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Find all numbers $n$ such that $S_7$ contains an element of order $n.$

Identity is the only element of order $1.$So $n=1$ is possible.

Case 1: Elements that can be written as a unique cycle of length$≥2$:Cycles of length 2 to 7 can exists whence the corresponding values of n are 2 to 7.

Case 2: Elements that can be written as a product of two disjoint cycles of length$≥2$:Here the possible elements are $σ_1 σ_2$ where $σ_1$ and $σ_2$ are disjoint cycles of length$≥2$,such that $4≤|σ_1 |+|σ_2 |≤7$. So in this case the possible values of n are {lcm(|σ_1 |,|σ_2 | ): σ_1 & σ_2 are disjoint cycles of length≥2 with 4 ≤|σ_1 |+|σ_2 |≤7}={lcm(n_1,n_2 ): n_1,n_2∈Z^+-{1} with 4≤n_1+n_2≤7} (Since there always exist disjoint cycles of length n_1 and n_2 such that n_1,n_2∈Z^+-{1} with 4≤n_1+n_2≤7)={2,6,4,10,3,12}.

Case 3: Elements that can be written as a product of three disjoint cycles of length≥2:Here the possible elements are σ_1 σ_2 σ_3 where σ_1 and σ_2 are disjoint cycles of length≥2,such that 6≤|σ_1 |+|σ_2 |+|σ_3 |≤7. So in this case the possible values of n are {lcm(|σ_1 |,|σ_2 |,|σ_3 | ): σ_1,σ_2,σ_3 are disjoint cycles of length≥2 with 6≤|σ_1 |+|σ_2 |+|σ_3 |≤7}={lcm(n_1,n_2,n_3 ): n_1,n_2,n_3∈Z^+-{1} with 4≤n_1+n_2+n_3≤7} (Since there always exist disjoint cycles of length n_1,n_2,n_3 with n_1,n_2,n_3∈Z^+-{1} with 4≤n_1+n_2+n_3≤7)={2,6}. We further note that there^' s no element in S_5 whose representation as a product of disjoint cycles contains 4 or more cycles of length≥2.Consequently the above three cases exhaust all the posibilites for the order of the elements of S_5.Thus n=1,2,3,4,5,6,7,10,12.

Am I correct?

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    $\begingroup$ Looks good to me. $\endgroup$ – Jonathan Y. Aug 31 '13 at 10:35
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    $\begingroup$ @JonathanY.: Don't u think that finding the numbers $n$ in which the group doesn't have any elements of that order is easier? $\endgroup$ – mrs Aug 31 '13 at 10:57
  • $\begingroup$ If you want to do in this way, this is fine. there are other easier ways for this. Good job anyways :) $\endgroup$ – user87543 Aug 31 '13 at 10:58
  • $\begingroup$ @BabakS. That didn't occurred to me. I will take a try with your clue. $\endgroup$ – Sriti Mallick Aug 31 '13 at 14:27
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I, by using a computational software in Group Theory called GAP, could checked that:

gap> e:=Elements(SymmetricGroup(7));;
 s:=DivisorsInt(Size(e));;
 Print("The Group does not contain any elements of order", ":", "  ");
 for n in s do if Size(Filtered(e,t->Order(t)=n))=0 then Print( "n=", 
     n,",  "); fi;
 od;

The Group does not contain any elements of order:  n=8,  n=9,  n=14,  n=
15,  n=16,  n=18,  n=20,  n=21,  n=24,  n=28,  n=30,  n=35,  n=36,  n=40,  n=
42,  n=45,  n=48,  n=56,  n=60,  n=63,  n=70,  n=72,  n=80,  n=84,  n=90,  n=
105,  n=112,  n=120,  n=126,  n=140,  n=144,  n=168,  n=180,  n=210,  n=
240,  n=252,  n=280,  n=315,  n=336,  n=360,  n=420,  n=504,  n=560,  n=
630,  n=720,  n=840,  n=1008,  n=1260,  n=1680,  n=2520,  n=5040,
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  • $\begingroup$ Size(Set([]) is a very strange way to write zero! You could also write NumberOfSolutions(a^7+b^7==c^7) :-) $\endgroup$ – Mariano Suárez-Álvarez Sep 25 '13 at 16:39
  • $\begingroup$ @MarianoSuárez-Alvarez: I fixed it. :D $\endgroup$ – mrs Sep 25 '13 at 16:41
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    $\begingroup$ Notice that you can say simply Set(List(SymmetricGroup(7), Order)); to get the list of orders of elements (this computes the order of each element of the group exactly once) $\endgroup$ – Mariano Suárez-Álvarez Sep 25 '13 at 16:43
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    $\begingroup$ @MarianoSuárez-Alvarez: you are right assuming that GAP exploits here the well-known description of classes of $S_n$. For example, my code does $S_{20}$ instantly. That may not be the case for an arbitrary group, of course. For 2-groups of orders <=512 from Small Groups Library it's faster to run over all their elements than to calculate classes first, but as soon as the group size grows, using classes may start to win. If calculating classes is too hard, likely the group is so large that iterating over all its elements it time-consuming too. $\endgroup$ – Alexander Konovalov Sep 26 '13 at 19:56
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    $\begingroup$ P.S. and of course your mileage may vary dependently on whether the group is given as fp-group, or pc-group, or permutation group, or matrix group... $\endgroup$ – Alexander Konovalov Sep 27 '13 at 7:21

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