# Stabilizers in a permutation group

abstract-algebragroup-actionsgroup-theorypermutation-cyclespermutations

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?

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