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Suppose we have a permutation group $G = \langle (245), (123),(124) \rangle$ with generators. It seems quite obvious that the stabilizer of 3 is the subgroup $\langle (245),(124) \rangle$, but I am not sure how to prove it. It's obvious that $\langle (245), (124) \rangle \leq{\rm Stab}_G(3)$, but how does one show that a sequence of permutations that move $3$ somewhere in between and then bring it back to its starting position is still part of the subgroup?

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    $\begingroup$ It is not too hard to show that $G = {\rm Alt}(5)$ (the alternating group), and the two generators that stabilize $3$ generate ${\rm Alt}(4)$, so that must be the complete stabilizer of $3$. But general problems of this type are moderately difficult. There is a general method called the Schreier-Sims Algorithm. $\endgroup$
    – Derek Holt
    Dec 27, 2021 at 22:33

2 Answers 2

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First we notice that $G\leq A_5$, the alternating group of degree five. Let's make use of this post. Since $(123)$ and $(245)$ don't fix a common element of $\{1,2,3,4,5\}$ and $(123)\neq (245)^{-1}$, we have that $\langle (123),(245)\rangle=A_5$. Therefore $G=A_5$.

Remember that $A_5$ is made of the identity element, 3-cycles, 5-cycles and products of two 2-cycles. By doing some combinatorics, we get that $\text{Stab}_{A_5}(3)$ consists of the identity element, eight 3-cycles and three products of two 2-cycles. This is exactly the same element structure of $A_4$.

We conclude that $\text{Stab}_{A_5}(3)\cong A_4$.

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The previous solution was incorrect because I used GAP incorrectly. Here is a correct sequence of GAP commands. In general GAP is very good for dealing with small finite groups (and even for some infinite groups):

gap> G:=SymmetricGroup(5);

Sym( [ 1 .. 5 ] )

gap> H:=Subgroup(G,[(1,2,4), (2,4,5),(1,2,3)]);

Group([ (1,2,4), (2,4,5), (1,2,3) ])

gap> Elements(H);

[ (), (3,4,5), (3,5,4), (2,3)(4,5), (2,3,4), (2,3,5), (2,4,3), (2,4,5), (2,4)(3,5), (2,5,3), (2,5,4), (2,5)(3,4), (1,2)(4,5), (1,2)(3,4), (1,2)(3,5), (1,2,3), (1,2,3,4,5), (1,2,3,5,4), (1,2,4,5,3), (1,2,4), (1,2,4,3,5), (1,2,5,4,3), (1,2,5), (1,2,5,3,4), (1,3,2), (1,3,4,5,2), (1,3,5,4,2), (1,3)(4,5), (1,3,4), (1,3,5), (1,3)(2,4), (1,3,2,4,5), (1,3,5,2,4), (1,3)(2,5), (1,3,2,5,4), (1,3,4,2,5), (1,4,5,3,2), (1,4,2), (1,4,3,5,2), (1,4,3), (1,4,5), (1,4)(3,5), (1,4,5,2,3), (1,4)(2,3), (1,4,2,3,5), (1,4,2,5,3), (1,4,3,2,5), (1,4)(2,5), (1,5,4,3,2), (1,5,2), (1,5,3,4,2), (1,5,3), (1,5,4), (1,5)(3,4), (1,5,4,2,3), (1,5)(2,3), (1,5,2,3,4), (1,5,2,4,3), (1,5,3,2,4), (1,5)(2,4) ]

gap> Number(Elements(H));

$60$

So $H$ is equal to $A_5$.

The stabilizer of $3$ in $H$ is then the subgroup $A_4$ acting on $\{1,2,4,5\}$. It has $12$ elements ($4!/2$).

Check that the subgroup generated by $(1,2,4), (2,4,5)$ has $12$ elements:

gap> P:=Subgroup(G,[(1,2,4), (2,4,5)]);

Group([ (1,2,4), (2,4,5) ])

gap> Number(Elements(P));

12

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  • $\begingroup$ Isn't (245)*(245)*(124) = (125)? $\endgroup$
    – AlexC
    Dec 27, 2021 at 21:24
  • $\begingroup$ Indeed, $(2,4,5)(2,4,5)=(2,5,4), (2,5,4)(1,2,4)=(1,2,5)$. Most probably I used GAP incorrectly. $\endgroup$
    – markvs
    Dec 27, 2021 at 21:48

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