What Is the Largest Cyclic Subgroup of S_n?

JoanBraidy · May 26, 2010

I became interested in this question a few weeks ago, I couldn't find much on it

basically I've realized it's equivalent to finding for each n a partition of n say

x_1,x_2,...,x_k such that x_1+x_2+...+x_k=n and lcm(x_1,...,x_k) is maximum

(because you can then take the subgroup <(123...x_1)(x_2...x_3)...(x_(k-1)...x_k)>

however computing this seems pretty much impossible

does anyone have any insights on this problem?

eok20 · May 27, 2010

I don't think there is a closed form. Check out this paper: http://www.cs.ust.hk/mjg_lib/Library/Miller87.pdf

Martin Rattigan · May 27, 2010

EDIT: Both the formula and programme in this post are incorrect (see sequel)

For the numbers P_n=\sum_{r=1}^{n}p_r the largest order of an element in S_{P_n} would be \Pi_{r=1}^{n}p_r. It's also clear that the largest order for S_n is monotonic non decreasing with n.

In the gaps between the P_n the best places to distribute "remainder" between the existing cycles and how effective that can be will depend on both the remainder and the differences between p_{n+i}, i=1,2,\dots and p_{n-i}, i=0,1,2,\dots. Because of the awkwardness of the sequence of primes I would be surprised if a handy formula could be found for these.

I've put together a cheap and nasty javascript routine (below) to calculate the first few values, but this bows out after n=270 because the javascript integer accuracy is exceeded. Even if an unlimited accuracy routine for the arithmetic were included, I wouldn't hold out much hope of a BFI routine like this getting very far owing to the eventual size of the numbers if nothing else. (I haven' t attempted to quantify this either.)

Code:

<script>
big=Math.pow(2,53)
toobig=false
function hcf(p,q){
  var u=Math.max(p,q)
  var v=Math.min(p,q)
  while(1){
    h=v
    v=u%v  
    if(!v)break
    u=h
  }
  return h  
}
function lcm(p,q){
  var wk=Math.min(p,q)/hcf(p,q)
  if(big/wk>=Math.max(p,q))return Math.max(p,q)*wk
  return 0
}
function cpy(u){
  if(!(u instanceof Array)){if(typeof(u)!='object')return u;wobbler()}
  var r=[]
  for(var i in u){
    r[i]=cpy(u[i])
  }
  return r
}
function part(n){
  if(!('done' in this))done=[]
  if(n in done)return cpy(done[n])
  var max=[n]
  max.v=n
  for(var i=n-1;i>=1;i--){
    var rest=n-i 
    rest=part(rest) 
    if(rest)rest.v=lcm(i,rest.v)             
    if(!(rest&&rest.v)){toobig=true;return null}
    if(rest.v>max.v){
      max.v=rest.v
      max.length=0
      var mx=0
      var idone=0
      for(var j=0;j<rest.length;j++){
        if(!idone&&i>=rest[j]){
          max[mx]=i
          mx++
          idone=1          
        }
        if(i!=rest[j]){
          max[mx]=rest[j]
          mx++
        }
      }
      if(!idone)max[mx]=i
    }    
  }
  done[n]=cpy(max) 
  return max
}
function doit(n,testlim,parm){
  if(!parm)parm=[100,10000]
  var pglim=parm[0]?parm[0]:100
  var gap=parm[1]?parm[1]:10000  
  if(!testlim)testlim=n
  var upper=n-(n%pglim)+pglim
  if(!('upto' in this))upto=document.getElementById('upto')
  if(!('blurb' in this))blurb=document.getElementById('blurb')
  var show=[]
  sp='&nbsp;';sp+=sp;sp+=sp;
  for(;n<Math.min(upper,testlim+1);n++){
    var maxpart=part(n)
    if(toobig){show[show.length]='<br>accuracy limit exceeded - terminating';break}
    show[show.length]=n+sp+maxpart.v+sp+'('+maxpart+')'
  }
  upto.innerHTML=n
  blurb.innerHTML=(n>pglim?'...<br>':'')+show.join('<br>')+(n<testlim?'<br>...':'')
  if(n<testlim&&!toobig)setTimeout('doit('+(n+1)+','+testlim+','+gap+')',gap)
}
</script>
<body onload="doit(1,1000000,[1000,1])">
<div>up to:</div>
<div id=upto></div>
<div>===========<br>n&nbsp;&nbsp;&nbsp;max_order&nbsp;&nbsp;&nbsp;max_cycle</div>
<div id=blurb></div>

JoanBraidy · May 27, 2010

interesting

i don't get why it's obvious that s_sum(p_i) has productp_i as its largest order

couldnt there exist another partion of sum(p_i) whose lowest common multiple is bigger

say 2+3+5=10=7+3

but 2*3*5=30 and 7*3=21

so S_(7+3) would have largest order 21 according to you but

then it also has an element of order 30 just take (12)(345)(678910)

so anytime you can write n as a sum of primes 2 different ways this would probably

happen

Martin Rattigan · May 28, 2010

By p_i I meant the i^{th} prime, not just any old prime, so P_3=10 in that notation, and the max cycle order for 10 would be given by {\Pi_{i=1}^{3}p_i=2.3.5=30}.

If n=14, P_3<n<P_4, so you have a "remainder" of 4 to adjust the factors of 2.3.5. The best way is add 2 to each of 2,5 or, equivalently, 1 to each of 2,3 and 2 to 5. In each case the new factors are relatively prime to the rest, but this rests partly on the fact that 7-5=2, and similar relations between primes for higher numbers, so I wouldn't be optimistic about finding a simple formula for values of n between the P_n.

But you're right that the formula is not obvious. In fact it breaks down for P_k when k\geq 13, possibly before. In this case splitting 2+41 as 1+8+9+25 still gives factors that are relatively prime to anything except 2,3,5, but 1.8.9.25>2.41\times 3.5, hence {8.9.25\times 7.11.13\dots 37>2.3.5\dots 41}. So all I can actually say is:
n\geq P_k\Rightarrow\text{max cycle length in }S_n\geq \Pi_{i=1}^{k}p_i

and this bound will no doubt get progressively worse as n increases beyond {8+9+25+7+11+13+\dots +37=229} (at most).

The programme posted is also flawed (I think for similar reasons).

Goes to show it would probably be a good idea if I thought before posting.

Martin Rattigan · May 28, 2010

I'm attaching a (hopefully) corrected version of the program previously posted. This version slows to a snails pace after about n=60 and would probably take all day to get up to the limit of javascript integer accuracy (I haven't tried it).

First difference from previous program is at n=29.

Challenge is to get it to run at reasonable speed or prove it never will.

Code:

<script>
big=Math.pow(2,53)
toobig=false
done=[];done.maxix=1
show=[]
function hcf(p,q){
  var u=Math.max(p,q)
  var v=Math.min(p,q)
  while(h=v){    
    v=u%v  
    if(!v)break
    u=h
  }
  return h  
}
function lcm(p,q){
  var wk=Math.min(p,q)/hcf(p,q)
  if(big/wk>=Math.max(p,q))ret=Math.max(p,q)*wk
  else{
    toobig=true
    ret=0
  }
  return ret
}
function part(n){ 
  if(!done[n]){
    var cycs=[[n]]
    cycs[0].v=n
    cycs.max=0
    for(var i=n-1;i>=Math.floor((n+1)/2);i--){
      var icycs=part(i),jcycs=part(n-i)
      for(var ic in icycs)for(var jc in jcycs){
        if(ic!='max'&&jc!='max'){
          var icyc=icycs[ic],jcyc=jcycs[jc]
          var vtry=lcm(icyc.v,jcyc.v)            
          if(toobig)break;
          else{
            cycins=cycs.length
            for(var cx in cycs){
              if(cx!='max'){
                var cyc=cycs[cx]
                if(!(cyc.v%vtry)){cycins=undefined;break}              
                if(!(vtry%cyc.v)){cycins=cx;delete cycs[cx]}
              }
            }
            if(cycins!=undefined){ 
              if(cycins>done.maxix)done.maxix=cycins   
              if(!cycs[cycs.max]||cycs[cycs.max].v<vtry)cycs.max=cycins
              cyc=cycs[cycins]=[]
              cyc.v=vtry
              cyc.length=0
              var ix,jx; ix=jx=cx=0
              while(ix<icyc.length||jx<jcyc.length){
                if(ix<icyc.length&&(jx>=jcyc.length||icyc[ix]>=jcyc[jx]))cyc[cx++]=icyc[ix++]
                else cyc[cx++]=jcyc[jx++]
              }
            }
          }
        }    
      }
      if(toobig)break;
    }    
    done[n]=cycs    
  }
  return done[n]
}
function doit(n,lim,showix){
  var cyc=part(n)
  cyc=cyc[cyc.max]
  sp='&nbsp;';sp+=sp;sp+=sp;    
  if(toobig)show[show.length]='<br>accuracy limit exceeded - terminating'
  else show[show.length]=n+sp+cyc.v+sp+'('+cyc+')'    
  upto=document.getElementById('upto')
  upto.innerHTML=n+(showix?' max cycle index='+done.maxix:'')  
  blurb=document.getElementById('blurb')  
  blurb.innerHTML=show.join('<br>')
  n++
  if(!(toobig||(lim&&n>lim)))setTimeout('doit('+n+','+lim+','+showix+')',1)
}
</script>
<body onload="doit(1,1000)">
<div>up to:</div>
<div id=upto></div>
<div>===========<br>n&nbsp;&nbsp;&nbsp;max_order&nbsp;&nbsp;&nbsp;max_cycle</div>
<div id=blurb></div>

What Is the Largest Cyclic Subgroup of S_n?

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