Recognition of M × M by Its Complex Group Algebra Where M Is a Simple K₃-Group

Abstract

In this paper we prove that if M is a simple $K_3$-group, then $M\times M$ is uniquely determined by its order and some information on irreducible character degrees and as a consequence of our results we show that $M\times M$ is uniquely determined by the structure of its complex group algebra.

1. Introduction

Let G be a finite group, $\mathrm{Irr}\left(G\right)$ be the set of irreducible characters of G, and denote by $\mathrm{cd}\left(G\right)$, the set of irreducible character degrees of G. A finite group G is called a $K_3$-group if $\left|G\right|$ has exactly three distinct prime divisors. By [1], simple $K_3$-groups are $A_5$, $A_6$, $L_2\left(7\right)$, $L_2\left(8\right)$, $L_2\left(17\right)$, $L_3\left(3\right)$, $U_3\left(3\right)$ and $U_4\left(2\right)$. Chen et al. in [2,3] proved that all simple $K_3$-groups and the Mathieu groups are uniquely determined by their orders and one or both of their largest and second largest irreducible character degrees. In [4], it is proved that $L_2\left(q\right)$ is uniquely determined by its group order and its largest irreducible character degree when q is a prime or when $q=2^a$ for an integer $a\ge 2$ such that $2^a-1$ or $2^a+1$ is a prime.

Let p be an odd prime number. In [5,6,7,8], it is proved that the simple groups $L_2\left(q\right)$ and some extensions of them, where $q\mid p^3$ are uniquely determined by their orders and some information on irreducible character degrees.

In ([9], Problem 2${}^{*}$)R. Brauer asked: Let G and H be two finite groups. If for all fields $\mathbb{F}$, two group algebras $\mathbb{F}G$ and $\mathbb{F}H$ are isomorphic can we get that G and H are isomorphic? This is false in general. In fact, E. C. Dade [10] constructed two nonisomorphic metabelian groups G and H such that $\mathbb{F}G\cong \mathbb{F}H$ for all fields $\mathbb{F}$. In [11], Tong-Viet posed the following question:

Question. 

Which groups can be uniquely determined by the structure of their complex group algebras?

In general, the complex group algebras do not uniquely determine the groups, for example, $\mathbb{C}{D}_8\cong \mathbb{C}{Q}_8$. It is proved that nonabelian simple groups, quasi-simple groups and symmetric groups are uniquely determined up to isomorphism by the structure of their complex group algebras (see [12,13,14,15,16,17,18]). Khosravi et al. proved that $L_2\left(p\right)\times L_2\left(p\right)$ is uniquely determined by its complex group algebra, where $p\ge 5$ is a prime number (see [19]). In [20], Khosravi and Khademi proved that the characteristically simple group $A_5\times A_5$ is uniquely determined by its order and its character degree graph (vertices are the prime divisors of the irreducible character degrees of G and two vertices p and q are joined by an edge if $pq$ divides some irreducible character degree of G). In this paper, we prove that if M is a simple $K_3$-group, then $M\times M$ is uniquely determined by its order and some information about its irreducible character degrees. In particular, this result is the generalization of ([19], Theorem 2.4) for $p=5,7$ and 17. Also as a consequence of our results we show that $M\times M$ is uniquely determined by the structure of its complex group algebra.

2. Preliminaries

If $\chi ={\sum }_{i=1}^k{e}_i{\chi }_i$, where for each $1\le i\le k$, ${\chi }_i\in \mathrm{Irr}\left(G\right)$ and $e_i$ is a natural number, then each ${\chi }_i$ is called an irreducible constituent of $\chi$.

Lemma 1.

(Itô’s Theorem) ([21], Theorem 6.15) Let $A⊴G$ be abelian. Then $\chi \left(1\right)$ divides $|G:A|$, for all $\chi \in \mathrm{Irr}\left(G\right)$.

Lemma 2.

([21], Corollary 11.29) Let $N⊴G$ and $\chi \in \mathrm{Irr}\left(\mathrm{G}\right)$. If θ is an irreducible constituent of ${\chi }_N$, then $\chi \left(1\right)/\theta \left(1\right)\left|\right|G:N|$.

Lemma 3.

([2], Lemma 1) Let G be a nonsolvable group. Then G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic nonabelian simple groups and $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$.

Lemma 4.

(Itô-Michler Theorem) [22] Let $\rho \left(G\right)$ be the set of all prime divisors of the elements of $\mathrm{cd}\left(G\right)$. Then $p\notin \rho \left(G\right)=\{p:pis\mathrm{a}primenumber,p\mid \chi (1),\chi \in Irr(G\left)\right\}$ if and only if G has a normal abelian Sylow p-subgroup.

Lemma 5.

([3], Lemma 2) Let G be a finite solvable group of order $p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_n^{{\alpha }_n}$, where $p_1$, $p_2$, ..., $p_n$ are distinct primes. If $(k{p}_n+1)\nmid p_i^{{\alpha }_i}$, for each $i\le n-1$ and $k>0$, then the Sylow $p_n$-subgroup is normal in G.

Lemma 6.

([19], Theorem 2.4) Let $p\ge 5$ be a prime number. If G is a finite group such that (i) $\left|G\right|=|L_2{\left(p\right)|}^2$, (ii) $p^2\in \mathrm{cd}\left(G\right)$, (iii) there does not exist any element $a\in \mathrm{cd}\left(G\right)$ such that $2p^2\mid a$, (iv) if p is a Mersenne prime or a Fermat prime, then ${(p\pm 1)}^2\in \mathrm{cd}\left(G\right)$, then $G\cong L_2\left(p\right)\times L_2\left(p\right)$.

3. The Main Results

Lemma 7.

Let S be a simple $K_3$-group and let G be an extension of S by S. Then $G\cong S\times S$.

Proof. 

There exists a normal subgroup of G which is isomorphic to S and we denote it by the same notation. By [23], we know that $\left|\mathrm{Out}\right(S\left)\right|\le 4$ and $G/C_G\left(S\right)\hookrightarrow \mathrm{Aut}\left(S\right)$, which implies that $C_G\left(S\right)\ne 1$. As S is a nonabelian simple group, $S\cap C_G\left(S\right)=1$ and it follows that $S{C}_G\left(S\right)\cong S\times C_G\left(S\right)$. Also $C_G\left(S\right)\cong S{C}_G\left(S\right)/S⊴G/S\cong S$ which implies that G is isomorphic to $S\times S$. ☐

Theorem 1.

Let G be a finite group. Then $G\cong A_5\times A_5$ if and only if $\left|G\right|=|A_5{|}^2$ and $5^2\in \mathrm{cd}\left(G\right)$.

Proof. 

Obviously by Itô’s theorem, we get that $O_5\left(G\right)=1$. First we show that G is not a solvable group. If G is a solvable group, then let H be a Hall subgroup of G of order $2^4{5}^2$. Since $G/H_G\hookrightarrow S_9$, we get that $5\mid |H_G|$. If $5^2\mid \left|H_G\right|$, then $25\in \mathrm{cd}\left(H_G\right)$. On the other hand, ${25}^2<\left|H_G\right|\le 2^4{5}^2$, a contradiction. If $|H_G|=2^{4}5$, then $|G/H_G|=45$. Let $L/H_G$ be a Sylow 5-subgroup of $G/H_G$. Then $L/H_G⊴G/H_G$ and so $L⊴G$ and $\left|L\right|=5^2{2}^4$. Then $25\in \mathrm{cd}\left(L\right)$, which is a contradiction. If $|H_G|\mid 2^{3}5$, then P, a Sylow 5-subgroup of $H_G$ is a normal subgroup of G, which is a contradiction by Lemma 4. Therefore G is a nonsolvable group.

Since G is nonsolvable, by Lemma 3, G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic nonabelian simple groups and $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$. As $\left|G\right|=2^4{3}^2{5}^2$, we have $K/H\cong A_5,A_6$ or $A_5\times A_5$ by [23]. If $K/H\cong A_6$, then $\left|H\right|=5$ or 10. Using Lemma 2, $5\in \mathrm{cd}\left(H\right)$, a contradiction. If $K/H\cong A_5$, then $\left|H\right|=60$ or $\left|H\right|=30$. By Lemma 5, $5\in \mathrm{cd}\left(H\right)$. If H is a solvable group, then by Lemma 5, $P⊴H$, where $P\in {\mathrm{Syl}}_5\left(H\right)$, which is a contradiction. Therefore $\left|H\right|=60$ and so $H\cong A_5$. Hence G is an extension of $A_5$ by $A_5$ and by Lemma 7, $G\cong A_5\times A_5$. If $K/H\cong A_5\times A_5$, then $\left|H\right|=1$ and $G\cong A_5\times A_5$. ☐

Theorem 2.

Let G be a finite group. Then $G\cong L_2\left(17\right)\times L_2\left(17\right)$ if and only if $\left|G\right|=|L_2{\left(17\right)|}^2$ and ${17}^2\in \mathrm{cd}\left(G\right)$.

Proof. 

Obviously $O_{17}\left(G\right)=1$. On the contrary let G be a solvable group. First we show that there exists no normal subgroup N of G such that

(a) $\left|N\right|=2^i{3}^j{17}^k$, where $k\ne 0$ and $i<8$; or (b) $\left|N\right|=2^8{17}^2$; or (c) $\left|N\right|=2^{8}17$.

Let N be a normal subgroup of G. If $\left|N\right|=2^i{3}^j{17}^k$, where $k\ne 0$ and $i<8$, then by Lemma 5, $P⊴G$, where $P\in {\mathrm{Syl}}_{17}\left(G\right)$. Hence $O_{17}\left(G\right)\ne 1$, which is a contradiction. If $\left|N\right|=2^8{17}^2$, then ${17}^2\in \mathrm{cd}\left(N\right)$, which is impossible. If $\left|N\right|=2^{8}17$, then $|G/N|=3^{4}17$. If $T/N\in {\mathrm{Syl}}_{17}(G/N)$, then $T/N⊴G/N$. Therefore $T⊴G$, where $\left|T\right|={17}^2{2}^8$ and this is a contradiction as we stated above.

Let M be a minimal normal subgroup of G, which is an elementary abelian p-group. Obviously $p\ne 17$. Let $p=2$. Then $\left|M\right|=2^i$, where $0<i\le 8$ and so $|G/M|=2^{8-i}{3}^4{17}^2$. Then $T/M⊴G/M$, where $T/M\in {\mathrm{Syl}}_{17}(G/M)$. Therefore $T⊴G$ and $\left|T\right|={17}^2{2}^i$, which is a contradiction. Hence $p=3$ and $\left|M\right|=3^i$, where $1\le i\le 4$.

If $i=4$, then $G/C_G\left(M\right)\hookrightarrow \mathrm{Aut}\left(M\right)\cong \mathrm{GL}(4,3)$ and $\left|\mathrm{GL}(4,3)\right|=2^9\times 3^6\times 5\times 13$. Hence ${17}^2\mid \left|C_G\left(M\right)\right|$. Since M is an abelian subgroup of G, thus $3^4\mid \left|C_G\left(M\right)\right|$. If $|C_G\left(M\right)|={17}^2{3}^4{2}^j$, where $j\ne 8$, then by the above discussion we get a contradiction. Otherwise, $C_G\left(M\right)=G$ and so by Burnside normal p-complement theorem, G has a normal 3-complement of order ${17}^2{2}^8$, which is a contradiction.

If $i=3$, then $|G/M|=2^8{17}^{2}3$. Let $H/M$ be a Hall subgroup of $G/M$ of order $2^8{17}^2$. Then $\left|H\right|=2^8{3}^3{17}^2$. Since $G/H_G\hookrightarrow S_3$, thus $3^3{17}^2\mid \left|H_G\right|$. If $2^8\nmid \left|H_G\right|$, then by the above discussion we get a contradiction. Therefore $|H_G|=2^8{3}^3{17}^2$, i.e., $H⊴G$. Let B be a Hall subgroup of H of order $\left|B\right|=2^8{17}^2$. Then similarly to the above $2^{8}17\mid \left|B_H\right|$. If $|B_H|=2^8{17}^2$, then we get a contradiction. If $|B_H|=2^{8}17$, then $T/B_H⊴B/B_H$ where $T/B_H\in {\mathrm{Syl}}_{17}(B/B_H)$. Therefore $\left|T\right|=2^8{17}^2$, which is a contradiction.

If $i=2$, then $|G/M|=2^8{3}^2{17}^2$. Let $H/M$ be a Hall subgroup of $G/M$ of order $2^8{17}^2$. Then $\left|H\right|=2^8{3}^2{17}^2$. Thus similarly to the above, ${17}^2\mid \left|H_G\right|$ and ${17}^2\in \mathrm{cd}\left(H_G\right)$. Then by the same argument as above we get that $H_G$ has a normal subgroup of order $2^i{17}^2$, which is a contradiction.

If $i=1$, then $|G/M|=2^8{3}^3{17}^2$. Let $H/M$ be a Hall subgroup of $G/M$ of order $2^8{17}^2$. Then $\left|H\right|=2^8{17}^{2}3$. Since $G/H_G\hookrightarrow S_{27}$ we get that $17\mid |H_G|$. If $2^8\nmid \left|H_G\right|$ or $|H_G|=2^8{17}^k,$ where $k\ne 0$, then we get a contradiction. If $|H_G|=2^8{17}^{2}3$, then $H_G$ has a normal subgroup of order $2^i{17}^2$, which is a contradiction. If $|H_G|=2^8\times 17\times 3$, then $|G/H_G|=3^{3}17$. Therefore $T/H_G⊴G/H_G$, where $T/H_G\in {\mathrm{Syl}}_{17}(G/H_G)$. Hence $T⊴G$ and $\left|T\right|=2^8{17}^{2}3$, which is a contradiction as we stated above.

Therefore G is nonsolvable and by Lemma 3, G has a normal series $1⊴H⊴K⊴G$ such that $K/H\cong L_2\left(17\right)$ or $L_2\left(17\right)\times L_2\left(17\right)$ and $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$.

If $K/H\cong L_2\left(17\right)$, then $\left|H\right|=2^3{3}^{2}17$ or $2^4{3}^{2}17$ and so $17\in \mathrm{cd}\left(H\right)$. If H is a solvable group, then by Lemma 5, $P⊴H$, where $P\in {\mathrm{Syl}}_{17}\left(H\right)$, which is a contradiction by Lemma 4. Otherwise by Lemma 3 and [23] we get that $H\cong L_2\left(17\right)$. Therefore G is an extension of $L_2\left(17\right)$ by $L_2\left(17\right)$ and by Lemma 7, $G\cong L_2\left(17\right)\times L_2\left(17\right)$.

Obviously if $K/H\cong L_2\left(17\right)\times L_2\left(17\right)$, then $G\cong L_2\left(17\right)\times L_2\left(17\right)$. ☐

In the sequel, we show that if G is a finite group of order $|L_2\left(7\right)\times L_2\left(7\right)|$, such that G has an irreducible character of order $7^2$ or $2^6$, then we can not conclude that $G\cong L_2\left(7\right)\times L_2\left(7\right)$. So we need more assumptions to characterize $L_2\left(7\right)\times L_2\left(7\right)$.

Remark 1.

Using the notations of GAP [24], if $A=\mathrm{SmallGroup}(56,11)$ and $H=A\times A\times {\mathbb{Z}}_9$, then $\left|H\right|=|L_2\left(7\right)\times L_2\left(7\right)|$ and H has an irreducible character of degree $7^2$.

Similarly if $B=\mathrm{SmallGroup}(784,160)$ and $K=B\times S_3\times S_3$, then $\left|H\right|=|L_2\left(7\right)\times L_2\left(7\right)|$ and H has an irreducible character of degree $2^6$.

Theorem 3.

Let G be a finite group. Then $G\cong L_2\left(7\right)\times L_2\left(7\right)$ if and only if $\left|G\right|=2^6{3}^2{7}^2$ and $2^6,7^2\in \mathrm{cd}\left(G\right)$.

Proof. 

If G is a solvable group, then let H be a Hall subgroup of G of order $2^6{7}^2$. Since $G/H_G\hookrightarrow S_9$, we have $|H_G|=2^i{7}^j$, where $0\le i\le 6$ and $1\le j\le 2$. Using Lemma 2, $2^i,7^j\in \mathrm{cd}\left(H_G\right)$. If $O_2\left(H_G\right)\ne 1$, then by Lemma 2, $|O_2\left(H_G\right)|\in \mathrm{cd}\left(O_2\left(H_G\right)\right)$, which is a contradiction. Similarly $O_7\left(H_G\right)=1$, which shows that G is a nonsolvable group.

Therefore G has a normal series $1⊴H⊴K⊴G$ such that $K/H\cong L_2\left(8\right),L_2\left(7\right)$ or $L_2\left(7\right)\times L_2\left(7\right)$ and $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$.

If $K/H\cong L_2\left(8\right)$, then $\left|H\right|=56$. Using Lemma 2, $8\in \mathrm{cd}\left(H\right)$ and since $64>56$, we get a contradiction.

If $K/H\cong L_2\left(7\right)$, then $\left|H\right|=2^2\times 3\times 7$ or $2^3\times 3\times 7$. If $\left|H\right|=2^2\times 3\times 7$, then by Lemma 2, $7\in \mathrm{cd}\left(H\right)$. Since there exists no nonabelian simple group S such that $\left|S\right|\mid \left|H\right|$, we get that H is a solvable group. then by Lemma 5, $P⊴H$ where $P\in {\mathrm{Syl}}_7\left(H\right)$, which is a contradiction by Lemma 4. So $\left|H\right|=2^3\times 3\times 7$, by the same argument for the proof of Theorem A in [2], we get that $H\cong L_2\left(7\right)$. Therefore G is an extension of $L_2\left(7\right)$ by $L_2\left(7\right)$ and by Lemma 7, $G\cong L_2\left(7\right)\times L_2\left(7\right)$.

If $K/H\cong L_2\left(7\right)\times L_2\left(7\right)$, obviously we have $G\cong L_2\left(7\right)\times L_2\left(7\right)$. ☐

Remark 2.

We note that Theorems 1, 2 and 3 are generalizations of Lemma 6 for special cases $p=5,7,17$.

Lemma 8.

Let G be a finite group. If $\left|G\right|=2^i{3}^{j}5$, where $i\ge 3$ or $j\ge 1$, and $2^i,3^j\in \mathrm{cd}\left(G\right)$, then G is not solvable. If $\left|G\right|=2^i{3}^j{5}^2$, where $i\ge 6$ or $j\ge 2$, and $2^i,3^j\in \mathrm{cd}\left(G\right)$, then G is not solvable.

Proof. 

On the contrary let G be a solvable group.

Let $O_2\left(G\right)\ne 1$ and $|O_2\left(G\right)|=2^t$, where $1\le t\le i$. By the assumption, there exists $\chi \in \mathrm{Irr}\left(G\right)$ such that $\chi \left(1\right)=2^i$. If $\sigma \in \mathrm{Irr}\left({\mathrm{O}}_2\left(\mathrm{G}\right)\right)$ such that $[{\chi }_{O_2\left(G\right)},\sigma ]\ne 0$, then by Lemma 2, $2^i/\sigma \left(1\right)$ is a divisor of $|G:O_2\left(G\right)|=2^{i-t}$. Since $\sigma \left(1\right)\mid |O_2\left(G\right)|$, we get that $\sigma \left(1\right)=2^t$, which is a contradiction. Similarly $O_3\left(G\right)=1$.

Therefore $\mathrm{Fit}\left(G\right)=O_5\left(G\right)\ne 1$. We know that $G/C_G\left(\mathrm{Fit}\left(G\right)\right)\hookrightarrow \mathrm{Aut}\left(\mathrm{Fit}\left(G\right)\right)$ and since G is a solvable group, $C_G\left(\mathrm{Fit}\left(G\right)\right)⩽\mathrm{Fit}\left(G\right)$. Therefore $\left|G\right|$ is a divisor of $\left|\mathrm{Fit}\right(G\left)\right|·\left|\mathrm{Aut}\right(\mathrm{Fit}\left(G\right)\left)\right|$ and easily we can see that in each case we get a contradiction. ☐

Similarly to the above we have the following result:

Lemma 9.

Let G be a finite group.

(a) 

If $\left|G\right|=2^i{3}^{j}7$, where $i\ge 2$ or $j\ge 2$, and $2^i,3^j\in \mathrm{cd}\left(G\right)$, then G is not solvable.

(b) 

If $\left|G\right|=2^i{3}^j{7}^2$, where $i\ge 6$ or $j\ge 3$, and $2^i,3^j\in \mathrm{cd}\left(G\right)$, then G is not solvable.

Theorem 4.

Let G be a finite group.

(a) 

If $\left|G\right|=2^6{3}^4{5}^2$ and $2^6,3^4\in \mathrm{cd}\left(G\right)$, then $G\cong A_6\times A_6$ or $G\cong {\mathbb{Z}}_5\times U_4\left(2\right)$;

(b) 

If $\left|G\right|=2^{12}{3}^8{5}^2$ and $2^{12},3^8\in \mathrm{cd}\left(G\right)$, then $G\cong U_4\left(2\right)\times U_4\left(2\right)$.

Proof. 

Lemma 8 gives us that G is not solvable and so G has a normal series $1⊴H⊴K⊴G$ such that $K/H$ is a direct product of isomorphic nonabelian simple groups and $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$.

(a) By assumptions $K/H$ is isomorphic to $A_5,A_6,U_4\left(2\right),A_5\times A_5$ or $A_6\times A_6$.

If $K/H\cong A_5$, then $\left|H\right|=2^4{3}^{3}5$ or $\left|H\right|=2^3{3}^{3}5$. By Lemma 8, H is not solvable and H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a direct product of m copies of a nonabelian simple group S and $|H/B|\mid \left|\mathrm{Out}\right(B/A\left)\right|$. If $\left|H\right|=2^4{3}^{3}5$, we have $B/A\cong A_5$ or $A_6$. Then $\left|A\right|=36$, 18, 6 or 3, which is a contradiction. If $\left|H\right|=2^3{3}^{3}5$, then similarly we get a contradiction.

If $K/H\cong A_6$, then $\left|H\right|=2^i{3}^{2}5$, where $1\le i\le 3$. By Lemma 2, $2^i,3^2\in \mathrm{cd}\left(H\right)$. Using Lemma 8, H is not a solvable group and so $i\ne 1$. Also H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a direct product of m copies of a nonabelian simple group S and $|H/B|\mid \left|\mathrm{Out}\right(B/A\left)\right|$. If $\left|H\right|=2^3{3}^{2}5$, by Theorem B in [2], we get that $H\cong A_6$, and so by Lemma 7, $G\cong A_6\times A_6$. If $\left|H\right|=2^2{3}^{2}5$, then $\left|A\right|=3$, which is a contradiction.

If $K/H\cong U_4\left(2\right)$, then $\left|H\right|=5$ and $G=K$. Therefore G is an extension of ${\mathbb{Z}}_5$ by $U_4\left(2\right)$. We know that $G/C_G\left(H\right)\hookrightarrow \mathrm{Aut}\left(H\right)$ and $(G/H)/(C_G\left(H\right)/H)\cong G/C_G\left(H\right)$. So G is a central extension of H by $U_4\left(2\right)$. Since the Schur multiplier of $U_4\left(2\right)$ is 2, we get that $G\cong {\mathbb{Z}}_5\times U_4\left(2\right)$.

Let $K/H\cong A_5\times A_5$. We know that $\mathrm{Out}(K/H)\cong \mathrm{Out}\left(A_5\right)\wr S_2$, and so $|G/K|\mid 8$. Thus $\left|H\right|=2^i{3}^2$, where $0\le i\le 2$, which is a contradiction.

Finally, if $K/H\cong A_6\times A_6$, then $G\cong A_6\times A_6$.

(b) In this case, we have $K/H\cong A_5,A_6,U_4\left(2\right),A_5\times A_5,A_6\times A_6$ or $U_4\left(2\right)\times U_4\left(2\right)$.

If $K/H\cong A_5$, then $\left|H\right|=2^{10}{3}^{7}5$ or $2^9{3}^{7}5$. By Lemma 8, H is not a solvable group and H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a nonabelian simple group. Therefore A is a $\{2,3\}$-group such that $O_2\left(A\right)=O_3\left(A\right)=1$ and this is a contradiction.

If $K/H\cong A_6$, then similarly to the above we get a contradiction.

If $K/H\cong U_4\left(2\right)$, then $\left|H\right|=2^i{3}^{4}5$, where $5\le i\le 6$. By Lemma 2, $2^i,3^4\in \mathrm{cd}\left(H\right)$. Therefore H is not a solvable group and H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a nonabelian simple group. If $\left|H\right|=2^5{3}^{4}5$, then A is a $\{2,3\}$-group such that $O_2\left(A\right)=O_3\left(A\right)=1$ and this is a contradiction. If $\left|H\right|=2^6{3}^{4}5$, by Theorem A in [2], we get that $H\cong U_4\left(2\right)$ and by Lemma 7, $G\cong U_4\left(2\right)\times U_4\left(2\right)$.

Let $K/H\cong A_5\times A_5$. We know that $\mathrm{Out}(K/H)\cong \mathrm{Out}\left(A_5\right)\wr S_2$. Therefore $|G/K|\mid 8$ and thus $\left|H\right|=2^i{3}^6$, where $5\le i\le 8$, which is a contradiction.

If $K/H\cong A_6\times A_6$, then $\left|\mathrm{Out}(K/H)\right|=2^5$ and thus $\left|H\right|=2^i{3}^4$, where $1\le i\le 6$, which is a contradiction.

Therefore $K/H\cong U_4\left(2\right)\times U_4\left(2\right)$, and so $G\cong U_4\left(2\right)\times U_4\left(2\right)$. ☐

Corollary 1.

If $\left|G\right|=2^6{3}^4{5}^2$ and $2^6,3^4\in \mathrm{cd}\left(G\right)$ and $6\notin \mathrm{cd}\left(G\right)$, then $G\cong A_6\times A_6$.

Theorem 5.

If $\left|G\right|=2^{10}{3}^6{7}^2$ and $2^{10},3^6\in \mathrm{cd}\left(G\right)$, then $G\cong U_3\left(3\right)\times U_3\left(3\right)$.

Proof. 

By Lemma 9 it follows that G is not solvable and G has a normal series $1⊴H⊴K⊴G$ such that $K/H\cong L_2\left(7\right),L_2\left(8\right),U_3\left(3\right),L_2\left(7\right)\times L_2\left(7\right),L_2\left(8\right)\times L_2\left(8\right)$ or $U_3\left(3\right)\times U_3\left(3\right)$ and $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$.

If $K/H\cong L_2\left(7\right)$, then $\left|H\right|=2^7{3}^{5}7$ or $2^6{3}^{5}7$. By Lemma 9, H is not solvable and H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a nonabelian simple group. Therefore A is a $\{2,3\}$-group such that $O_2\left(A\right)=O_3\left(A\right)=1$, which is a contradiction. If $K/H\cong L_2\left(8\right)$, then similarly to the above we get a contradiction.

If $K/H\cong L_2\left(7\right)\times L_2\left(7\right)$ or $K/H\cong L_2\left(8\right)\times L_2\left(8\right)$, then H is a $\{2,3\}$-group, and we get a contradiction similarly.

If $K/H\cong U_3\left(3\right)$, then $\left|H\right|=2^5{3}^{3}7$ or $2^4{3}^{3}7$. By Lemma 9, H is not a solvable group and H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a nonabelian simple group.

If $\left|H\right|=2^4{3}^{3}7$, then A is a $\{2,3\}$-group such that $O_2\left(A\right)=O_3\left(A\right)=1$, which is a contradiction. If $\left|H\right|=2^5{3}^{3}7$, by Theorem C in [2], we get that $H\cong U_3\left(3\right)$ and by Lemma 7, $G\cong U_3\left(3\right)\times U_3\left(3\right)$.

Finally, if $K/H\cong U_3\left(3\right)\times U_3\left(3\right)$, then obviously $G\cong U_3\left(3\right)\times U_3\left(3\right)$. ☐

Theorem 6.

If G is a finite group such that

(i) 

$\left|G\right|=2^6{3}^4{7}^2$,

(ii) 

$2^6,3^4\in \mathrm{cd}\left(G\right)$,

(iii) 

$6,12,18\notin \mathrm{cd}\left(G\right)$,

then $G\cong L_2\left(8\right)\times L_2\left(8\right)$.

Proof. 

By Lemmas 3 and 9, we get that G has a normal series $1⊴H⊴K⊴G$ such that $K/H\cong L_2\left(7\right),L_2\left(8\right),U_3\left(3\right),L_2\left(7\right)\times L_2\left(7\right)$ or $L_2\left(8\right)\times L_2\left(8\right)$, and $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$.

If $K/H\cong L_2\left(7\right)$, then $\left|H\right|=2^3{3}^{3}7$ or $2^2{3}^{3}7$. By Lemma 9, H is not a solvable group and H has a normal series $1⊴A⊴B⊴H$ such that $B/A$ is a nonabelian simple group and $|H/B|\mid \left|\mathrm{Out}\right(B/A\left)\right|$. If $\left|H\right|=2^3{3}^{3}7$, we have $B/A\cong L_2\left(7\right)$ or $L_2\left(8\right)$. If $B/A\cong L_2\left(7\right)$, then $\left|A\right|=3^2$, a contradiction. If $B/A\cong L_2\left(8\right)$, then by Itô’s theorem, $\left|A\right|=1$ and $1⊴B\cong L_2\left(8\right)⊴H$, where $|H:B|=3$. By the proof of Lemma 1 in [2] (Lemma 3 in the present paper), $H/B$ is isomorphic to a subgroup of $\mathrm{Out}(B/A)$ and by [23] we have $H\cong L_2\left(8\right).3$. Using GAP $\mathrm{cd}\left(H\right)=\{1,7,8,21,27\}$, $Z\left(H\right)=1$ and $\mathrm{Aut}\left(H\right)\cong H$. Now similarly to the proof of Lemma 7, $G\cong (L_2\left(8\right).3)\times L_2\left(7\right)$. Then $6\in \mathrm{cd}\left(G\right)$, which is a contradiction by (iii). If $\left|H\right|=2^2{3}^{3}7$, then by Lemma 9, H is not a solvable group, and this is a contradiction by [23].

If $K/H\cong L_2\left(8\right)$, then $\left|H\right|=2^3·3^2·7$ or $2^3·3·7$. Using Lemma 9, H is not a solvable group. If $\left|H\right|=2^3·3^2·7$, by the same argument as Theorem C in [2], we get that $H\cong L_2\left(8\right)$ and by Lemma 7, $G\cong L_2\left(8\right)\times L_2\left(8\right)$. If $\left|H\right|=2^3·3·7$, then by Theorem A in [2], $H\cong L_2\left(7\right)$. Since $K/H\cong L_2\left(8\right)$, similarly to the proof of Lemma 7, we get that $K\cong L_2\left(7\right)\times L_2\left(8\right)$. So G is a an extension of $Z_3$ by $L_2\left(7\right)\times L_2\left(8\right)$. Since $6\in \mathrm{cd}\left(G\right)$ or $18\in \mathrm{cd}\left(G\right)$, we get a contradiction by (iii).

If $K/H\cong U_3\left(3\right)$, then $\left|H\right|=42$ or $\left|H\right|=21$.

If $\left|H\right|=42$, then H is solvable and $H^{\prime }$ is a cyclic group, since $\left|H\right|$ is square-free. Therefore $|H^{\prime }|=7$ and $|H/H^{\prime }|=6$. Now easily we see that the equation ${\sum }_{\phi \in \mathrm{Irr}\left(H\right)}{\phi }^2\left(1\right)=\left|H\right|$, where $\phi \left(1\right)\mid \left|H\right|$, has no solution and so we get a contradiction.

If $\left|H\right|=21$, then by Lemma 2, we get that $3\in \mathrm{cd}\left(H\right)$ and so H is a Frobenius group of order 21, which is denoted by $7:3$. Also $Z\left(H\right)=1$ and $\mathrm{Aut}\left(H\right)\cong H.2$. Now similarly to the proof of Lemma 7, we get that $K\cong (7:3)\times U_3\left(3\right)$. Since $|G:K|=2$, we have $G\cong (7:3)\times U_3\left(3\right)).2$ and so $6\in \mathrm{cd}\left(G\right)$ or $12\in \mathrm{cd}\left(G\right)$, which is a contradiction by (iii).

If $K/H\cong L_2\left(7\right)\times L_2\left(7\right)$. We know that $\mathrm{Out}(K/H)\cong \mathrm{Out}\left(L_2\left(7\right)\right)\wr S_2$. Then $|G/K|\mid 8$ and thus $\left|H\right|=3^2$, which is a contradiction.

Finally $K/H\cong L_2\left(8\right)\times L_2\left(8\right)$, and so $G\cong L_2\left(8\right)\times L_2\left(8\right)$. ☐

Theorem 7.

If $\left|G\right|=|L_3{\left(3\right)|}^2$ and $2^8,3^6\in \mathrm{cd}\left(G\right)$, then $G\cong L_3\left(3\right)\times L_3\left(3\right)$.

Proof. 

First we show that G is not a solvable group. If G is a solvable group, then $O_2\left(G\right)=O_3\left(G\right)=1$ and so $\mathrm{Fit}\left(G\right)=O_{13}\left(G\right)\ne 1$. Since $|\mathrm{Aut}\left({\mathbb{Z}}_{13}\right)|=2^{2}3$, $|\mathrm{Aut}\left({\mathbb{Z}}_{169}\right)|=2^2·3·13$ and $|\mathrm{Aut}({\mathbb{Z}}_{13}\times {\mathbb{Z}}_{13})|=2^5·3^2·7·13$, therefore $\left|G\right|\nmid \left|\mathrm{Fit}\right(G\left)\right|·\left|\mathrm{Aut}\right(\mathrm{Fit}\left(G\right)\left)\right|$, which is a contradiction. Therefore G is nonsolvable and G has a normal series $1⊴H⊴K⊴G$ such that $K/H\cong L_3\left(3\right)$ or $L_3\left(3\right)\times L_3\left(3\right)$, where $|G/K|\mid \left|\mathrm{Out}\right(K/H\left)\right|$. If $K/H\cong L_3\left(3\right)\times L_3\left(3\right)$, then $G=L_3\left(3\right)\times L_3\left(3\right)$. If $K/H\cong L_3\left(3\right)$, then $|G/K|=1$ or 2, and thus $\left|H\right|=2^4{3}^{3}13$ or $\left|H\right|=2^3{3}^{3}13$. If H is a solvable group, then $\mathrm{Fit}\left(H\right)\cong {\mathbb{Z}}_{13}$ and $\left|H\right|\nmid \left|\mathrm{Fit}\right(H\left)\right|·\left|\mathrm{Aut}\right(\mathrm{Fit}\left(H\right)\left)\right|$, which is a contradiction. Hence H is not a solvable group and so $H\cong L_3\left(3\right)$ and by Lemma 7, $G\cong L_3\left(3\right)\times L_3\left(3\right)$. ☐

As a consequence of the above theorem, by ([25], Theorem 2.13), we have the following result which is a partial answer to the question arose in [11].

Corollary 2.

Let M be a simple $K_3$-group and $H=M\times M$. If G is a group such that $\mathbb{C}G\cong \mathbb{C}H$, then $G\cong H$. Thus $M\times M$, where M is a simple $K_3$-group, is uniquely determined by the structure of its complex group algebra.