Let $G$ be a finite group and $p$ a prime. The number $n_p(G)$ of Sylow $p$-subgroups of $G$ is an important invariant pertaining to $G$ and we call it the Sylow $p$-number of $G$. By a Sylow number for $G$, we mean an integer which is a Sylow $p$-number of $G$ for some prime $p$. The Sylow number was investigated by many authors such as Hall, Brauer, Hall, Zhang, and Moret$\acute{\rm o}$ (see for instance [1, 3, 4, 6-9]). Zhang [9] launched a systematic study on the influence of arithmetical properties on the group structure.

We set $sn(G)=\{n_p(G)| p\mid |G|\}$. Zhang [9] posed the following problem, namely what can we see about the finite groups $G$ in terms of $|sn(G)|$? And he made the following claim: it seems true that $G$ is solvable if $|sn(G)|= 2$. The above claim was proved in [7]. Now we consider the influence of $sn(G)$ on simple $K_3$-groups.

A finite simple group $G$ is called a simple $K_3$-group if $|G|$ has exactly three distinct prime divisors. We know that $|A_5|=2^2\cdot3\cdot5$ and $sn(A_5)=\{5, 2\cdot5, 2\cdot3\}$. So the following problem is interesting: if $|G|=p^2qr$ and $sn(G)=\{ r, pr, pq\}$, where $p<q<r$ are different primes, then $G\cong A_5$ holds? The answer of the problem is yes. In this paper, we get the following results by using an elementary and skillful method of applying Sylow's theorem.

Main Theorem (1) Let $|G|=p^2qr$ and $sn(G)=\{ r, pr, pq\}$, where $p<q<r$ are different primes, then $G\cong A_5$.

(2) Let $|G|=p^3q^2r$ and $sn(G)=\{ q^2r, pr, p^2q^2\}$, where $p<q<r$ are different primes, then $G\cong A_6$.

(3) Let $|G|=p^3qr$ and $sn(G)=\{ qr, p^2r, p^3\}$, where $p<q<r$ are different primes, then $G\cong L_2(7)$.

(4) Let $|G|=p^3q^2r$ and $sn(G)=\{ q^2, p^2r, p^2q^2\}$, where $p<q<r$ are different primes, then $G\cong L_2(8)$.

(5) Let $|G|=p^4q^2r$ and $sn(G)=\{ q^2r, p^3r, pq^2\}$, where $p<q<r$ are different primes, then $G\cong L_2(17)$.

(6) Let $|G|=p^4q^3r$ and $sn(G)=\{ q^3r, p^2r, p^4q^2\}$, where $p<q<r$ are different primes, then $G\cong L_3(3)$.

(7) Let $|G|=p^5q^3r$ and $sn(G)=\{ q^3r, p^2r, p^5q^2\}$, where $p<q<r$ are different primes, then $G\cong U_3(3)$.

(8) Let $|G|=p^6q^4r$ and $sn(G)=\{ r, pr, pq\}$, where $p<q<r$ are different primes, then $G\cong U_4(2)$.

In this paper, all groups are finite and by simple groups we mean non-abelian simple groups. All further unexplained notations are standard (cf. [2] for example).

We need the following two simple lemmas to show our results.

Lemma 2.1 (see [5]) If $G$ is a simple $K_3$-group, then $G$ is isomorphic to one of the following groups: $A_5$, $A_6$, $L_2(7)$, $L_2(8)$, $L_2(17)$, $L_3(3)$, $U_3(3)$ or $U_4(2)$.

Lemma 2.2 (see [9]) Let $G$ be a finite group and $M$ a normal subgroup of $G$, then the product of $n_p(G)$ and $n_p(G/M)$ divides $n_p(G)$.

Now we will prove the main theorem case by case.

Proof (1) If $G$ is solvable, then $G$ has an elementary abelian minimal normal subgroup $N$. Note that $sn(G)=\{ r, pr, pq\}$, thus $|N|=p$ and $|G/N|=pqr$. And it follows that $G/N$ is supersolvable. Therefore $G$ is supersolvable and $n_r(G)=1$, which is a contradiction. And so $G$ is unsolvable and $G\cong A_5$ by Lemma 2.1.

(2) Assume that $G$ is solvable, then $G$ has a $\{q, r\}$-Hall subgroup $H$ and $|H|=q^2r$. By Sylow's theorem, we know that $n_r(H)\mid q^2$. If $n_r(H)=q$, then $q\equiv 1$ (mod $r$), which is a contradiction since $q<r$. If $n_r(H)=q^2$, then $q^2\equiv 1$ (mod $r$), which implies that $r\mid q+1$. Consequently $q=2$ and $r=3$, a contradiction since $p<q$. Thus $n_r(H)=1$. Note that $|G:N_G(H)|\mid p^3$, thus $n_r(G)$ is at most $p^3$, which is impossible since $n_r(G)=p^2q^2$. Therefore $G$ is unsolvable.

We obtain that $G$ has a chief factor $H/N$ such that $H/N \cong A_5$, $A_6$, $L_2(7)$ or $L_2(8)$ by Lemma 2.1, where $N$ is a maximal solvable normal subgroup of $G$. Set $\overline{H}:=H/N\cong A_5$, $\overline{G}:= G/N$, we have

$ A_5\cong \overline{H} \cong \overline{H}C_{\overline{G}}(\overline{H})/C_{\overline{G}}(\overline{H})\leq \overline{G}/C_{\overline{G}}(\overline{H}) =N_{\overline{G}}(\overline{H})/C_{\overline{G}}(\overline{H})\lesssim {\rm Aut}(\overline{H}). $

Let $K:=\{x\in G |xN\in C_{\overline{G}}(\overline{H})\}$, then $A_5\leq G/K \leq {\rm Aut}(A_5)\cong S_5$. Hence $G/K\cong A_5$ or $G/K \cong S_5$. If $G/K \cong A_5$, then $|K|=6$ and so $K=N$. Let $P_5\in Syl_5(G)$, it follows that $|G/N: N_G(P_5)N/N|=6$. Note that $|N_G(P_5)|=10$ and $|N_G(P_5)N/N|=10$, thus $N\bigcap N_G(P_5)=1$ and consequently $n_5(N_G(P_5)N)=6$. Since $N_G(P_5)N$ is of order 60 and not 5-closed, we get that $N_G(P_5)N\cong A_5$, which is contradict to the solvability of $N_G(P_5)N$. If $G/K\cong S_5$, similarly we can get a contradiction. If $H/N\cong L_2(7)$, then $n_3(H/N)\mid n_3(G)$ by Lemma 2.2, namely $28\mid 14$, a contradiction. In fact 14 is not a Sylow 3-number. Similarly $H/N\ncong L_2(8)$. If $H/N\cong A_6$, then by Lemma 2.1, we have $N=1$ and $H=G\cong A_6$ since $G=p^3q^2r$ and $|A_6|=2^3\cdot3^2\cdot5$.

(3) If $G$ is solvable, then $G$ has a maximal subgroup $M$ such that $M\unlhd G$ and $|G:M|$ is a prime. If $|M|=p^3q$, then $n_q(G)=n_q(M)\mid p^3$, a contradiction since $n_q(G)=p^2r$. By the same reason $|M|\neq p^3r$. Hence $|M|=p^2qr$. Let $N$ be a minimal normal subgroup of $M$, then $|N|=p$ or $p^2$. If $|N|=p$, then $|M/N|=pqr$ and so $M/N$ is supersovable. Therefore $M$ is supersolvable, which implies $n_r(G)=n_r(M)=1$, a contradiction. If $|N|=p^2$, then $N$ Char $M$ and so $N \unlhd G$. Since $|G/N|=pqr$, we obtain that $G/N$ is supersolvable. Let $R\in Syl_r(G)$, then $RN/N\unlhd G/N$. Since $N_{G/N}(RN/N)=N_G(R)N/N=G/N$, we have $G=N_G(R)N$. Note that $|G|=p^3qr$ and $|N|=p^2$, we get that $p\mid|N_G(R)|$, contradict to $n_r(G)=p^3$. Therefore $G$ is unsolvable.

By Lemma 2.1, it follows that $G$ has a chief factor $H/N$ such that $H/N\cong A_5$ or $L_2(7)$. If $H/N\cong A_5$, then $n_r(G)=n_5(G)=p^3=8$, which is a contradiction. If $H/N\cong L_2(7)$, then by Lemma 2.1, we have $N=1$ and $H=G\cong L_2(7)$ since $G=p^3qr$ and $|L_2(7)|=2^3\cdot3\cdot7$.

(4) Suppose that $G$ is solvable, then $G$ has a $\{q, r\}$-Hall subgroup $H$ and $|G:N_G(H)| \mid p^3$. It is easy to show that $n_r(H)=1$ by Sylow's theorem. Therefore $n_r(G)$ is at most $p^3$, contradict to $n_r(G)=p^2q^2$. So $G$ is unsolvable.

By Lemma 2.1 $G$ has a chief factor $H/N$ such that $H/N\cong A_5$, $A_6$, $L_2(7)$ or $L_2(8)$. If $H/N\cong A_5$, then $n_2(G)=n_p(G)=q^2=9$. By Lemma 2.2, we get that $n_2(H/N)\mid n_2(G)$, namely $5\mid9$, which is a contradiction. By the same reason, $H/N\ncong A_6$, $L_2(7)$. If $H/N\cong L_2(8)$, then by Lemma 2.1 we have $N=1$ and $H=G\cong L_2(8)$ since $G=p^3q^2r$ and $|L_2(8)|=2^3\cdot3^2\cdot7$.

(5) Suppose that $G$ is solvable, then $G$ has a maximal subgroup $M$ such that $M\unlhd G$ and $|G:M|$ is a prime. If $|G:M|=q$, then the Sylow $p$-subgroup $P$ of $M$ is also the the Sylow $p$-subgroup of $G$. Since $n_p(G)=q^2r$ we have $N_G(P)=P\leq M$, which implies that $N_G(M)=M$, contradict to $M\unlhd G$. Similarly $|G:M|\neq r$. Consequently $|G:M|=p$ and $|M|=p^3q^2r$. Now we consider the $\{q, r\}$-Hall subgroup $N$ of $M$. It is easy to show that $n_r(N)=1$. Therefore $n_r(M)$ is at most $p^3$, a contradiction since $n_r(M)=n_r(G)=pq^2$. So $G$ is unsolvable.

We get that $G$ has a chief factor $H/N$ such that $H/N\cong A_5$, $A_6$, $L_2(7)$, $L_2(8)$ or $L_2(17)$ by Lemma 2.1. Similarly to the above, we can show that $G\cong L_2(17)$.

(6) Suppose that $G$ is solvable, then there exist a maximal subgroup $M$ of $G$ such that $M\unlhd G$ and $|G:M|$ is a prime. If $|G:M|=r$, then $n_q(G)=n_q(M)\mid p^4$ by Sylow's theorem, contradict to $n_q(G)=p^2r$. If $|G:M|=q$, then $n_p(G)=n_p(M)=\mid q^2r$, a contradiction since $n_p(G)=q^3r$. If $|G:M|=p$, then $n_r(G)=n_r(M)\mid p^3q^3$, which is impossible since $n_r(G)=p^4q^2$. Therefore $G$ is unsolvable.

We can see from Lemma 2.1 that $G$ has a chief factor $H/N$ such that $H/N\cong A_5$, $A_6$, $L_2(7)$, $L_2(8)$, $L_2(17)$ or $L_3(3)$. Now similarly to the above, we can show that $G\cong L_3(3)$.

(7) Assume that $G$ is solvable, then $G$ has a maximal subgroup $M$ such that $M\unlhd G$ and $|G:M|$ is a prime. If $|G:M|=r$, then $n_q(G)=n_q(M)\mid p^5$ by Sylow's theorem, which is contradict to $n_q(G)=p^2r$. If $|G:M|=q$, then $n_p(G)=n_p(M)\mid q^2r$, a contradiction since $n_p(G)=q^3r$. If $|G:M|=p$, then $n_r(G)=n_r(M)\mid p^4q^3$, which is impossible since $n_r(G)=p^5q^2$. Therefore $G$ is unsolvable.

It is easy to see that $G$ has a chief factor $H/N$ such that $H/N\cong A_5$, $A_6$, $L_2(7)$, $L_2(8)$, $L_2(17)$, $L_3(3)$ or $U_3(3)$ by Lemma 2.1. Now similarly to the above, we can show that $G\cong U_3(3)$.

(8) Suppose that $G$ is solvable, then $G$ has a maximal subgroup $M$ such that $M\unlhd G$ and $|G:M|$ is a prime. If $|G:M|=r$, then $n_p(G)=n_p(M)\mid q^4$ by Sylow's theorem, which is contradict to $n_p(G)=q^3r$. If $|G:M|=q$, then $n_r(G)=n_r(M)\mid p^6q^3$, a contradiction since $n_r(G)=p^4q^4$. It follows that $|G:M|=p$ and $|M|=p^5q^4r$. Now we consider a $\{q, r\}$-Hall subgroup $H$ of $M$. It is evident that $H$ is also a $\{q, r\}$-Hall subgroup of $G$. Note that $n_r(G)=p^4q^4$ and $|G:N_G(H)|\mid p^5$, thus $n_r(H)=q^4$ by Sylow's theorem. In fact, if $n_r(H)\leq q^3$, then $n_r(G)$ is at most $p^5q^3$, a contradiction since $p^5q^3<n_r(G)=p^4q^4$. Hence $r\mid q^4-1=(q^2+1)(q^2-1)$. If $r\mid q^2-1$, then $r\mid q+1$ since $q<r$. Consequently $q=2$ and $r=3$, which is contradict to $p<q<r$. Therefore $r\mid q^2+1$. Since $n_r(G)=p^4q^4$, we get that $r\mid p^4q^4-1$. Hence $r\mid p^4-1$ since $r\mid q^4-1$. Therefore $r\mid p^2+1$. And it follows that $r\mid (q^2+1)-(p^2+1)$, namely $r\mid(q-p)(q+p)$, which implies that $r\mid p+q$. Now we get a contradiction since $p+q<2r$. Therefore $G$ is unsolvable.

By Lemma 2.1, we know that $G$ has a chief factor $H/N$ such that $H/N\cong A_5$, $A_6$, $L_2(7)$, $L_2(8)$, $L_2(17)$, $L_3(3)$, $U_3(3)$ or $U_4(2)$. Now similarly to above, we can show that $G\cong U_4(2)$.

Now the proof of the theorem is complete.