A new characterization of Mathieu-groups by the order and one irreducible character degree

Abstract

The main aim of this article is to characterize the finite simple groups by less character quantity. In fact, we show that each Mathieu-group G can be determined by their largest and second largest irreducible character degrees.

MSC:20C15.

1 Introduction and preliminary results

Classifying finite groups by the properties of their characters is an interesting problem in group theory. In 2000, Huppert conjectured that each finite non-abelian simple group G is characterized by the set $cd(G)$ of degrees of its complex irreducible characters. In [1–4], it was shown that many non-abelian simple groups such as $L_2(q)$ and $S_z(q)$ satisfy the conjecture. In this paper, we manage to characterize the finite simple groups by less character quantity. Let G be a finite group; $L(G)$ denotes the largest irreducible character degree of G and $S(G)$ denotes the second largest irreducible character degree of G. We characterize the five Mathieu groups G by the order of G and its largest and second largest irreducible character degrees. Our main results are the following theorems.

Theorem A Let G be a finite group and let M be one of the following Mathieu groups: $M_{11}$, $M_{12}$ and $M_{23}$. Then $G\cong M$ if and only if the following conditions are fulfilled:

• (1) $|G|=|M|$;

• (2) $L(G)=L(M)$.

Theorem B Let G be a finite group. Then $G\cong M_{24}$ if and only if $|G|=|M_{24}|$ and $S(G)=S(M_{24})$.

Theorem C Let G be a finite group. If $|G|=|M_{22}|$ and $L(G)=L(M_{22})$, then either G is isomorphic to $M_{22}$ or $H\times M_{11}$, where H is a Frobenius group with an elementary kernel of order 8 and a cyclic complement of order 7.

We need the following lemmas.

Lemma 1 Let G be a non-solvable group. Then G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic non-abelian simple groups and $|G/K|\mid |Out(K/H)|$.

Proof Let G be a non-solvable group. Then G has a chief factor $M/N$ such that $M/N$ is a direct product of isomorphic non-abelian simple groups. Hence $C_{G/N}(M/N)\cap M/N=Z(M/N)=1$, and so

$$M/N\cong \frac{C_{G/N}(M/N)\times M/N}{C_{G/N}(M/N)}\le \frac{G/N}{C_{G/N}(M/N)}\lesssim Aut(M/N).$$

Let $K/N=C_{G/N}(M/N)\times M/N$ and $H/N=C_{G/N}(M/N)$. Then $G/K\le Out(M/N)$ and $K/H\cong M/N$ is a direct product of isomorphic non-abelian simple groups. Thus $1⊴H⊴K⊴G$ is a normal series, as desired. □

Lemma 2 Let G be a finite solvable group of order $p_1^{a_1}{p}_2^{a_2}\cdots p_n^{a_n}$, where $p_1,p_2,\dots ,p_n$ are distinct primes. If $k{p}_n+1\nmid p_i^{a_i}$ for each $i\le n-1$ and $k>0$, then the Sylow $p_n$-subgroup is normal in G.

Proof Let N be a minimal normal subgroup of G. Then $|N|=p^m$ for G is solvable. If $p=p_n$, by induction on $G/N$, we see that normality of the Sylow $p_n$-subgroup in G. Now suppose that $p=p_i$ for some $i<n$. Now consider $G/N$. By induction, the Sylow $p_n$-subgroup $P/N$ of $G/N$ is normal in $G/N$. Thus $P⊴G$. Let Q be a Sylow $p_n$-subgroup of P. Then $P=NQ$. By Sylow’s theorem, $|P:N_P(Q)|=p_i^l$ ($l\le m\le a_i$) and $p_n\mid p_i^l-1$. But this means that $k{p}_n+1\mid p^{a_i}$, and then $k=0$ by assumption. Hence $Q⊴P$ and $Q⊴G$. □

2 Proof of theorems

Proof of Theorem A We only need to prove the sufficiency. We divide the proof into three cases.

Case 1.1 $M=M_{11}$

In this case, we have $|G|=2^4\cdot 3^2\cdot 5\cdot 11$ and $L(G)=55$. We first show that G is non-solvable. Assume the contrary. By Lemma 2, we know that the Sylow 11-subgroup of G is normal in G. Let N be the 11-Sylow subgroup of G. Since N is abelian, we have $\chi (1)\mid |G/N|$ for all $\chi \in Irr(G)$. But $L(G)=55$ and $55\nmid |G/N|$, a contradiction. Therefore, G is non-solvable.

Since G is non-solvable, by Lemma 1, we get that G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic non-abelian simple groups and $|G/K|\mid |Out(K/H)|$. As $|G|=2^4\cdot 3^2\cdot 5\cdot 11$, we have $K/H\cong A_5,A_6,L_2(11)\text{ or }{M}_{11}$.

We first assume that $K/H\cong A_5$. Since $|Out(A_5)|=2$, we have $|G/K|\mid 2$ and $|H|=2^t\cdot 3\cdot 11$, where $t=1$ or 2. Let $\chi \in Irr(G)$ such that $\chi (1)=L(G)=55$ and $\theta \in Irr(H)$ such that $[{\chi }_H,\theta ]\ne 0$. Then $\theta (1)=11$ by the Clifford theorem (see Theorem 6.2 in [5]). On the other hand, since $|H|=2^t\cdot 3\cdot 11$, we have H is solvable. Let N be a Sylow 11-subgroup of H. Then $N⊴H$ by Lemma 2. Hence $\theta (1)\mid |H/N|=2^t\cdot 3$, a contradiction.

By the same reason as above, one has that $K/H\ncong A_6$.

Suppose that $K/H\cong L_2(11)$. Since $|Out(L_2(11))|=2$, we have $|G/K|\mid 2$ and so $|H|=2^a\cdot 3$, where $a=1$ or 2. Let $\theta \in Irr(H)$ such that $e=[{\chi }_H,\theta ]\ne 0$ and let $t=|G:I_G(\theta )|$. Then $\theta (1)=1$ and $et=\chi (1)/\theta (1)=55$. Since $|H|=2^t\cdot 3$, where $a=1$ or 2, we have that $55\nmid |Aut(H/H^{\mathrm{\prime }})|$. Hence $t=1$, $e=55$. But ${(55)}^2=e^{2}t=[{\chi }_H,{\chi }_H]>|G:H|=2^b\cdot 3\cdot 5\cdot 11$, where $2\le b\le 3$, a contradiction.

If $K/H\cong M_{11}$, by comparing the orders of G and $M_{11}$, we have $|H|=1$. Therefore $G=K\cong M_{11}$.

Case 1.2 $M=M_{23}$

In this case, we have $|G|=2^7\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 23$ and $L(G)=2^3\cdot 11\cdot 23$. Then $O_{23}(G)=1$. If not, then $|O_{23}(G)|=23$ and $O_{23}(G)$ is abelian. Hence $L(G)=2^3\cdot 11\cdot 23\mid |G/O_{23}(G)|$, a contradiction.

If G is solvable, then the Sylow 23-subgroup of G is normal in G by Lemma 2, which leads to a contradiction as above. Therefore G is non-solvable.

Since G is non-solvable, by Lemma 1, we get that G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic non-abelian simple groups and $|G/K|\mid |Out(K/H)|$. As $|G|=2^7\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 23$, we have that $K/H$ can be isomorphic to one of the simple groups: $A_5$, $L_2(7)$, $A_6$, $L_2(8)$, $L_2(11)$, $A_7$, $M_{11}$, $L_3(4)$, $A_8$, $M_{22}$ and $M_{23}$.

We first assume that $K/H\cong A_5$. Since $|Out(A_5)|=2$, we have $|G/K|\mid 2$ and $|H|=2^m\cdot 3\cdot 7\cdot 11\cdot 23$, where $m=4$ or 5. Suppose that H is non-solvable. By Lemma 1, H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a direct product of isomorphic non-abelian simple groups and $|H/B|\mid |Out(B/A)|$. Since $|H|=2^m\cdot 3\cdot 7\cdot 11\cdot 23$, we have $B/A\cong L_2(7)$ and $|H/B|\mid 2$. Thus $|A|=2^a\cdot 11\cdot 23$, where $0\le a\le 2$. Let N be a Sylow 23-subgroup of A. Then $N⊴A$ by Lemma 2. Hence we get a subnormal series of G, $N\mathit{char}A⊴B⊴H⊴K⊴G$, which implies that $N⊴G$. But $O_{23}(G)=1$, a contradiction. If H is solvable, then the Sylow 23-subgroup of H is normal in H by Lemma 2, which leads to a contradiction as before.

By the same arguments as the proofs of $K/H\cong A_5$, we show that $K/H$ cannot be isomorphic to one of the simple groups: $A_6$, $L_2(7)$, $L_2(8)$, $L_2(11)$, $A_7$, $M_{11}$, $L_3(4)$, $A_8$ and $M_{22}$.

If $K/H\cong M_{23}$, since $|G|=2^7\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 23$, we have that $|H|=1$ and $G=K\cong M_{23}$.

Case 1.3 $M=M_{12}$

In this case, $|G|=2^6\cdot 3^3\cdot 5\cdot 11$ and $L(G)=2^4\cdot 11$. Since $11\mid L(G)$, by the same arguments as the proofs of Case 1.2, we have that $O_{11}(G)=1$.

We will show that G is non-solvable. If G is solvable, then the Sylow 11-subgroup of G is normal in G by Lemma 2, a contradiction. Therefore, G is non-solvable.

By Lemma 1, we get that G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic non-abelian simple groups and $|G/K|\mid |Out(K/H)|$. As $|G|=2^6\cdot 3^3\cdot 5\cdot 11$, we have $K/H\cong A_5,A_6,L_2(11),M_{11}\text{ or }{M}_{12}$.

By the same arguments as the proofs of Case 1.2, we can prove that $K/H$ cannot be isomorphic to $A_5$ or $A_6$.

Assume that $K/H\cong L_2(11)$. Since $|Out(L_2(11))|=2$, we have $|G/K|\mid 2$ and $|H|=2^a\cdot 3$, where $a=3$ or 4. Suppose that $|G/K|=1$. Then $|H|=2^4\cdot 3^2$. Let $\chi \in Irr(G)$ such that $\chi (1)=L(G)=2^4\cdot 11$ and $\theta \in Irr(H)$ such that $e=[{\chi }_H,\theta ]\ne 0$. Then $\chi (1)=et\theta (1)=176$, where $t=|G:I_G(\theta )|$. Since $\chi (1)/\theta (1)\mid |G/H|$, we have that $\theta (1)=4$ or 8. If $\theta (1)=4$, then $et=44$. Since $|H|=2^4\cdot 3^2$, we have that H has at most eight irreducible characters of degree 4. Hence $t\le 4$. We assert that $I_G(\theta )=G$. If not, then $I_G(\theta )<G$. Let U containing $I_G(\theta )$ be a maximal subgroup of G. Then $1\le |G:U|\mid |G:I_G(\theta )|=4$. By checking the maximal subgroups of $L_2(11)$ (see ATLAS table in [6]), it is easy to get a contradiction. Hence $I_G(\theta )=G$, and so $t=1$ and $e=44$. But $e^2\cdot t=[{\chi }_H,{\chi }_H]>|G:H|$, a contradiction. If $\theta (1)=8$, then $|O_3(H)|=9$ and $I_G(\theta )=G$. Since $H⊴G$, we have that $O_3(H)⊴G$. Let $\lambda \in Irr(O_3(H))$ such that $[{\theta }_{O_3(H)},\lambda ]\ne 0$. Since $\theta (1)=8$, we have $4\le |H:I_H(\lambda )|\le 8$. But $I_G(\theta )=G$, which implies that $4\le |G:I_G(\lambda )|=|H:I_H(\lambda )|\le 8$. Let $S={\bigcap }_{g\in G}{I}_G{(\lambda )}^g$. Then $S⊴G$ and $G/S\lesssim S_8$. By the Jordan-Hölder theorem, S has a normal series $1⊴O_3(H)⊴C⊴D⊴S$ such that $D/C\cong L_2(11)$ and $|C/O_3(H)|=1,2\text{ or }4$. Let $\alpha \in Irr(S)$ such that $[{\chi }_S,\alpha ]\ne 0$. Since $\chi (1)/\alpha (1)\mid |G/S|$, we have that $22\mid \alpha (1)$. Since ${\lambda }^g$ is invariant in S, for each $g\in G$ and $4\le |G:I_G(\lambda )|$, we have that each irreducible character is invariant in S and $O_3(H)\le Z(S)$. Therefore, the following conclusions hold:

• (a) $S\cong L_2(11)\times O_3(H)$ if $|C/O_3(H)|=1$;

• (b) $S\cong (2\cdot L_2(11))\times O_3(H)$ or $(Z_2\times L_2(11))\times O_3(H)$ if $|C/O_3(H)|=2$.

By checking the character table of $2\cdot L_2(11)$ and $L_2(11)$, we see that both conclusions (a) and (b) are not satisfied with the above conditions. Now, we suppose that $|C/O_3(H)|=4$. Then $44\mid \alpha (1)$. Since $O_3(H)\le Z(S)$, one has that $C\cong O_3(H)\times B$, where B is a group of order 4. Let β be an irreducible component of ${\alpha }_C$ and $t_1=|S:I_S(\beta )|$. Then $\beta (1)=1$ and $t_1\mid \alpha (1)/\beta (1)\mid 44$. Since the indexes of the maximal subgroups of S containing $I_S(\beta )$ divide $t_1$ and $t_1\mid |Aut(C)|$, we have that $t_1=1$. Hence $[{\alpha }_C,{\alpha }_C]>|S:C|=2^2\cdot 3\cdot 5\cdot 11$, a contradiction.

Similarly, we can show that $|G/K|\ne 2$.

Suppose that $K/H\cong M_{11}$. Since $|Out(M_{11})|=Mult(M_{11})=1$, we have $G\cong H\times M_{11}$, where $|H|=2^2\cdot 3$. By checking the character table of $M_{11}$, we see that G has no irreducible character of degree $L(G)=2^4\cdot 11$, a contradiction.

If $K/H\cong M_{12}$, since $|G|=2^6\cdot 3^3\cdot 5\cdot 11$, we conclude that $|H|=1$ and $G=K\cong M_{12}$, which completes the proof of Theorem A. □

Proof of Theorem B We only need to prove the sufficiency.

In this case, we have $|G|=2^{10}\cdot 3^3\cdot 5\cdot 7\cdot 11\cdot 23$ and $S(G)=2^2\cdot 3^2\cdot 7\cdot 23$. Let $\chi \in Irr(G)$ such that $\chi (1)=S(G)$. If $O_{23}(G)\ne 1$, then $|O_{23}(G)|=23$, which implies that $\chi (1)\mid |G:N|$, a contradiction. Hence $O_{23}(G)=1$.

We have to show that G is non-solvable. Assume the contrary, by Lemma 2, we have that the Sylow 23-subgroup is normal in G, a contradiction. Therefore, G is non-solvable.

Since G is non-solvable, by Lemma 1, one has that G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic non-abelian simple groups and $|G/K|\mid |Out(K/H)|$. As $|G|=2^{10}\cdot 3^3\cdot 5\cdot 7\cdot 11\cdot 23$, then $K/H$ can be isomorphic to one of the following simple groups: $A_5$, $L_2(7)$, $A_6$, $L_2(8)$, $L_2(11)$, $A_7$, $U_3(3)$, $M_{11}$, $L_3(4)$, $A_8$, $M_{12}$, $M_{22}$, $M_{23}$ and $M_{24}$.

We first assume that $K/H\cong A_5$. Since $|Out(A_5)|=2$, we have $|G/K|\mid 2$ and $|H|=2^t\cdot 3^2\cdot 7\cdot 11\cdot 23$, where $t=7$ or 8. Let $\theta \in Irr(H)$ such that $[{\chi }_H,\theta ]\ne 0$. Since $\chi (1)/\theta (1)\mid |G/H|$, it implies that $23\mid \theta (1)$. If H is solvable, then $O_{23}(H)\ne 1$ by Lemma 2, which implies that $O_{23}(H)=O_{23}(G)\ne 1$, a contradiction. Thus H is non-solvable. Then there exists a normal series of H: $1⊴N⊴M⊴H$ such that $M/N$ is a direct product of isomorphic non-abelian simple groups and $|H/M|\mid |Out(M/N)|$. As $|H|=2^t\cdot 3^2\cdot 7\cdot 11\cdot 23$, we have $M/N\cong L_2(7)$ or $L_2(8)$, which implies that $23\mid |N|$. Hence $O_{23}(N)\ne 1$ by Lemma 2, which implies that $O_{23}(N)=O_{23}(G)\ne 1$, a contradiction.

By the same arguments as the proof of $K/H\cong A_5$, we show that $K/H$ cannot be isomorphic to one of the simple groups: $L_2(7)$, $A_6$, $L_2(8)$, $L_2(11)$, $A_7$, $U_3(3)$, $M_{11}$, $L_3(4)$, $A_8$, $M_{12}$ and $M_{22}$.

Suppose that $K/H\cong M_{23}$. Since $|Out(M_{23})|=Mult(M_{23})=1$, we have that $G\cong H\times M_{23}$, where $|H|=2^3\cdot 3$. By checking the character table of $M_{23}$, it is easy to see that there exists no irreducible character of degree $2^2\cdot 3^2\cdot 7\cdot 23$ in G, a contradiction.

If $K/H\cong M_{24}$, since $|G|=2^{10}\cdot 3^3\cdot 5\cdot 7\cdot 11\cdot 23$, one has that $|H|=1$ and $G=K\cong M_{24}$, which completes the proof of Theorem B. □

Proof of Theorem C We only need to prove the sufficiency.

In this case, $|G|=2^7\cdot 3^2\cdot 5\cdot 7\cdot 11$ and $L(G)=385$. Let $\chi \in Irr(G)$ such that $\chi (1)=L(G)=5\cdot 7\cdot 11$. We assert that $O_{11}(G)=1$. Otherwise, we have that $|O_{11}(G)|=11$ and $O_{11}(G)$ is abelian. Hence $\chi (1)\mid |G/O_{11}(G)|$, a contradiction. Similarly, $O_5(G)=O_7(G)=1$.

If G is solvable, then the Sylow 11-subgroup of G is normal in G by Lemma 2. But $O_{11}(G)=1$, a contradiction. Therefore, G is non-solvable.

Since G is non-solvable, by Lemma 1, we get that G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic non-abelian simple groups and $|G/K|\mid |Out(K/H)|$. As $|G|=2^7\cdot 3^2\cdot 5\cdot 7\cdot 11$, we see that $K/H$ is isomorphic to one of the simple groups: $A_5$, $L_2(7)$, $A_6$, $L_2(8)$, $L_2(11)$, $A_7$, $M_{11}$, $L_3(4)$, $A_8$ and $M_{22}$.

We first assume that $K/H\cong A_5$. Since $|Out(A_5)|=2$, we have $|G/K|\mid 2$ and $|H|=2^t\cdot 3\cdot 7\cdot 11$, where $t=4$ or 5. If H is solvable, then $O_{11}(H)=O_{11}(G)\ne 1$ by Lemma 2, a contradiction. Hence H is non-solvable and H has a normal series $1⊴N⊴M⊴H$ such that $M/N$ is a direct product of isomorphic non-abelian simple groups and $|H/M|\mid |Out(M/N)|$. As $|H|=2^t\cdot 3\cdot 7\cdot 11$, one has that $M/N\cong L_3(2)$ and $|N|=2^s\cdot 11$, where $0\le s\le 2$. Let P be the Sylow 11-subgroup of N. Then P is normal in N by Sylow theorem. Since P is also a Sylow 11-subgroup in G and N is subnormal in G, we have $P⊴G$, a contradiction.

Similarly, $K/H$ cannot be isomorphic to the simple groups: $L_2(7)$, $A_6$, $L_2(8)$, $L_2(11)$, $A_7$, $L_3(4)$ or $A_8$.

Assume that $K/H\cong L_2(11)$. Since $|Out(L_2(11))|=2$, we have $|G/K|\mid 2$ and $|H|=2^{\alpha }\cdot 3\cdot 7$, where $\alpha =4$ or 5. Suppose that $H=H^{\mathrm{\prime }}$. Then H has a normal subgroup S such that $H/S\cong L_2(7)$, where $|S|=2$ or 4. Obviously, we know that $S\le Z(H)$, and then $S⊴G$. Let $\theta \in Irr(S)$ such that $[{\chi }_S,\theta ]\ne 0$. Then $\theta (1)=1$ since S is abelian. Let $e=[{\chi }_S,\theta ]$ and $t=|G:I_G(\theta )|$. Then $t=1$ and $e=\chi (1)=385$ by the Clifford theorem (see Theorem 6.2 in [5]). But $e^2\cdot t=[{\chi }_H,{\chi }_H]>|G:H|$, a contradiction. Hence $H^{\mathrm{\prime }}<H$. Suppose that $|H/H^{\mathrm{\prime }}|=2$. Then $H/H^{\mathrm{\prime }}$ is central in $G/H$. Let β be an irreducible component of ${\chi }_H$, and let θ be an irreducible component of ${\beta }_{H^{\mathrm{\prime }}}$. Then $\theta (1)=\beta (1)=7$ and θ is extendible to β. Hence λβ is invariant in G for every $\lambda \in Irr(H/H^{\mathrm{\prime }})$ if β is invariant in G. Since $|H|=2^{\alpha }\cdot 3\cdot 7\cdot 11$, where $\alpha =4$ or 5, H has at most 12 irreducible characters of degree 7. Let $t=|G:I_G(\beta )|$. Then $t\le 12$. Since the index of the maximal subgroup of U containing $I_G(\theta )$ divides t, we have that $t=1$ or 11 by checking maximal subgroups of $L_2(11)$ (see ATLAS table in [6]). If $t=11$, then H has exactly 12 irreducible characters of degree 7, and one of them, say δ, is invariant in G. Hence, λδ is also invariant in G for $\lambda \in Irr(H/H^{\mathrm{\prime }})$, which forces $t\le 10$, a contradiction. Therefore $t=1$ and $e=55$. But ${(55)}^2=[{\chi }_H,{\chi }_H]>|G:H|$, a contradiction. By the same reasoning as before, we can prove that $|H/H^{\mathrm{\prime }}|\ne 2^m\cdot 3^n$, where $1\le m\le 3$ and $0\le n\le 1$. If $|H/H^{\mathrm{\prime }}|=2^m\cdot 3^n$, where $m=4$ or 5, then the Sylow 7 subgroup of $H^{\mathrm{\prime }}$ is normal in $H^{\mathrm{\prime }}$, and so it is normal in G, a contradiction. Now we assume that $7\mid |H/H^{\mathrm{\prime }}|$. Let $H^{\mathrm{\prime }}\le A⊴H$ such that $|H/A|=7$, then $H/A\le Z(G/A)$. Since $Mult(L_2(11))=2$, we have $G/A\cong H/A\times L_2(11)$. Hence G has a normal series $1⊴N⊴M⊴G$ such that $M/N\cong L_2(11)$ and $|N|=2^{\alpha }\cdot 3$, where $\alpha =4$ or 5. Let φ be an irreducible component of ${\chi }_M$, and let η be an irreducible component of ${\phi }_N$. Then $\phi (1)=55$ and $\eta (1)=1$ by the Clifford theorem (see Theorem 6.2 in [5]). Since $|Aut(N)|$ is not divided by 5 and 11, one has that $t=|M:I_M(\eta )|=1$. Therefore ${(55)}^2=[{\phi }_N,{\phi }_N]>|M:N|$, a contradiction.

Suppose that $K/H\cong M_{11}$. Since $|Out(M_{11})|=Mult(M_{11})=1$, we have $G\cong H\times M_{11}$, where $|H|=2^3\cdot 7$. Let $\theta \in Irr(H)$ such that $[{\chi }_H,\theta ]\ne 0$. Since $\theta (1)\mid \chi (1)$ and $\theta (1)\mid |H|$, one has that $\theta (1)=7$, which implies that H is a Frobenius group with an elementary kernel of order 8 and a cyclic complement of order 7.

Suppose that $K/H\cong M_{22}$. Since $|G|=2^7\cdot 3^2\cdot 5\cdot 7\cdot 11$, we have that $|H|=1$ and $G=K\cong M_{22}$, which completes the proof of Theorem C. □