# Orbits - Dummit and Foote and Fraleigh

• Dec 23rd 2011, 09:38 PM
Bernhard
Orbits - Dummit and Foote and Fraleigh
In section 4.1 Group Actions and Permutation Representations Dummit and Foote define an orbit as follows:

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"Let G be a group acting on the nonempty set A.

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

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However, Fraleigh [A First Course in Abstract Algebra] defines Orbits in Section 9: Orbits, Cycles and Alternative Groups as follows: [calling it an orbit of a permutation]

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"Each permutation $\sigma$ of a set A determines a natural partition of A into cells with the property that a,b $\in$ A are in the same cell if an only if b = $\sigma^n(a)$ for some n $\in$ Z"

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Are these two definitions equivalent. If so, can someone please explicity & formally demonstrate how/why they are equivalent?

Peter
• Dec 24th 2011, 01:15 AM
girdav
Re: Orbits - Dummit and Foote and Fraleigh
If $G$ is a group acting on $A$, we have a morphism between $G$ and the group of permutations of $A$.
• Dec 24th 2011, 01:27 AM
Bernhard
Re: Orbits - Dummit and Foote and Fraleigh
Thanks

I do understand that but I am not sure how this fact makes a link between the two definitions.

Can you be more explicit about how the two definitions are equivalent?

Peter
• Dec 24th 2011, 06:48 AM
Deveno
Re: Orbits - Dummit and Foote and Fraleigh
there are two parallel ways to view an action of a group G on a set A:

1. a homomorphism from G to Sym(A), the set of bijective functions on A. if A is finite, then this is a homomorphism from $G \to S_{|A|}$.

this homomorphism is usually writen $g \to \sigma_g$.

2. a mapping from $G \times A \to A$ (written g.a, or g(a)) such that:

g.(h.a) = gh.a
e.a = a

these two definitions of a group action are equivalent, given an action in the second sense, we can define a map from G to T(A), the set of all functions A-->A by:

$g \to \sigma_g$ where $\sigma_g(a) = g.a$. is $\sigma_g$, defined in this way, a bijection?

well, suppose $\sigma_g(a) = \sigma_g(b)$. then g.a = g.b, so $a = e.a = (g^{-1}g).a = g^{-1}.(g.a) = g^{-1}.(g.b) = (g^{-1}g).b = e.b = b$

so $\sigma_g$ is injective. is $\sigma_g$ surjective? that is, given any b in A, can we find a in A with $\sigma_g(a) = b$?

this amounts to asking, given g, is there a in A so that g.a = b? let a = $g^{-1}.b$ then it's clear that $g.a = g.(g^{-1}.b) = b$.

so $\sigma_g$ is a bijection, for every g in G. so $\sigma_g \in S_A$. is the map $g \to \sigma_g$ indeed a homomorphism of G into $S_A$?

let's call this map φ. we need to verify that φ(gh) = φ(g)φ(h), that is:

$\sigma_{gh} = \sigma_g \circ \sigma_h$. but, for any a in A:

$\sigma_{gh}(a) = (gh).a = g.(h.a) = \sigma_g(\sigma_h(a)) = (\sigma_g \circ \sigma_h)(a)$.

so given way number 2 of looking at a group action, we have way number 1. now if we start with way number 1, and we define g.a to be $\sigma_g(a)$, then the rules:

g.(h.a) = (gh).a
e.a = a

are a consequence of φ being a homomorphism (since e.(_) is clearly the identity of Sym(A)). the two ways are equivalent, we can regard an action as "an induced permutation group" on A, or as a way to "multiply a set by a group".

now, Dummit and Foote, in their definition of orbit, are viewing an orbit of a as the set of "G-multiples of a". this is way #2 of looking at an action. Fraleigh, is looking at the orbit of a, as the possible images of a under repeated permutation by σ (way #1).

you see, a group action may only yield a subgroup of $S_A$ (as a factor group of G). Fraleigh's definition can be used to refer to any permutation group, whether or not we get it from an action. it might seem that these are two different concepts, but remember, given a permutation group G on A (a subgroup of $S_A$), there is a "natural" action of G on A (the homomorphism is just the inclusion map of G in $S_A$, that is g.a = g(a)).

if G is a subgroup of $S_A$, then the two definitions of orbit coincide (via the natural action of G on A). certainly if g.a = b, then g(a) = b (we can take n = 1). on the other hand, if $g^k(a) = b$, then $g^k$ is in G, so $b = (g^k).a$. we don't get any "new" members of [a] from Fraleigh's definition, because <g> is always contained in G, whenever g is.
• Dec 24th 2011, 01:20 PM
Bernhard
Re: Orbits - Dummit and Foote and Fraleigh
Thanks for the extremely helpful post!!!

Just working through it now!

Peter