Outer automorphisms in $S_6$

In permutations: The Symmetric Group: Permutations of a Finite Set

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In 1895, Otto Hölder proved the following theorem: Among symmetric groups, only $S_6$ has a non-trivial outer automorphism. In this short document I prepare a computational version of an outer autormorphism from $S_6$ to itself, and numerically verify some of its properties. @janusz1982 consider an outer automorphism $\phi$ of $S_6$ and offer the following properties:

$$ \phi(12) = (12)(36)(45)\ \phi(13) = (16)(24)(35)\ \phi(14) = (13)(25)(46)\ \phi(15) = (15)(26)(34)\ \phi(16) = (14)(23)(56) $$

[from which we can see directly that $\phi$ is not an inner automorphism: inner automorphisms preserve a permutation's shape]. Here, the general plan will be as follows. Firstly, express any element of $S_6$ as a product of cycles of the form $(1n)$ for $2\leqslant n\leqslant 6$; secondly, apply $\phi$ as defined above and, assuming $\phi$ to be a homomorphism, exhibit an explicit outer autormorphism of $S_6$. First we define f() that gives transpositions of the form $(1n)$:

f <- function(n){as.cycle(unique(c(1,n)))}
f(5)
f(1)

We may generate any cycle using the following identity:

$$ (abcd) = \left((1b)(1c)(1d)\right)^{(1a)} = (1a)(1b)(1c)(1d)(1a) $$

where the last equality follows from $(1a)^{-1}=(1a)$. We will verify the case $a=4,b=5,c=2,d=9$, viz $(529)^{(14)}=(4529)$:

(f(5)*f(2)*f(9))^f(4) == as.cycle("(4529)")

We may implement this in R, here to express cycle $(73456)$ as a product of transpositions of the form $(1n)$. First we define cyctopair(), a (trivial) helper function:

cyctopair <- function(vec){ c(vec,vec[1]) }
v <- c(7,3,4,5,6)
cyctopair(v)

Then to verify the identity in package idiom, we need lapply():

Reduce("*",lapply(cyctopair(c(7,6,4,9)),f))

Above we see that $(7649)=(17)(16)(14)(19)(17)$. It doesn't look like it has worked [because we see $(4976)$ rather than the expected $(7649)$], but it has:

Reduce("*",lapply(cyctopair(v),f)) == as.cycle(v)

It might be worth checking that the identities operate correctly if the cycle includes a `1':

v <- c(1,7,6,4,9)  # the "1" might break things
cyctopair(v)
Reduce("*",lapply(cyctopair(v),f)) == as.cycle(v)

Now define holder(), which goes through its argument replacing each element with a series of integers to be passed to f():

holder <- function(p){
    p <- as.cycle(p)
    out <- rep(list(NULL),length(p))
    for(i in seq_along(p)){
        pi <- p[[i]]
        if(!is.id(p[i])){
            for(j in seq_along(pi)){
                out[[i]] <- c(out[[i]],cyctopair(pi[[j]]))
            }
        } else {
            out[[i]] <- 1
        }
    }
    return(out)
}

(p <- c(id,as.cycle(rperm(5))))
(hp <- holder(p))

OK so now we can translate any vector of permutations into a product of transpositions of the form $(1n)$ for $2\leqslant n\leqslant 6$. Verify:

all(p == do.call("c",lapply(hp,function(v){as.cycle(Reduce("*",lapply(v,f)))})))

Now a function to translate an integer into a triple of cycles as per Janusz's formulation above:

offer <- function(n){
    if(n == 1){
        return(as.word(id))
    } else if(n == 2){
        return(as.cycle("(12)(36)(45)"))
    } else if(n == 3){
        return(as.cycle("(16)(24)(35)"))
    } else if(n == 4){
        return(as.cycle("(13)(25)(46)"))
    } else if(n == 5){
        return(as.cycle("(15)(26)(34)"))
    } else if(n == 6){
        return(as.cycle("(14)(23)(56)"))
    } else{
        stop()
    }
}

single <- function(v){ Reduce("*",lapply(v, offer)) }
janusz <- function(p){do.call("c",lapply(holder(p),single))}

(p <- c(as.word(id),rperm(10,6)))
janusz(p)

To check that this is a homomorphism we must verify that $\phi(ab)=\phi(a)\phi(b)$:

a <- c(rperm(10,6),as.word(id))
b <- c(as.word(id),rperm(10,6))
all(janusz(a*b) == janusz(a) * janusz(b))

To verify that $\phi$ is an isomorphism we need to check that it is 1-1. The best way I have found to do it is to systematically compare each element with each other element. This takes about an hour to run, too long for a package vignette.

n <- 720
M <- matrix(0,n,n)
x <- allperms(6)

system.time(for(i in 1:n){for(j in 1:n){M[i,j] <- janusz(x[i]) == janusz(x[j])}})
all(M[1:n,1:n] == diag(n))  # evaluates to TRUE but takes about an hour

There are many outer automorphisms, all conjugates of $\phi$. For example:

h <- function(x){janusz(x)^as.cycle("(162)(35)")}
h(as.cycle("(12"))
a <- c(rperm(10,6))
b <- c(as.word(id))
all(h(a*b) == h(a) * h(b))

References {-}

permutations documentation built on April 3, 2025, 7:09 p.m.