Maximum Order of elements in $GL(n,Z)$ Hi,
I know that $\mathrm{GL}(n,\mathbb{Z})$ has an element of order $m$ iff $\Phi(m)\leq n$, where $\Phi(m) = \varphi(m)$ if $p_1^{\alpha_1}\neq 2$ or $m=2$, $\Phi(m) = \varphi(m)-1$ if $p_1^{\alpha_1}= 2$ or $m\not=2$, and $\varphi$ is Euler totient.
From there I can show that the maximum order of a element of finite order in $\mathrm{GL}(n,\mathbb{Z})$, $f(n)$ statisfies $\displaystyle \lim_{n\to \infty} \frac{\ln f(n)}{\sqrt{n\ln n}} =1$.
Here is my question:
Can we find the asymptotic behavior of $f(n)$ (and not $\ln \bigl(f(n)\bigr)$)?
 A: See MR1655470 (99m:20111) on MathSciNet. 
A: From the fact you're citing, it looks like 
$$ f(n) = \max \{ m : \Phi(m) \le n \}. $$
For example, $\Phi(30) = 7$ and $\Phi(m) \ge 11$ for all $m \ge 31$ (note that since $\phi(n) \ge \sqrt{n}$ we only need to check finitely many values!) -- so $f(7) = f(8)  = f(9) = f(10) = 30$.
Now, consider the fact that
$$ \lim \inf \phi(n) {\log \log n \over n} = e^{-\gamma} $$
which is equation (20) in this Mathworld article.  Of course this holds if we replace $\phi$ by $\Phi$.
So $f(n)$ should grow like the inverse of the function
$$ n \to {e^{-\gamma} n  \over \log \log n} $$.  It appears, then, that $f(n) \sim e^\gamma n \log \log n$ as $n \to \infty$.
Unfortunately this disagrees with your estimate.  One of us is wrong somewhere.
EDIT: I believe my  argument is basically right, but the original fact was stated incorrectly.  From the paper of Levitt that Stanley pointed to, we should actually have 
$$ \Phi( p_1^{\alpha_1} \cdots p_k^{\alpha_k}) = \phi(p_1^{\alpha_1}) + \cdots + \phi(p_k^{\alpha_k}) - [k \equiv 2 \mod 4] $$
and so $\Phi(x)$ is usually much smaller than $\phi(x)$ -- therefore $f$ grows much faster than I said it did.
