# 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 $cd\left(G\right)$ of degrees of its complex irreducible characters. In , it was shown that many non-abelian simple groups such as ${L}_{2}\left(q\right)$ and ${S}_{z}\left(q\right)$ satisfy the conjecture. In this paper, we manage to characterize the finite simple groups by less character quantity. Let G be a finite group; $L\left(G\right)$ denotes the largest irreducible character degree of G and $S\left(G\right)$ 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. (1)

$|G|=|M|$;

2. (2)

$L\left(G\right)=L\left(M\right)$.

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

Theorem C Let G be a finite group. If $|G|=|{M}_{22}|$ and $L\left(G\right)=L\left({M}_{22}\right)$, then either G is isomorphic to ${M}_{22}$ or $H×{M}_{11}$, 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 $|G/K|\mid |Out\left(K/H\right)|$.

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 ${C}_{G/N}\left(M/N\right)\cap M/N=Z\left(M/N\right)=1$, and so

$M/N\cong \frac{{C}_{G/N}\left(M/N\right)×M/N}{{C}_{G/N}\left(M/N\right)}\le \frac{G/N}{{C}_{G/N}\left(M/N\right)}\lesssim Aut\left(M/N\right).$

Let $K/N={C}_{G/N}\left(M/N\right)×M/N$ and $H/N={C}_{G/N}\left(M/N\right)$. Then $G/K\le Out\left(M/N\right)$ 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. 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, $|P:{N}_{P}\left(Q\right)|={p}_{i}^{l}$ ($l\le m\le {a}_{i}$) 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 $L\left(G\right)=55$. 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 $\chi \left(1\right)\mid |G/N|$ for all $\chi \in Irr\left(G\right)$. But $L\left(G\right)=55$ 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 $|G/K|\mid |Out\left(K/H\right)|$. As $|G|={2}^{4}\cdot {3}^{2}\cdot 5\cdot 11$, we have .

We first assume that $K/H\cong {A}_{5}$. Since $|Out\left({A}_{5}\right)|=2$, we have $|G/K|\mid 2$ and $|H|={2}^{t}\cdot 3\cdot 11$, where $t=1$ or 2. Let $\chi \in Irr\left(G\right)$ such that $\chi \left(1\right)=L\left(G\right)=55$ and $\theta \in Irr\left(H\right)$ such that $\left[{\chi }_{H},\theta \right]\ne 0$. Then $\theta \left(1\right)=11$ by the Clifford theorem (see Theorem 6.2 in ). 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 $\theta \left(1\right)\mid |H/N|={2}^{t}\cdot 3$, a contradiction.

By the same reason as above, one has that $K/H\ncong {A}_{6}$.

Suppose that $K/H\cong {L}_{2}\left(11\right)$. Since $|Out\left({L}_{2}\left(11\right)\right)|=2$, we have $|G/K|\mid 2$ and so $|H|={2}^{a}\cdot 3$, where $a=1$ or 2. Let $\theta \in Irr\left(H\right)$ such that $e=\left[{\chi }_{H},\theta \right]\ne 0$ and let $t=|G:{I}_{G}\left(\theta \right)|$. Then $\theta \left(1\right)=1$ and $et=\chi \left(1\right)/\theta \left(1\right)=55$. Since $|H|={2}^{t}\cdot 3$, where $a=1$ or 2, we have that $55\nmid |Aut\left(H/{H}^{\mathrm{\prime }}\right)|$. Hence $t=1$, $e=55$. But ${\left(55\right)}^{2}={e}^{2}t=\left[{\chi }_{H},{\chi }_{H}\right]>|G:H|={2}^{b}\cdot 3\cdot 5\cdot 11$, 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 $L\left(G\right)={2}^{3}\cdot 11\cdot 23$. Then ${O}_{23}\left(G\right)=1$. If not, then $|{O}_{23}\left(G\right)|=23$ and ${O}_{23}\left(G\right)$ is abelian. Hence $L\left(G\right)={2}^{3}\cdot 11\cdot 23\mid |G/{O}_{23}\left(G\right)|$, 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 $|G/K|\mid |Out\left(K/H\right)|$. 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: ${A}_{5}$, ${L}_{2}\left(7\right)$, ${A}_{6}$, ${L}_{2}\left(8\right)$, ${L}_{2}\left(11\right)$, ${A}_{7}$, ${M}_{11}$, ${L}_{3}\left(4\right)$, ${A}_{8}$, ${M}_{22}$ and ${M}_{23}$.

We first assume that $K/H\cong {A}_{5}$. Since $|Out\left({A}_{5}\right)|=2$, 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 $|H/B|\mid |Out\left(B/A\right)|$. Since $|H|={2}^{m}\cdot 3\cdot 7\cdot 11\cdot 23$, we have $B/A\cong {L}_{2}\left(7\right)$ 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 ${O}_{23}\left(G\right)=1$, 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: ${A}_{6}$, ${L}_{2}\left(7\right)$, ${L}_{2}\left(8\right)$, ${L}_{2}\left(11\right)$, ${A}_{7}$, ${M}_{11}$, ${L}_{3}\left(4\right)$, ${A}_{8}$ 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 $L\left(G\right)={2}^{4}\cdot 11$. Since $11\mid L\left(G\right)$, by the same arguments as the proofs of Case 1.2, we have that ${O}_{11}\left(G\right)=1$.

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 $|G/K|\mid |Out\left(K/H\right)|$. As $|G|={2}^{6}\cdot {3}^{3}\cdot 5\cdot 11$, we have .

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

1. (a)

$S\cong {L}_{2}\left(11\right)×{O}_{3}\left(H\right)$ if $|C/{O}_{3}\left(H\right)|=1$;

2. (b)

$S\cong \left(2\cdot {L}_{2}\left(11\right)\right)×{O}_{3}\left(H\right)$ or $\left({Z}_{2}×{L}_{2}\left(11\right)\right)×{O}_{3}\left(H\right)$ if $|C/{O}_{3}\left(H\right)|=2$.

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

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

Suppose that $K/H\cong {M}_{11}$. Since $|Out\left({M}_{11}\right)|=Mult\left({M}_{11}\right)=1$, we have $G\cong H×{M}_{11}$, where $|H|={2}^{2}\cdot 3$. By checking the character table of ${M}_{11}$, we see that G has no irreducible character of degree $L\left(G\right)={2}^{4}\cdot 11$, 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 $S\left(G\right)={2}^{2}\cdot {3}^{2}\cdot 7\cdot 23$. Let $\chi \in Irr\left(G\right)$ such that $\chi \left(1\right)=S\left(G\right)$. If ${O}_{23}\left(G\right)\ne 1$, then $|{O}_{23}\left(G\right)|=23$, which implies that $\chi \left(1\right)\mid |G:N|$, a contradiction. Hence ${O}_{23}\left(G\right)=1$.

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 $|G/K|\mid |Out\left(K/H\right)|$. 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: ${A}_{5}$, ${L}_{2}\left(7\right)$, ${A}_{6}$, ${L}_{2}\left(8\right)$, ${L}_{2}\left(11\right)$, ${A}_{7}$, ${U}_{3}\left(3\right)$, ${M}_{11}$, ${L}_{3}\left(4\right)$, ${A}_{8}$, ${M}_{12}$, ${M}_{22}$, ${M}_{23}$ and ${M}_{24}$.

We first assume that $K/H\cong {A}_{5}$. Since $|Out\left({A}_{5}\right)|=2$, 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 $\theta \in Irr\left(H\right)$ such that $\left[{\chi }_{H},\theta \right]\ne 0$. Since $\chi \left(1\right)/\theta \left(1\right)\mid |G/H|$, it implies that $23\mid \theta \left(1\right)$. If H is solvable, then ${O}_{23}\left(H\right)\ne 1$ by Lemma 2, which implies that ${O}_{23}\left(H\right)={O}_{23}\left(G\right)\ne 1$, 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 $|H/M|\mid |Out\left(M/N\right)|$. As $|H|={2}^{t}\cdot {3}^{2}\cdot 7\cdot 11\cdot 23$, we have $M/N\cong {L}_{2}\left(7\right)$ or ${L}_{2}\left(8\right)$, which implies that $23\mid |N|$. Hence ${O}_{23}\left(N\right)\ne 1$ by Lemma 2, which implies that ${O}_{23}\left(N\right)={O}_{23}\left(G\right)\ne 1$, 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: ${L}_{2}\left(7\right)$, ${A}_{6}$, ${L}_{2}\left(8\right)$, ${L}_{2}\left(11\right)$, ${A}_{7}$, ${U}_{3}\left(3\right)$, ${M}_{11}$, ${L}_{3}\left(4\right)$, ${A}_{8}$, ${M}_{12}$ and ${M}_{22}$.

Suppose that $K/H\cong {M}_{23}$. Since $|Out\left({M}_{23}\right)|=Mult\left({M}_{23}\right)=1$, we have that $G\cong H×{M}_{23}$, 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 $L\left(G\right)=385$. Let $\chi \in Irr\left(G\right)$ such that $\chi \left(1\right)=L\left(G\right)=5\cdot 7\cdot 11$. We assert that ${O}_{11}\left(G\right)=1$. Otherwise, we have that $|{O}_{11}\left(G\right)|=11$ and ${O}_{11}\left(G\right)$ is abelian. Hence $\chi \left(1\right)\mid |G/{O}_{11}\left(G\right)|$, a contradiction. Similarly, ${O}_{5}\left(G\right)={O}_{7}\left(G\right)=1$.

If G is solvable, then the Sylow 11-subgroup of G is normal in G by Lemma 2. But ${O}_{11}\left(G\right)=1$, 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 $|G/K|\mid |Out\left(K/H\right)|$. 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: ${A}_{5}$, ${L}_{2}\left(7\right)$, ${A}_{6}$, ${L}_{2}\left(8\right)$, ${L}_{2}\left(11\right)$, ${A}_{7}$, ${M}_{11}$, ${L}_{3}\left(4\right)$, ${A}_{8}$ and ${M}_{22}$.

We first assume that $K/H\cong {A}_{5}$. Since $|Out\left({A}_{5}\right)|=2$, 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 ${O}_{11}\left(H\right)={O}_{11}\left(G\right)\ne 1$ 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 $|H/M|\mid |Out\left(M/N\right)|$. As $|H|={2}^{t}\cdot 3\cdot 7\cdot 11$, one has that $M/N\cong {L}_{3}\left(2\right)$ 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: ${L}_{2}\left(7\right)$, ${A}_{6}$, ${L}_{2}\left(8\right)$, ${L}_{2}\left(11\right)$, ${A}_{7}$, ${L}_{3}\left(4\right)$ or ${A}_{8}$.

Assume that $K/H\cong {L}_{2}\left(11\right)$. Since $|Out\left({L}_{2}\left(11\right)\right)|=2$, 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 $H/S\cong {L}_{2}\left(7\right)$, where $|S|=2$ or 4. Obviously, we know that $S\le Z\left(H\right)$, and then $S⊴G$. Let $\theta \in Irr\left(S\right)$ such that $\left[{\chi }_{S},\theta \right]\ne 0$. Then $\theta \left(1\right)=1$ since S is abelian. Let $e=\left[{\chi }_{S},\theta \right]$ and $t=|G:{I}_{G}\left(\theta \right)|$. Then $t=1$ and $e=\chi \left(1\right)=385$ by the Clifford theorem (see Theorem 6.2 in ). But ${e}^{2}\cdot t=\left[{\chi }_{H},{\chi }_{H}\right]>|G:H|$, a contradiction. Hence ${H}^{\mathrm{\prime }}. 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 $\theta \left(1\right)=\beta \left(1\right)=7$ and θ is extendible to β. Hence λβ is invariant in G for every $\lambda \in Irr\left(H/{H}^{\mathrm{\prime }}\right)$ 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 $t=|G:{I}_{G}\left(\beta \right)|$. Then $t\le 12$. Since the index of the maximal subgroup of U containing ${I}_{G}\left(\theta \right)$ divides t, we have that $t=1$ or 11 by checking maximal subgroups of ${L}_{2}\left(11\right)$ (see ATLAS table in ). 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 $\lambda \in Irr\left(H/{H}^{\mathrm{\prime }}\right)$, which forces $t\le 10$, a contradiction. Therefore $t=1$ and $e=55$. But ${\left(55\right)}^{2}=\left[{\chi }_{H},{\chi }_{H}\right]>|G:H|$, 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 $H/A\le Z\left(G/A\right)$. Since $Mult\left({L}_{2}\left(11\right)\right)=2$, we have $G/A\cong H/A×{L}_{2}\left(11\right)$. Hence G has a normal series $1⊴N⊴M⊴G$ such that $M/N\cong {L}_{2}\left(11\right)$ 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 $\phi \left(1\right)=55$ and $\eta \left(1\right)=1$ by the Clifford theorem (see Theorem 6.2 in ). Since $|Aut\left(N\right)|$ is not divided by 5 and 11, one has that $t=|M:{I}_{M}\left(\eta \right)|=1$. Therefore ${\left(55\right)}^{2}=\left[{\phi }_{N},{\phi }_{N}\right]>|M:N|$, a contradiction.

Suppose that $K/H\cong {M}_{11}$. Since $|Out\left({M}_{11}\right)|=Mult\left({M}_{11}\right)=1$, we have $G\cong H×{M}_{11}$, where $|H|={2}^{3}\cdot 7$. Let $\theta \in Irr\left(H\right)$ such that $\left[{\chi }_{H},\theta \right]\ne 0$. Since $\theta \left(1\right)\mid \chi \left(1\right)$ and $\theta \left(1\right)\mid |H|$, one has that $\theta \left(1\right)=7$, 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. □

## References

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## Acknowledgements

Article is supported by the Natural Science Foundation of China (Grant No. 11271301; 11171364), The Fundamental Research Funds for the Central Universities (Grant No. XDJK2009C074), Graduate-Innovation Funds of Science of SWU (Grant No. ky2009013) and Natural Science Foundation Project of CQ CSTC (Grant No. cstc2011jjA00020).

## Author information

Authors

### Corresponding author

Correspondence to Guiyun Chen.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

HX carried out the study of the Mathieu groups ${M}_{11}$, ${M}_{12}$ and ${M}_{23}$. YY carried out the study of the Mathieu group ${M}_{24}$. GC carried out the study of the Mathieu group ${M}_{22}$. All authors read and approved the final manuscript.

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Xu, H., Yan, Y. & Chen, G. A new characterization of Mathieu-groups by the order and one irreducible character degree. J Inequal Appl 2013, 209 (2013). https://doi.org/10.1186/1029-242X-2013-209 