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

- Haijing Xu
^{1}, - Yanxiong Yan
^{1, 2}and - Guiyun Chen
^{1}Email author

**2013**:209

https://doi.org/10.1186/1029-242X-2013-209

© Xu et al.; licensee Springer 2013

**Received: **29 December 2012

**Accepted: **15 April 2013

**Published: **26 April 2013

## 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.

### Keywords

finite group simple group character degree## 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* $1\u22b4H\u22b4K\u22b4G$ *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

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 $1\u22b4H\u22b4K\u22b4G$ 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 $P\u22b4G$. 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 $Q\u22b4P$ and $Q\u22b4G$. □

## 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 $1\u22b4H\u22b4K\u22b4G$ 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 $N\u22b4H$ 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 $1\u22b4H\u22b4K\u22b4G$ 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 $1\u22b4A\u22b4B\u22b4H$ 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 $N\u22b4A$ by Lemma 2. Hence we get a subnormal series of *G*, $N\mathit{char}A\u22b4B\u22b4H\u22b4K\u22b4G$, which implies that $N\u22b4G$. 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 $1\u22b4H\u22b4K\u22b4G$ 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 ${A}_{6}$.

*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 $H\u22b4G$, we have that ${O}_{3}(H)\u22b4G$. 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 $S\u22b4G$ and $G/S\lesssim {S}_{8}$. By the Jordan-Hölder theorem,

*S*has a normal series $1\u22b4{O}_{3}(H)\u22b4C\u22b4D\u22b4S$ 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 $1\u22b4H\u22b4K\u22b4G$ 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*: $1\u22b4N\u22b4M\u22b4H$ 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 $1\u22b4H\u22b4K\u22b4G$ 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 $1\u22b4N\u22b4M\u22b4H$ 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 $P\u22b4G$, 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 $S\u22b4G$. 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 ${H}^{\mathrm{\prime}}\le A\u22b4H$ 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 $1\u22b4N\u22b4M\u22b4G$ 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. □

## Declarations

### 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).

## Authors’ Affiliations

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