Stabilizers in a permutation group

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?

• 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. Dec 27, 2021 at 22:33

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

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

• Isn't (245)*(245)*(124) = (125)? Dec 27, 2021 at 21:24
• 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. Dec 27, 2021 at 21:48