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?

# Stabilizers in a permutation group

abstract-algebragroup-actionsgroup-theorypermutation-cyclespermutations

#### Related Solutions

It's pretty hard to answer this question when 'familiar' isn't defined further.

Every finite group (and thus every permutation group) has a composition series, which is unique in the sense that the length and composition factors of any two composition series are the same up to permutation and isomorphism. If you define familiar groups to be simple groups, then these quotients are the familiar pieces you're looking for.

Defining familiar as simple is a stretch, though. You'd be hard pressed to find someone who found the Held group especially recognizable.

The other problem is that two nonisomorphic groups can have the same composition factors. The factors do help you break the group into smaller, more recongizable chunks, but they don't tell you how those chunks interact.

There are so many variations on combinations of smaller groups that it is difficult to imagine how we could recognize them all without group presentations. Another way of breaking down a group into recognizable groups is by looking at its Sylow subgroups and seeing how they interact (this is *local group theory*). But for this, we need to understand $p$-groups enough to call them recognizable. My answer to this question should give you an idea of the magnitude of what we're dealing with just in the world of $p$-groups.

Take for example $\text{SmallGroup}(16,3)$. There is no other name for that group, as far as I know. It is isomorphic to $(\mathbb{Z}_4\times\mathbb{Z}_2)\rtimes \mathbb{Z}_2$, which shows us it can be split into those recognizable pieces. This decomposition uses semidirect products, however, which is basically the same as the relations / presentation concept we are trying to avoid.

So in summary, there are many a lot of different ways to divide a group up into pieces that are as familiar as possible, to gain understanding about it and how it works. However, to achieve a full description of the group, most of the time we need to use specific relations in a group presentation to show the way those pieces fit together.

Consider $L=\{x\in K, h(x)=x,h\in H\}$, $L$ is a subfield of $K$. The primitive element implies that $L=F(a)$, $H\subset Stab(a)$ since $a\in L$.

Let $g:g(a)=a$, let $x\in L, x=c_0+c_1a+c_2a^2+..+c_na^n$ $c_i\in F$ since $L=F(a)$, this implies that $g(x)=g(c_0+c_1a+c_2a^2+..+c_na^n)=c_0+c_1g(a)+..+c_ng(a_n)=c_0+c_1a+c_2a^2+..+c_na^n=x$, we deduce that for every $x\in L, g(x)=x$, implies $g\in H$ and $Stab(a)=H$.

## Best Answer

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