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let $G$ be a finite non-abelian simple group.If there exist $p$ and $q$ which are different prime numbers of $|G|$ such that $n_p(G)=n_q(G)$?

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What about $A_5$ ? – js21 Oct 22 '12 at 11:36
$n_2(A_5)=5,n_3(A_5)=10$and$n_5(A_5)=6$ – Tom Oct 23 '12 at 4:55
up vote 2 down vote accepted

Not sure if this question really qualifies for MO.

Anyway, the answer very much depends on the group $G$. In most cases $n_p(G)\ne n_q(G)$ for distinct prime divisors of the group order. However, there are infinitely many examples where equality occurs: If $r$ is an odd prime, then $n_p(\text{PSL}(2,r))=r(r+1)/2$ for each odd prime divisor $p$ of $r-1$.

But there are other examples too. For instance the atlas of finite simple groups shows that in the Janko group $J_1$, the normalizers of the $3$-Sylows and $5$-Sylows have order $60$.

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I was assuming that $n_p(G)$ denotes the number of Sylow $p$-subgroups (although he didn't say so), in which case $n_p({\rm PSL}(2,r) = r(r+1)/2$ in your example. – Derek Holt Oct 22 '12 at 12:33
@Derek: Thanks, I fixed that. (Was thinking of $N_G(p-\text{Sylow})=|G|/n_p$.) – Peter Mueller Oct 22 '12 at 12:50
Then how about $A_n$(Alternating groups) – Tom Oct 23 '12 at 4:47
@Tom: That would be a new question. But it should be made clear why it is an interesting question. The number of $p$-Sylows of an alternating group can be explicitly written out, it's quite a complicated expression. Probably one can check that $n_p\ne n_q$ for distinct prime divisors of the group order. – Peter Mueller Oct 23 '12 at 9:20
If $G\in A_n$ is a $K_m$ group,then I know that there are $m$ different sylow numbers about $G$. – Tom Oct 23 '12 at 10:48

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