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If $G$ is a non abelian finite simple group, we know that $4$ divides $|G|$.

More precisely there are infinitely many finite simple groups $G$ such that $v_2(|G|)=2$, just consider $\mathrm{PSL}_2(\mathbb F_p)$ with $p\equiv 3 \pmod 8$, $p$ prime and $p>3$.

So my question tries to generalize the above statement:

Do we know the primes $p$ and positive integers $n$ such that there exists (or doesn't exist) infinitely many finite simple groups $G$ such that $v_p(|G|)=n$?

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up vote 5 down vote accepted

I think you don't need any new tricks, other than a Suzuki group. The short answer is that if $p$ is odd or $n\geq 2$, then there are infinitely many simple groups (of twisted Lie rank 1) whose order has $p$-adic valuation exactly $n$. $\newcommand{\PSL}{\operatorname{PSL}}\newcommand{\Sz}{\operatorname{Sz}}$

For $p=2$ and $n\geq 2$ choose $q \equiv 2^n \pm 1 \mod 2^{n+2}$ by Dirichlet. $$q^2-1 \equiv (2^n\pm1)^2 -1 \equiv 2^{2n}\pm2^{n+1} +1-1 \equiv 2^{n+1} \mod 2^{n+2},$$ so $v_2(q^2-1) = n+1$ and $v_2(|\PSL(2,q)|) = n$.

If $p$ is odd and $n \geq 1$, then choose $q\equiv p^n \pm 1 \mod p^{n+1}$, then $$q^2-1 \equiv (p^n \pm 1)^2-1 \equiv p^{2n} \pm 2\cdot p^{n} +1-1 \equiv \pm 2 p^n \mod p^{n+1},$$ and $v_p(\gcd(q-1,2)) = 0$, so $v_p( |\PSL(2,q)| ) = n$.

If $p > 3$ and $n=0$, then choose $q \not\equiv \pm1 \mod p$, then $$q^2-1 \not\equiv 0 \mod p,$$ and $v_p(|\PSL(2,q)|) = n = 0$.

If $p=3$ and $n=0$, then choose $q=2^{2n+1}$ and then $v_3(|\Sz(q)|) = 0$, where $\Sz(q)$ is the Suzuki simple group defined as a subgroup of the group of Lie type B2 over the field of $q$ elements.

If $p=2$ and $n \leq 1$, then no simple group has order with 2-adic valuation $n$ by Feit-Thompson and Cayley (for $n=0$ and $n=1$ respectively).

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At some point I wrote some code to find answers to this related question:… (and I guess lots of my other Sylow questions). At any rate, you that the p-adic valuation of the order of a simple group of lie type X(r,q) is exactly n exactly when q is equivalent to (long but finite list) mod p^(n+...). – Jack Schmidt Oct 15 '11 at 2:52
Jack, thank you very much. You answers to my questions (not just this one) are extremely useful. – Nathan Portland Oct 15 '11 at 5:04

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