# Thread: Orbits of an element and the Stabilizer of the element

1. ## Orbits of an element and the Stabilizer of the element

I am reading Dummit and Foote Section 4.1 on Group Actions and Permutation Representations.

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"Let G be a group acting on the non-empty set A. The relation on A defined by

a $\displaystyle \sim$ b if and only if a = g.b for some g $\displaystyle \in$ G

is an equivalence relation.

For each a $\displaystyle \in$ A, the number of elements in the equivalence class containing a is |G:$\displaystyle G_a$| , the index of the stabilizer of a."

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The proof that a = g.b is an equivalence relation is clear to me. I am interested (and struggling with) in some elements of D&F's proof regarding the second sentence. The second part of the proof proceeds as follows:

"To prove the last statement of the proposition we exhibit a bijection between the left cosets of $\displaystyle G_ a$ in G and the elements of the equivalence class of a. Let $\displaystyle \C_a$ be the class of a, so

$\displaystyle C_a$ = {g.a | g$\displaystyle \in$G}

Suppose b = g.a $\displaystyle \in$ $\displaystyle C_a$. Thus g$\displaystyle \G_a$ is a left coset of $\displaystyle G_a$ in G.

The map

b = g.a $\displaystyle \mapsto$ g$\displaystyle G_a$

is a map from $\displaystyle C_a$ to the set of left cosets of $\displaystyle G_a$ in G. This map is surjective since for any g $\displaystyle \in$ G the element g.a is an element of $\displaystyle C_a$. Since g.a = h.a if and only if $\displaystyle h^{-1}$g $\displaystyle \in$$\displaystyle G_a if and only if g\displaystyle G_a = h\displaystyle G_a, the map is injective, hence is a bijection. This completes the proof." ================================================== === The problems I am having is with the statements: a) This map is surjective since for any g \displaystyle \in G the element g.a is an element of \displaystyle C_a. b) Since g.a = h.a if and only if \displaystyle h^{-1}g \displaystyle \in$$\displaystyle G_a$ if and only if g$\displaystyle G_a$ = h$\displaystyle G_a$, the map is injective

Can anyone help by giving a very explicit proof or indication of why these statements are true?

Would be appreciative of help.

Peter

2. ## Re: Orbits of an element and the Stabilizer of the element

ok, we have a map from $\displaystyle C_a \to G/G_a$ from the orbit of a (that is all possible images g.a) to the set of left cosets of $\displaystyle G_a$.

that is, given b in the set A, such that b is in the equivalence class of a, then b~a, so b = g.a (g permutes A in such a way as to take a to b).

so we send b (=g.a) to the coset of $\displaystyle G_a$ that contains g, which is $\displaystyle gG_a$.

now, given an arbitrary coset of $\displaystyle G/G_a$, say $\displaystyle xG_a$, we know that x.a is some element of the orbit of a,

and our mapping takes x.a to $\displaystyle xG_a$, so x.a is a pre-image of $\displaystyle xG_a$ (pretty much because that's the way we defined the orbit of a).

now, suppose two elements of [a], let's call them b and c, get mapped to the same coset. we can write b = g.a, and c = h.a, so we have:

$\displaystyle gG_a = hG_a$. this means that $\displaystyle h^{-1}g \in G_a$.

but what is $\displaystyle G_a$? it's the stabilizer of a; that is, the set of g in G such that g.a = a. so, $\displaystyle h^{-1}g.a = a$.

therefore:

$\displaystyle h(h^{-1}g).a = h.a$
$\displaystyle (hh^{-1})g.a = h.a$
$\displaystyle eg.a = h.a$
$\displaystyle g.a = h.a$, that is, b = c.

perhaps this will be more clear if we see it "in action" (excuse the terrible pun). let's have A = {1,2,3,4}, and let G be the group D4 (where we are considering A as the set of vertices (corners) of the square).

now, just from looking at the rotations alone, we can see that the orbit of 1, is all of A. let's write the elements of D4 in permutation form:

{e, (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 2)(3 4), (1 4)(2 3), (1 3), (2 4)}.

what is Stab(1)?

well clearly e stabilizes (or "fixes") 1, and the only other element to do so is (2 4).

so Stab(1) = {e, (2 4)}. the left cosets are as follows:

Stab(1) = {e, (2 4)}
(1 2 3 4)Stab(1) = {(1 2 3 4), (1 2)(3 4)}
(1 3)(2 4)Stab(1) = {(1 3)(2 4), (1 3)}
(1 4 3 2)Stab(1) = {(1 4 3 2), (1 4)(2 3)}, and our bijection is:

1<--->Stab(1)
2<--->(1 2 3 4)Stab(1) (every element of (1 2 3 4)Stab(1) takes 1-->2)
3<--->(1 3)(2 4)Stab(1) (every element of (1 3)(2 4)Stab(1) takes 1-->3)
4<--->(1 4 3 2)Stab(1) (every element of (1 4 3 2)Stab(1) takes 1-->4).

and indeed we see that |[1]| = |A| = [D4:Stab(1)] = 4.

you can look at it this way: if g,h take a to different elements of A, then $\displaystyle gG_a, hG_a$ give rise to different cosets of $\displaystyle G_a$.

3. ## Re: Orbits of an element and the Stabilizer of the element

Thanks for that, Deveno

Peter

4. ## Re: Orbits of an element and the Stabilizer of the element

Quick question:

You write:

"and our mapping takes x.a to x$\displaystyle G_a$ , so x.a is a pre-image of x$\displaystyle G_a$(pretty much because that's the way we defined the orbit of a)."

Dummit and Foote define 'orbit' as follows:

"The equivalence class {g.a | g$\displaystyle \in$G} is called the orbit of G contaning a."

How does this definition relate to the fact that g.a is taken to g$\displaystyle G_a$?

Peter

5. ## Re: Orbits of an element and the Stabilizer of the element

if G = {g1,g2,g3,....} then the orbit of a is {g1.a,g2.a,g3.a,....} and the cosets of $\displaystyle G_a$ are $\displaystyle \{g_1G_a,g_2G_a,g_3G_a,\dots\}$.

it's obvious we can send the elements of the orbit of a to the corresponding coset, what isn't obvious is that the duplicates in each list correspond to each other.