one of

(4.9) $$ \begin{align} &L_2(19),\ U_3(3),\ L_2(31),\ U_4(2),\ L_2(32),\ M_{12},\nonumber\\ &\quad J_1,\ A_9,\ M_{22},\ S_6(2),\ L_3(7)\text{ or } U_3(8). \end{align} $$

If $H = \textit {Th}$ , since $|J_2|$ divides $|\textit {Th}|$ , then S is either one of the groups listed in (4.3), or one of

(4.10) $$ \begin{align} & L_2(19),\ L_3(3),\ L_2(25),\ L_2(27),\ L_2(31),\ U_4(2),\ Sz(8),\ L_2(49),\nonumber\\ &\quad U_3(4),\ U_3(5),\ A_9,\ L_2(64),\ L_3(5),\ J_2,\ L_2(125),\ S_4(4),\ S_6(2),\nonumber\\ &\quad A_{10},\ U_4(3), G_2(3),\ U_3(8),\ L_4(3), L_5(2),\ {}^2F_4(2)',\ L_3(9),\ O_8^+(2),\nonumber\\ &\quad {}^3D_4(2),\ G_2(4),\ S_4(8),\ O_7(3),\ S_6(3)\text{ or } L_6(2). \end{align} $$

If $H = HN$ , since $|M^cL|$ and $|HS|$ divide $|HN|$ , then S is either one of the groups listed in (4.1), (4.2), (4.3), (4.4) and (4.5), or one of

(4.11) $$ \begin{align} L_2(19),\ U_4(2),\ J_1,\ S_6(2),\ U_3(8),\ U_5(2),\ HS,\ O_8^+(2),\ A_{12}\text{ or } M^cL. \end{align} $$

According to [Reference Conway, Curtis, Wilson, Norton and Parker3], we always have $m^c(S) < m^c(H)$ . Since $m^c(G) \leq m^c(S)$ , it follows that $m^c(G) < m^c(H)$ , which leads to a contradiction.

Case 4. Assume that H is a Conway group. By Lemma 4.1 and [Reference Conway, Curtis, Wilson, Norton and Parker3], if $H = Co_3$ , since $|M^cL|$ , $|M_{24}|$ and $|HS|$ divide $|Co_3|$ , then S is either one of the groups listed in (4.2), (4.3), (4.4) and (4.5), or one of

(4.12) $$ \begin{align} L_2(23),\ S_6(2),\ M_{23},\ U_5(2),\ HS,\ A_{12},\ M_{24}\text{ or } M^cL. \end{align} $$

If $H = Co_2$ , since $|M^cL|$ , $|M_{24}|$ and $|HS|$ divide $|Co_2|$ , then S is either one of the groups listed in (4.2), (4.3), (4.4), (4.5) and (4.12), or one of

(4.13) $$ \begin{align} O_8^+(2)\text{ or } U_6(2). \end{align} $$

If $H = Co_1$ , since $|Co_3|$ , $|\textit {Suz}|$ and $|Co_2|$ divide $|Co_1|$ , then S is either one of the groups listed in (4.1), (4.2), (4.3), (4.4), (4.5), (4.6), (4.12) and (4.13), or one of

(4.14) $$ \begin{align} & L_2(16),\ L_2(49),\ S_4(5),\ {}^3D_4(2),\ S_4(8),\ S_6(3),\ O_7(3),\ U_6(2), \nonumber\\ & \quad U_4(5),\ A_n\ (14 \leq n \leq 16),\ \textit{Suz},\ Co_3,\ Co_2\text{ or } Fi_{22}. \end{align} $$

According to [Reference Conway, Curtis, Wilson, Norton and Parker3], we always have $m^c(S) < m^c(H)$ . Since $m^c(G) \leq m^c(S)$ , it follows that $m^c(G) < m^c(H)$ , which leads to a contradiction.

Case 5. Assume that H is one of the sporadic simple groups $Fi_{22}$ , $Fi_{23}$ or $Fi^{\prime }_{24}$ . By Lemma 4.1 and [Reference Conway, Curtis, Wilson, Norton and Parker3], if $H = Fi_{22}$ , since $|\textit {Suz}|$ , $|M_{22}|$ and $|J_2|$ divide $|Fi_{22}|$ , then S is either one of the groups listed in (4.1), (4.2), (4.3) and (4.6), or one of

(4.15) $$ \begin{align} S_6(3),\ O_7(3),\ U_6(2)\text{ or } \textit{Suz}. \end{align} $$

If $H = Fi_{23}$ , since $|M_{24}|$ , $|Fi_{22}|$ and $|J_2|$ divide $|Fi_{23}|$ , then S is either one of the groups listed in (4.1), (4.2), (4.3), (4.6) and (4.15), or one of

(4.16) $$ \begin{align} & L_2(17),\ L_2(16),\ L_2(23),\ S_4(4),\ M_{23},\ O_8^-(2), \nonumber \\ & \quad M_{24},\ L_4(4),\ L_3(16),\ S_8(2),\ O_8^+(3)\text{ or } Fi_{22}. \end{align} $$

If $H = Fi^{\prime }_{24}$ , since $|Fi_{23}|$ and $|He|$ divide $|Fi^{\prime }_{24}|$ , then S is either one of the groups listed in (4.1), (4.2), (4.3), (4.6), (4.7), (4.15) and (4.16), or one of

(4.17) $$ \begin{align} L_2(49),\ {}^3D_4(2),\ He,\ S_4(8),\ A_{14},\ O_{10}^-(2)\text{ or } Fi_{23}. \end{align} $$

According to [Reference Conway, Curtis, Wilson, Norton and Parker3], we always have $m^c(S) < m^c(H)$ . Since $m^c(G) \leq m^c(S)$ , it follows that $m^c(G) < m^c(H)$ , which leads to a contradiction.

Case 6. Assume that H is either the Monster group M or the Baby Monster group B. By Lemma 4.1 and [Reference Conway, Curtis, Wilson, Norton and Parker3], if $H = B$ , since $|M^cL|$ , $|M_{24}|$ , $|Fi_{23}|$ and $|Co_1|$ divide $|B|$ , then S is either one of the groups listed in (4.1), (4.2), (4.3), (4.4), (4.5), (4.6), (4.10), (4.11), (4.12), (4.13), (4.14), (4.15) and (4.16), or one of

$$ \begin{align*} \begin{aligned} &L_2(32),\ L_2(47),\ J_3,\ U_4(4),\ G_2(5),\ L_4(5),\ L_3(25),\ S_6(4),\ O_{10}^+(2), \\ &\quad O_{10}^-(2),\ U_4(8),\ A_n\ (17 \leq n \leq 20),\ L_5(4),\ HN,\ F_4(2),\ S_{10}(2),\ \textit{Th}, \\ &\quad Fi_{23},\ Co_1,\ O_{12}^+(2),\ O_{12}^-(2),\ L_6(4),\ {}^2E_6(2)\text{ or } S_{12}(2). \end{aligned} \end{align*} $$

If $H = M$ , since $|B|$ , $|Fi^{\prime }_{24}|$ , $|\textit {O'N}|$ and $|Ru|$ divide $|M|$ , then S is either one of the groups listed in (4.1), (4.2), (4.3), (4.4), (4.5), (4.6), (4.7), (4.8), (4.10), (4.11), (4.12), (4.13), (4.14), (4.15), (4.16) and (4.17), or one of

$$ \begin{align*} \begin{aligned} & A_n\ (21 \leq n \leq 32),\ L_2(2^{10}),\ L_2(13^2),\ L_2(71),\ L_2(81),\ \textit{Sz}(32), \\& \quad \ S_4(7),\ S_4(9),\ Ru, \ L_5(3),\ O\text{'}N,\ L_4(7),\ O_8^-(3),\ L_4(9),\ U_5(4), \\& \quad O_7(5),\ S_6(5),\ L_6(3),\ O_9(3),\ S_8(3),\ U_6(4)\text{ or } O_{10}^+(3). \end{aligned} \end{align*} $$

According to [Reference Conway, Curtis, Wilson, Norton and Parker3], we always have $m^c(S) < m^c(H)$ . Since $m^c(G) \leq m^c(S)$ , it follows that $m^c(G) < m^c(H)$ , which leads to a contradiction.

In summary, G must be simple. By Lemma 2.6, it follows that $G \cong H$ and, hence, the theorem holds.