Timeline for Maximum Order of elements in $GL(n,Z)$
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Feb 12, 2010 at 20:02 | comment | added | Portland | Obviously we have $f(n)\geqslant a(n)$. We can also show that $\lim_{n\to \infty}\frac{f(n)}{a(n)}= + \infty$ (that's not obvious at all). | |
| Feb 12, 2010 at 19:11 | comment | added | Michael Lugo | The symmetric group $S_n$ can be represented by the permutation matrices, so your $f(n)$ grows at least as fast as the maximal order $a(n)$ of a permutation of $n$ elements; it turns out that $\lim_{n \to \infty} \log a(n)/\sqrt{n \log n} = 1$ (see research.att.com/~njas/sequences/A000793). So another logical question is to ask how $a(n)$ and $f(n)$ are related. | |
| Feb 12, 2010 at 17:43 | answer | added | Richard Stanley | timeline score: 5 | |
| Feb 12, 2010 at 15:39 | answer | added | Michael Lugo | timeline score: 0 | |
| Feb 12, 2010 at 15:30 | comment | added | Greg Kuperberg | He wants to estimate $f(n)$ to within a constant factor. The given estimate is much looser than that. | |
| Feb 12, 2010 at 15:23 | comment | added | Harald Hanche-Olsen |
What is $\Phi(m)$ when $m$ is odd or a multiple of 4? Both branches of the definition seem to apply in these cases. And your result already is about asymptotic behaviour of $f(n)$. I suppose you want a better estimate? Can you indicate how good an estimate you need?
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| Feb 12, 2010 at 15:10 | comment | added | Portland | Yes Michael, this is correct $p_1$ is the smallest prime dividing $m$ and $\alpha_1$ is its exponent. | |
| Feb 12, 2010 at 15:06 | comment | added | Michael Lugo | What's $p_1^{\alpha_1}$? My guess is that $p_1$ is the smallest prime dividing $m$ and $\alpha_1$ is its exponent. So $\Phi(m) = \phi(m) - 1$ if m is congruent to 2 mod 4, and $\Phi(m) = \phi(m)$ otherwise. | |
| Feb 12, 2010 at 14:55 | history | asked | Portland | CC BY-SA 2.5 |