Fast detection of giant permutation groups

TL;DR

This paper introduces a faster randomized algorithm for detecting if a permutation group contains the alternating group, improving upon existing methods and previous algorithms by Cameron and Cannon.

Contribution

A novel 1-sided randomized algorithm that efficiently detects the presence of the alternating group within permutation groups given by generators.

Findings

Algorithm is faster than standard methods.

Outperforms previous algorithms by Cameron and Cannon.

Provides reliable detection with probabilistic guarantees.

Abstract

We give a 1-sided randomised algorithm to detect when a permutation group of degree n, given by generators, contains the alternating group of degree n. This improves on standard methods, and on an algorithm of P. Cameron and J. Cannon.

Tables

Table 1. Table 1: Experimental values for k ​ ( N , ε ) 𝑘 𝑁 𝜀 k(N,\varepsilon) .

	$\epsilon$
$N$	$0.1$	$0.05$	$0.02$	$0.01$
$10$	6	6	8	9
$20$	4	5	6	6
$30$	4	4	5	6
$50$	3	3	4	5
$100$	3	3	4	4
$1000$	2	2	3	3
${10}^4$	2	2	2	2
${10}^5$	1	2	2	2
${10}^6$	1	1	2	2
${10}^8$	1	1	1	2
${10}^{12}$	1	1	1	1

Equations

$$(1-\delta )\log n<L<(1+\delta )\log n.$$

$$O(L^2(\log n{)}^3+L{n}^{\delta }(\log n{)}^2),$$

$$O(L{n}^{\delta }\log n(L\log n+2^{L/2})),$$

Full text

1 Introduction

The giant permutation groups are the alternating and symmetric groups. When we are given permutations generating a group, we need to determine quickly whether or not they generate a giant, as they are so much larger than other permutation groups of the same degree that we need to treat them specially. Standard texts, such as [20] and [9], offer the following solution.

Theorem 1

Let $G$ be a transitive permutation group of degree $n$, and let $p$ be a prime with $n/2<p<n-2$. If $G$ contains an element of order divisible by $p$ then $A_{n}\leq G$. 2. 2.

The proportion of elements of $A_{n}$ and $S_{n}$ that have order divisible by some prime $p$ where $n/2<p<n-2$, is asymptotic to $\frac{\log 2}{\log n}$ .

Determining whether or not $G$ is transitive from the generators is a standard test, so we will now assume that $G$ is transitive.

We use this theorem to test for a giant by generating some random elements of G and testing the element orders, looking for a large prime factor. If such a prime turns up, then return that $G$ is a giant. If not, then return that $G$ is not a giant.

Note that the answer “giant” is always correct, while “not a giant” could be incorrect. This is a one-sided Monte Carlo test.

For a true Monte Carlo algorithm, there is a question to answer: how many elements do we test? The number depends on how tolerant we feel about the possibility of “$G$ is not a giant” being incorrect. We will use $\varepsilon$ to denote the probability of error we will allow. Assuming that each random permutation is chosen independently of any other, the second part of the theorem lets us choose the number of elements to test. If $n=10^{6}$. then $\log 2/\log n$ is about $0.05$. Testing 45 elements gives error probability $\varepsilon<0.1$, 90 elements gives $\varepsilon<0.01$. These numbers increase linearly with $\log n$ and with $-\log\varepsilon$.

The time taken is dominated by the time to generate random elements of $G$ and to determine cycle lengths and the element order. We want a method that uses few elements.

We say that $x\in S_{n}$ is a Jordan element whenever, for all $G\leq S_{n}$, $G$ primitive and $x\in G$ implies $A_{n}\leq G$.

We use the name Jordan in honour of C. Jordan’s work of the 1870s, where he showed that 2-cycles, 3-cycles, and $p$-cycles, with $p$ prime and $p<n-2$, are Jordan elements.

Jordan’s result on $p$-cycles has been used for fast detection of giant permutation groups for many years. In the late 1950s, Parker & Nikolai [17] used Jordan’s result as a test, knowing that the groups under consideration were primitive. Cannon [2] describes a more general method, where the group is first tested for primitivity, and, if so, Jordan’s result is used.

It is unclear to the author when the method given above, i.e. to attempt to prove both primitivity and the Jordan property from random elements, emerged. The permutation group programs of Cayley and Magma contained such an algorithm that performs better than the Theorem 1 method. This algorithm is due to P. Cameron and J. Cannon and was alluded to in [1], but no details were published. As $n\to\infty$, the Cameron-Cannon method has to test at least 6 (for $\varepsilon<0.1$) and 11 ($\varepsilon<0.01$) elements, and experiments did not rule out needing increasing numbers with degree. Experimental results were 6 and 12 elements at degree $10^{6}$, very slowly increasing with degree. See section 6.3 for more discussion of this algorithm and the derivation of these lower bounds.

In this article, we improve Cameron and Cannon’s algorithm, again using cycle structure of random elements to prove the group primitive, and to prove that the primitive group contains the alternating group. This problem is closely related to the question of invariable generation of the symmetric group, see [3, 14], and has applications to detecting when the Galois group of a polynomial is a giant.

2 Jordan elements

Since the 1870s there has been considerable history of proving that certain permutations are Jordan elements. As the Jordan property is invariant under conjugation in $S_{n}$, it is the cycle structure of elements that determines the property.

The next result is Corollary 1.3 of [11]. This is a version of Jordan’s result, following the classification of finite simple groups, which removes the primality condition.

Theorem 2

If $x\in S_{n}$ is an $\ell$-cycle, with $1<\ell<n-2$, then $x$ is a Jordan element.

The following collects results ranging from Jordan in the 1870s to Saxl 1977.

Theorem 3** (Jordan, Manning, Weiss, Saxl)**

If $x\in S_{n}$ has prime order $p$, with cycle type $m$ $p$-cycles and $k$ fixed points, where $m<p$, then any of the following implies $x$ is a Jordan element.

• •

$m=2$: $k>3$, or $p>3$ and $k>2$;

• •

$m=3$: $k>3$;

• •

$m=4$: $k>5$, or $p>5$ and $k>4$;

• •

$m=5$: $k>2$;

• •

$m=6$: $k>7$, or $p>7$ and $k>6$;

• •

$m=7$: $k>8$;

The results for $m\leq 4$ are proved in [15], $m=5$ in [19], and $m=6,7$ in [21]. Results of this kind, for $m\leq 5$, were known to Jordan in 1873 (see [15]).

For larger $m$ we have the following.

Theorem 4** (Manning [16])**

If $x\in S_{n}$ has prime order $p$, with cycle type $m$ $p$-cycles and $k$ fixed points, where $m>5$, $p>2m-2$ and $k>4m-4$, then $x$ is a Jordan element.

Theorem 5** (Praeger [18])**

If $x\in S_{n}$ has prime order $p$, with cycle type $m$ $p$-cycles and $k$ fixed points, where $1<m<p$, $k>5m/2-2$, $n\neq 9$ and $n\neq{c\choose 2}$, where $c=m+(p+1)/2$, then $x$ is a Jordan element.

If necessary, stronger results could be derived from [13]. This work gives a classification of all primitive groups containing an element of prime order $p$, with cycle type $m$ $p$-cycles and $k$ fixed points, where $m<p$.

We note a simple fact.

Lemma 2.1

If $x\in S_{n}$ and some power of $x$ is a Jordan element, then $x$ is a Jordan element.

Consider an algorithm, JordanTest, which takes as input the cycle type of a permutation $x$, and returns true if some power of $x$ has cycle type as listed in one of Theorems 2, 3, 4, or 5, and returns false otherwise. When JordanTest returns true, it has shown that $x$ is a Jordan element; if false is returned, then we draw no conclusion.

3 Properties of Random Permutations

Suppose that $P$ is a property of elements of $S_{n}$, and let $p_{n}$ denote the proportion of elements of $S_{n}$ satisfying $P$. We say that $P$ is true for almost all permutations when $\lim_{n\to\infty}p_{n}=1$. If $q_{n}$ is the proportion of even permutations in $S_{n}$ satisfying $P$, and $\lim_{n\to\infty}q_{n}=1$, then we say that $P$ is true for almost all even permutations. The following lemmas are easily proved.

Lemma 3.1

If $P$ is true for almost all permutations, then $P$ is true for almost all even permutations.

Lemma 3.2

If both $P$ and $Q$ are true for almost all permutations, then $(P\mbox{ and }Q)$ is true for almost all permutations.

It has long been known that almost all permutations are Jordan elements. We now collect results on properties of random permutations, and prove that JordanTest detects almost all permutations as being Jordan elements.

We consider a random $x\in S_{n}$, chosen from the uniform distribution. Let the number of cycles of $x$ be $L$. A number of properties of the random variable $L$ are well-known. It is elementary to show that mean and variance are both asymptotic to $\log n$, see for instance [12, §1.2.10]. (All logarithms in this work are to base $e$.) This is enough to justify the following theorem.

Theorem 6

Fix $\delta>0$. For almost all permutations,

[TABLE]

Proof: It follows from Lemma 3 of [3] and Stirling’s approximation that the proportion of permutations of degree $n$ with number of cycles outside the range $(1-\delta)\log n<L<(1+\delta)\log n$ is $O(n^{\delta-(1+\delta)\log(1+\delta)})$, so going to 0 as a power of $n$.  

In fact, it is known that the distribution of $L$ approaches a normal distribution [7, 5]. This power of $n$, for small $\delta$, is $-\delta^{2}/2+O(\delta^{3})$. For future reference, the proportion of permutations with more than $2\log n$ cycles (i.e. $\delta=1$) is $O(n^{-0.38})$.

The next result is from [4].

Theorem 7

Let $\omega:{\bf N}\to{\bf R}$ be some function with $\lim_{n\to\infty}\omega(n)=\infty$. Then almost all permutations have order divisible by some prime greater than $n/\exp\left(\omega(n)\sqrt{\log n}\right)$.

Suppose that the cycles of $x\in S_{n}$ are ${\bf C}_{1},{\bf C}_{2},\dots,{\bf C}_{L}$, where the cycles are ordered by the least point in each cycle. Denote the length of cycle ${\bf C}_{i}$ by $C_{i}$. The following is proved in [14], as Claims 1 & 2.

Theorem 8

Let $x\in S_{n}$ have $L$ cycles, and cycle lengths $C_{1},C_{2},\dots,C_{L}$ as above. Put $m=\lfloor\sqrt{\log n}\rfloor$. Then, for almost all $x$ we have $L>m$, and

For $1\leq i<j\leq L$, $\gcd(C_{i},C_{j})<n^{0.9}$. 2. 2.

For $1\leq i\leq m$, $C_{i}>n^{0.99}$; 3. 3.

$\gcd(C_{1},C_{2},\dots,C_{m})=1$;

We can now prove an asymptotic result about JordanTest.

Theorem 9

For almost all permutations $x$, the JordanTest algorithm applied to the cycle type of $x$ will return true.

Proof: Put $\omega(n)=0.3\sqrt{\log n}$ in Theorem 7. Consider an $x\in S_{n}$ such that the properties of Theorems 7 and 8 hold for $x$. Almost all $x$ have these properties. We now show that the cycle type of $x$ is recognised by JordanTest as that of a Jordan element.

Let $p$ be the largest prime divisor of the order of $x$, so $p>n^{0.7}$ by Theorem 7. Also, by Theorem 8, there is some cycle of $x$ with length $\ell>n^{0.99}$ and $\ell$ not divisible by $p$.

Let $y$ be a power of $x$ so that $y$ has order $p$. The cycle type of $y$ is $m$ $p$-cycles and $k$ fixed points, where $m<n^{0.3}$ and $k\geq\ell>n^{0.99}$. We may assume that $n$ is large enough so that $n^{0.7}>2n^{0.3}$, which implies $p>2m$, and $n^{0.99}>4n^{0.3}$, so $k>4m$. This cycle type is recognised by JordanTest as a Jordan element using one of Theorems 2, 3, or 4.

4 Proving Primitivity

We consider the possibility that the transitive group $G\leq S_{n}$ is imprimitive, having a non-trivial block system consisting of $r$ blocks each of size $s$, so $rs=n$.

Such a block system imposes constraints on the cycle type of elements of $G$. These constraints are generally not satisfied by random elements of $S_{n}$. We may be able to prove $G$ primitive by showing that no $r$ is possible given the cycle type of an element of $G$. We will see that this is so for almost all permutations.

Suppose $x\in G$, $G\leq S_{n}$ is transitive, and $r$ divides $n$. How can we determine that $G$ has no block system with $r$ blocks?

Suppose that there is such a block system, and that $x$ has a cycle of length $\ell$. Then this cycle induces a cycle of blocks with length $c\geq 1$. We then have:

$c$ divides $\ell$, $\ell\leq cs$ and $c\leq r$; 2. 2.

There are other cycles of $x$, with lengths multiples of $c$, and lengths adding to $cs-\ell$.

Deciding whether or not item 2 holds is an instance of a subset sum decision problem. The subset sum decision problem is NP-complete (see [6]) so instances may be hard to solve. We will limit the subset sum problems we will consider in size so that an exponential algorithm to solve them will run in $o(n)$ time.

We use the above as follows: For each $r$ dividing $n$, if there is some cycle of length $\ell$ with no possible corresponding $c$ because the above criteria cannot be met, then a block system with $r$ blocks can be eliminated as a possibility. Once all $r$ have been eliminated, the group has been proved primitive.

We define an algorithm PrimitiveTest to take as input the cycle type of an element $x\in S_{n}$.

For each $r$ dividing $n$ with $1<r<n$, and for each cycle length $\ell$, it attempts to eliminate $r$ as above. If all $r$ are eliminated, then true is returned, otherwise false is returned. If true is returned by PrimitiveTest applied to some cycle type, then any transitive group containing an element of that cycle type is primitive.

Theorem 10

For almost all permutations $x$, the PrimitiveTest algorithm applied to the cycle type of $x$ will return true.

Proof: This proof follows a portion of the proof of [14] Theorem 1, where it is shown that almost all permutations have cycle structure such that the cycle structure cannot belong to an element of a transitive and imprimitive group. Set $r_{n}=\exp(\log\log n\sqrt{\log n})$. Observe that $r_{n}=o(n^{\delta})$ for all $\delta>0$. Possible $r$ dividing $n$ are split into 3 cases:

Case 1: $1<r\leq r_{n}$.

We may assume that the cycle structure satisfies the properties given in Theorem 8. For each $r$ in this range, there exists a cycle C, with length $\ell$ not a multiple of $r$. Assume there is a block system with $r$ blocks, and let $B$ be the union of all blocks that intersect C. Then $|B|<n$, since $r$ does not divide $\ell$, and $|B|=an/r$ for some integer $a$, $1\leq a<r\leq r_{n}$. It is shown in [14], proof of Theorem 1, Case 1, that the proportion of permutations that support such an invariant subset is $o(n^{-0.009})$, so almost all permutations do not support such an invariant subset. Whether or not such a subset is supported by a given cycle structure will be detected by solving the subset sum decision problem given above, so almost all permutations will allow PrimitiveTest to eliminate all $r$ in this range.

Case 2: $r_{n}<r\leq n/r_{n}$.

By Theorem 7 with $\omega(n)=\log\log n$, we may assume that the permutation has order divisible by a prime $p>n/r_{n}$, so $p>r$ and $p>s$. If $\ell$ is a cycle length divisible by $p$, then there is no $c$ which simultaneously satisfies: $c$ divides $\ell$, $c\leq r$, and $\ell/c\leq s$. This cycle length eliminates all $r$ in this range.

Case 3: $n/r_{n}<r<n$.

We may assume that the cycle structure satisfies the properties given in Theorem 8. For each $r$ in this range, there exists a cycle length $\ell$ with $\ell>n^{0.99}$ and $\ell$ not a multiple of $s$. Any $c$ must satisfy, for sufficiently large $n$, $c\geq\ell/s>\ell/r_{n}>n^{0.99}/r_{n}>n^{0.9}$. As $s$ does not divide $\ell$, we have $cs>\ell$, so for the subset sum problem to have a solution, there must be another cycle with length a multiple of $c>n^{0.9}$. However such a cycle contradicts the properties given in Theorem 8. So, in this case, for large $n$ and with a cycle structure satisfying the properties of Theorem 8, the subset sum decision problem will eliminate all $r>n/r_{n}$, as there will be no other cycle with length a multiple of $c$.

5 The full AltsymTest algorithm

Here we describe the full algorithm for detecting a giant permutation group. An assumption in this description is the ability to generate a random group element cheaply, when only generators of the group are given. For a discussion of this matter see section 3.2.2 of [9].

While the JordanTest algorithm is called directly by AltsymTest, PrimitiveTest is not called directly, but the mechanisms of PrimitiveTest are applied to a sequence of up to $k$ cycle structures in order to eliminate possible block systems.

For any transitive group $G$, and any $k\geq 1$, if AltsymTest applied to $(G,k)$ returns true, then $A_{n}\leq G$. As with the standard algorithm, it is possible to have $A_{n}\leq G$ and for AltsymTest to return false. The probability of this happening is the error probability, which can be reduced by increasing $k$. We have seen that almost all permutations have fewer than $2\log n$ cycles, and AltsymTest treats a permutation with more than $2\log n$ cycles as evidence against the group given being a giant. Using the previous results for JordanTest and PrimitiveTest, we have:

Theorem 11

For all $\varepsilon>0$ and $N>0$ there exists $k=k(N,\varepsilon)$ such that if the input to AltsymTest is $(G,k)$, with $n\geq N$ and $A_{n}\leq G$, then the probability that AltsymTest returns false is $<\varepsilon$. 2. 2.

For fixed $\varepsilon>0$, we may take $k(N,\varepsilon)$ to be a non-increasing function of $N$, with limit 1 as $N\to\infty$.

We now consider the time taken for the algorithms when degree is $n$ and there are $L$ cycles in the permutation. In the following we assume classical algorithms for integer arithmetic, factorization of integers $\leq n$ in time $O(n^{\delta})$ for all $\delta>0$, and use of the algorithm of [10] for the subset sum decision problem. Theorem 315 of [8] tell us that the number of divisors of $n$ is $O(n^{\delta})$ for all $\delta>0$, so we may compute a list of all divisors of any $\ell\leq n$ in time $O(n^{\delta})$ for all $\delta>0$.

With the above assumptions, we can implement JordanTest to run in time

[TABLE]

and PrimitiveTest to run in time

[TABLE]

for all $\delta>0$. The component $O(2^{L/2}\log n)$ is the time to solve a subset sum decision problem.

Turning to AltsymTest: In calls to JordanTest and PrimitiveTest we have $L<2\log n$, so we find that calls to these functions use time $O(kn^{0.8})$. Time to generate $k$ random permutations and determine cycle type is at least $Akn$, for some constant $A$, which dominates the times for JordanTest and PrimitiveTest operations.

The asymptotic results given above rely on theorems that do not apply at the currently feasible degrees for permutation group computation. The performance at computable degree was checked by experiments.

The time to generate random permutations and work out cycle type greatly exceeds the time for JordanTest and PrimitiveTest operations. 2. 2.

The dominant time within JordanTest and PrimitiveTest at low degree is factorising cycle lengths and getting the list of divisors. 3. 3.

The time for the subset sum problem seems negligible. 4. 4.

In summary, $k(N,\varepsilon)$ is the key timing parameter.

Experiments with the algorithm suggest the values for $k(N,\varepsilon)$ given in Table 1.

6 Other matters

6.1 The alternating group

The asymptotic results given here apply equally to detecting the alternating and symmetric groups, see Lemma 3.1. At very low degrees, less than 20 say, the AltsymTest algorithm detects the symmetric group more quickly than it detects the alternating group, mainly due to a scarcity of detectable Jordan elements. This effect quickly becomes insignificant as the degree rises.

6.2 The influence of divisors of the degree

The factorization and list of divisors of the degree is used in the primitivity test. At one end of the scale, when $n$ is prime, the test is over once primality is established.

There is a very obvious difference in performance of PrimitiveTest between odd and even degrees. Even degrees need distinctly larger $k$ for the same $\varepsilon$. The experimental values given are for even degrees. There are also detectable differences when the degree is divisible by 3 and 4, but the effects are much smaller.

6.3 On the Cameron-Cannon algorithm

The Cameron-Cannon algorithm mentioned earlier is similar to the AltsymTest algorithm described here, but without the use of the subset sum criterion111In early versions of Magma the implementation was slightly weaker than this, in that $r$’s were eliminated sequentially, leading to cycle length information being wasted.. Examining the proof of Theorem 10, we expect it to need more random permutations to eliminate small and large $r$.

In particular, when $n$ is even, to eliminate $r=2$ the Cameron-Cannon algorithm must find an element having a cycle of odd length $>n/2$. The limiting proportion of such elements in $S_{n}$ and $A_{n}$ is $\frac{1}{2}\log 2$, slightly over 1 in 3. This leads to the lower bounds mentioned earlier. In experiments this case was the case that determined the number of random elements needed by this algorithm.

In contrast, the AltsymTest algorithm eliminates $r=2$ by finding an element which has no fixed set of size $n/2$ and does not have all cycles of even length. The proportion of permutations with all cycles of even length is asymptotic to $\sqrt{\frac{2}{\pi n}}=O(n^{-0.5})$ (by Lemma 4 of [3] and Stirling’s approximation). The main lemma of [14] shows that the proportion of elements that have fixed set of size $n/2$ is $O(n^{-0.01})$. This exponent is probably rather pessimistic: the final discussion of [14] suggests that the same methods could give $O(n^{-0.3})$.

6.4 Constructing random cycle types

As the majority of time in the AltsymTest algorithm is taken up generating random permutations and getting their cycle type, a lot of time is saved when conducting the experiments needed to determine $k(N,\varepsilon)$ by just constructing cycle types. This also allows us to extend the degrees considered beyond the degrees where we are willing to write down a permutation.

Lemma 1 of [3] may be used to generate random cycle types of elements of $S_{n}$, with distribution corresponding to the uniform distribution on elements of $S_{n}$. This was used in the experiments for Table 1, which only required cycle types of elements of $A_{n}$ and $S_{n}$.

6.5 On Monte Carlo-ness

The AltsymTest algorithm as stated here is a randomised algorithm, but not a Monte Carlo algorithm, as I have not given a method for taking an error probability $\varepsilon>0$ and processing this to find the $k$ parameter.

The experiments do give usable values for $k$ that are much less than other methods commonly used. We may extend the range of $\varepsilon$ by using a geometric approximation to get an upper bound on $k$ giving smaller $\varepsilon$ than those in the table. For instance, doubling $k$ will extend from $\varepsilon=0.01$ to $\varepsilon=0.0001$ even if such $k$ are a little larger than needed.

6.6 Implementation

The algorithm described here is implemented as part of Magma’s routines for permutation groups. All the results of Theorems 2, 3, 4, and 5 are used to check for Jordan elements, as well as some special cases at degrees $\leq 8$. The result of Praeger, Theorem 5, which was not used in the asymptotic results, is useful in the range of degrees where computation is practical. Within the AltsymTest algorithm, the time for PrimitiveTest dominates the time for JordanTest, and using results of [13] to improve the recognition of Jordan elements has not seemed worthwhile.

Random elements are generated using the product-replacement algorithm. It appears to be worthwhile to use AltsymTest on the random elements generated before the mixing time for product-replacement is reached.