The Largest Prime Dividing the Maximal Order of an Element of $S_n$

TL;DR

This paper investigates the prime factorization of the maximal order of elements in symmetric groups, providing bounds on the largest prime divisor that improve previous estimates.

Contribution

It introduces new bounds on the largest prime dividing the maximal order of elements in symmetric groups, refining previous results.

Findings

Largest prime divisor of g(n) is bounded by 1.328√(n log n)

Provides a reduction in the upper bound for prime divisors

Enhances understanding of the structure of symmetric groups

Abstract

We define $g(n)$ to be the maximal order of an element of the symmetric group on $n$ elements. Results about the prime factorization of $g(n)$ allow a reduction of the upper bound on the largest prime divisor of $g(n)$ to $1.328\sqrt{n\log n}$.

Equations

$$\log g(n)\le a\sqrt{n\log n}a=1.05313\dots n\ge 1,$$

$$P(g(n))\le 1.328\sqrt{n\log n}.$$

$$g(n)=\text{max}\{M|\ell (M)\le n\}.$$

$$p^k+p^{\prime }\le q\le p^{k+1}+p^{\prime }-1.$$

$$\ell (M)\le \ell (g(n))+(p^k+p^{\prime }-q)\le \ell (g(n))\le n.$$

$$\begin{gathered} \text{multline}{p}^k{p}^{\prime }-q\ge p^k{p}^{\prime }-p^{k+1}-p^{\prime }+1=p^k(p^{\prime }-p)-p^{\prime }+1 \\ \ge p(p^{\prime }-p)-p^{\prime }+1=(p-1)(p^{\prime }-p-1)\ge 0.\text{endmultline}{p}^k{p}^{\prime }-q\ge p^k{p}^{\prime }-p^{k+1}-p^{\prime }+1=p^k(p^{\prime }-p)-p^{\prime }+1 \\ \ge p(p^{\prime }-p)-p^{\prime }+1=(p-1)(p^{\prime }-p-1)\ge 0. \end{gathered}$$

$$\ell (M)\le p+p^{\prime }-q-q^{\prime }+\ell (g(n))\le (1+\alpha )q-q-\alpha q+\ell (g(n))=\ell (g(n)).$$

$$g(n)\ge \frac{q\prod _{p\le .8357q}p}{\frac{q}{2}\prod \frac{1+{\alpha }_i}{2}q}.$$

$$\log g(n)\ge \theta (.8357q)-\log \frac{1}{2}-\sum \log \left(\frac{1+{\alpha }_i}{2}q\right),$$

$$\log g(n)\ge \theta (.8357q)-.01338q.$$

$$\log g(n)\ge .79307q.$$

$$1.328\sqrt{n\log n}\ge \frac{1.05314}{.79307}\sqrt{n\log n}\ge q.$$

Full text

††support: A portion of this research, including the computations, was done at the Supercomputing Research Center. The author would also like to thank the referee for helpful suggestions.

Let $S_{n}$ be the symmetric group on $n$ letters.

The first work on $g(n)$ was done by Landau [1] in 1903. He showed that $\log{g(n)}\sim\sqrt{n\log n}$ as $n\to\infty$. In 1984, Massias [2] showed an upper bound for $\frac{\log g(n)}{\sqrt{n\log n}}$,

[TABLE]

with $a$ attained for $n=1,319,166$.

Let $P(g(n))$ be the largest prime divisor of $g(n)$. In 1969, Nicolas [4] proved that $P(g(n))\sim\sqrt{n\log n}$ as $n\to\infty$. In 1989, Massias, Nicolas, and Robin [3] showed that $P(g(n))\leq 2.86\sqrt{n\log n}$, $n\geq 2$. They conjectured that $\frac{P(g(n))}{\sqrt{n\log n}}$ achieves a maximum ($1.265\dots$) for $n\geq 5$ at $n=215$, with $P(g(215))=43$. They note that improving this bound using the techniques of their proof would require “very extensive computation,” and even then would not be able to reduce the constant in the bound below $2$.

Using different techniques, however, we can improve this result to the following

[TABLE]

Our proof begins with the simple observation that $g(n)=\operatorname{max}\,\{\operatorname{ord}(\sigma)\,|\,\sigma\in S_{n}^{\prime}\},$ where $S_{n}^{\prime}$ is the subset of $S_{n}$ consisting of elements that are the product of disjoint cycles of prime power length.

To see this, recall the fact that we can write any $\sigma\in S_{n}$ as the product of disjoint cycles. Then $\operatorname{ord}(\sigma)$ is the least common multiple of the cycle lengths. Consider a cycle of length $ab$ with $(a,b)=1$, $a,b>1$. The product of a cycle of length $a$ with one of length $b$ also has order $ab$ and is a permutation on fewer elements. Thus, given any element of $S_{n}$, we may find another that has the same order and is a product of disjoint cycles of prime power length.

We observe that $\ell(M)$ is the shortest length of a permutation of order $M$. Thus, we can characterize $g(n)$ in terms of $\ell$ as follows:

[TABLE]

In particular, $\ell(g(n))\leq n$.

Nicolas [6] describes an algorithm for computing $g(n)$. Employing a variation of this algorithm, I computed exact values of $g(n)$ for $n\leq 500,000$ on a Sun 4/390. The accuracy of the computation was checked by calculating values of $g(n)$ using the set $G$ described in [3] and verifying that they matched those in the computations. Analysis of the computations confirmed that for $5\leq n\leq 500,000$, $\frac{P(g(n))}{\sqrt{n\log{n}}}$ attains a maximum at $n=215$.

[TABLE]

Let $M=\frac{p^{k}p^{\prime}g(n)}{q}$. Since $q\mid g(n)$, $M$ is an integer. Then

[TABLE]

Thus, an element of order $M$ can be written as a permutation on $n$ letters. Also,

[TABLE]

Therefore, $p^{k}p^{\prime}>q$, so $M>g(n)$. But $g(n)$ is the maximal order of a permutation on $n$ letters. Thus, we have a contradiction, and the lemma is proven.

Write $q=P(g(n))$. We immediately get the following

[TABLE]

But $pp^{\prime}-qq^{\prime}>(\sqrt{\beta}q)^{2}-q(\beta q)=0,$ so $M>g(n)$, giving a contradiction.

We note that these intervals cover $(.5q,.8357q)$. So at most ten primes less than $.8357q$ fail to divide $g(n)$, and they can be at most $\frac{q}{2}$, $\frac{(1+\alpha_{1})q}{2}$,

…, and $\frac{(1+\alpha_{9})q}{2}$.

Therefore,

[TABLE]

Taking logarithms, we get

[TABLE]

where $\theta$ is the Chebyshev function, $\theta(x)=\sum\limits_{p\leq x}{\log p}$.

For $q>3329$ the sum of the terms on the right is less than $.01338q$, so

[TABLE]

Using the estimates for $\theta(x)$ in [7], we get

[TABLE]

From [3], $1.05314\sqrt{n\log n}\geq\log{g(n)}$, so

[TABLE]

It is likely that further computation would be able to show that $\frac{P(g(n))}{\sqrt{n\log{n}}}$ attains a maximum at $n=215$ for all $n\geq 5$.