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Find the centralizer of $(123)$ in $S_6$.

Is there any software to calculate $C_{(123)}$ where $C_{(123)}$ denotes centralizer of $(123)$ in $S_6$.

Can anyone say how to write the code to find centralizer of $(123)$ in $S_6$?

Can it be done using SageMath?

I checked that it can be done by SageMath but it only lists the generators of the centralizer of $(123)$ in $S_6$.

Is there any code to list all the elements of the centralizer of $(123)$ in $S_6$?

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  • $\begingroup$ Is it $S_5$ or $S_6$? $\endgroup$ – Parcly Taxel Aug 3 at 9:13
  • $\begingroup$ @ParclyTaxel; edited the question $\endgroup$ – Math_Freak Aug 3 at 9:18
  • $\begingroup$ If you want the code to list all the elements, just add code at the end which calculates the subgroup explicitly from the generators you are given. $\endgroup$ – James Aug 3 at 10:08
  • $\begingroup$ @James;can you kindly give the code $\endgroup$ – Math_Freak Aug 3 at 10:15
  • $\begingroup$ Please remove one of the tags and insert the "GAP" tag. $\endgroup$ – Marc Bogaerts Aug 3 at 20:28
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The centralizer of $a = (1,2,3)$ is rather simple to describe. First assemble the permutations of $\{4,5,6\}$. There are $6$ of them. Then precede each of these with the permutations $a^0,a,a^3$ which gives you the $18$ elements of the centralizer. Here is the code in GAP:

gap> c1 := SymmetricGroup([4..6]);
Sym( [ 4 .. 6 ] )
gap> c2 := Group((1,2,3));
Group([ (1,2,3) ])
gap> c := Group(Union(c1,c2));
Group([ (), (5,6), (4,5), (4,5,6), (4,6,5), (4,6), (1,2,3), (1,3,2) ])

But Gap has the direct command 'Centralizer':

gap> S6 := SymmetricGroup(6);
Sym( [ 1 .. 6 ] )
gap> Centralizer(S6,(1,2,3)) = c;
true

GAP tag failed to add (more than 5 tags)

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  • $\begingroup$ How do you know that there will be 18 elements $\endgroup$ – Math_Freak Aug 4 at 2:52
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    $\begingroup$ Because there are $6$ permutations of $\{4,5,6\}$ and $3$ elements of the group generated by $(1,2,3)$. The centralizer consists of the products of an element of the first set with an element of the second set, giving $6 \times 3 = 18$ elements. $\endgroup$ – Marc Bogaerts Aug 4 at 4:33
  • $\begingroup$ My simple question is:: Why will the centralizer consist of "products of an element of the first set with an element of the second set" What is the mathematical explanation for this? $\endgroup$ – Math_Freak Aug 4 at 6:01
  • $\begingroup$ It are the only elements that commute with $(1,2,3)$, any other element of $S_6$, like e.g. $(3,4,5)$ or $(1,2)(3,6)$ does not commute. $\endgroup$ – Marc Bogaerts Aug 4 at 7:17
  • $\begingroup$ That does the give the required proof $\endgroup$ – Math_Freak Aug 4 at 12:24

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