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Let's say a group $G$ acts on a set $S$, a set does not necessarily have an order, so to make sense of the group action I would need to provide the set with some kind of initial state or configuration right?

for example, a Rubik cube could have any initial state, it could be in any position when I pick it up and try to act on it with a group action, so the same action will create different results in my cube depending on its initial state.

Why there seems to be a focus on group actions acting on sets but there is no mention anywhere of how the set is arranged before acting on it? is it implied that a set has some natural arrangement even if by definition a set does not need to be ordered or arranged in any particular way?

It would seem that without the initial state, any action on a group cant be fully defined, since you dont know what element you are acting on.

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  • $\begingroup$ There is no focus on this because there is no need to order a set in order to act on it. $\endgroup$ Aug 18, 2017 at 19:27
  • $\begingroup$ There is no need to impose any condition on $S$. It just has to be nonempty. $\endgroup$
    – George Law
    Aug 18, 2017 at 19:29
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    $\begingroup$ In that specific example, you want to define the action using an order, but that is a very restrictive setup. $\endgroup$ Aug 18, 2017 at 19:36
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    $\begingroup$ @Joaquin Brandan, that is why you have notion of orbit $G.x = \{g.x\mid g\in G\}$. You can think of some particular $x$ as initial state and observe $G.x$. $\endgroup$
    – Ennar
    Aug 18, 2017 at 19:47
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    $\begingroup$ Perhaps it would be clearer if the Rubik cube example was made mathematical. Let $S$ be the set of all possible configurations of the cube and let $G$ be the group generated by all possible "moves" you can do on the cube. Now for $g$ in $G$ and $s$ in $S$ the action of $g$ on $s$ is the configuration $gs$ of the cube after doing the sequence of moves corresponding to $g$ to the cube starting in configuration $s$. $\endgroup$
    – Nex
    Aug 19, 2017 at 7:01

1 Answer 1

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I don't know how much this might help, but I gave it my best! I don't usually write this long, so this probably doesn't look good.. It can be made shorter but I am too lazy to revise it.

The standard definition of a group action is what follows.

An action of a group $G$ on a set $S$ is a function $\star : G \times S \to S$ with the properties:

  • $1 \star x = x$ for every $x \in S$.

  • $(g_1g_2)\star x = g_1 \star (g_2 \star x)$ for every $g_1,g_2 \in G$ and every $x \in S$.

"Action" is not a physical-continuous-kinda thing (it wouldn't be a mathematical concept then). You simply have a set $S$, a group $G$, and a mapping which assigns to each pair $(g,s) \in G \times S$ an elements in $S$, with special properties.

It's best to think of a group action as follows (equivalently to the definition above). You have a group $G$ and a set $S$, and a function $f: G\times S \to S$. Given $g \in G$, consider the function $b_g(s) = g\star s$. This function is one-one (multiply both sides by $g^{-1}$), and onto: if $s' \in S$, we know $g^{-1} \star s' \in S$, and $b_g(g^{-1} s') = g\star (g^{-1} \star s') = (gg^{-1})\star s' = 1 \star s' = s'$. Hence $b_g$ is a bijective map from $S$ to $S$, i.e. an element of $\text{Sym}(S)$. Now define $\eta: G \to \text{Sym} (S)$ by $\eta(g) = b_g$. This map is a homomorphism as you can check. It's called the homomorphism associated to the action. It can be shown also that any homomorphism from $G$ to $\text{Sym}(S)$ gives rise to a group action (in the obvious way). Hence an action is, essentially, a way of turning the elements of a group into transformations of a set.

Are the terms "order" or "initial state" of relevance in this context? It might happen that they affect the situation, or that the way we define the action is related to the order, but what we are after is what the general case is: in general, what we need is a way to make the elements of $G$ into bijective maps $S \to S$, and have these form a group together, and nothing else.

Here's to be more precise.

When you talk about "order" of a set $S$ or when you want to talk about an "initial state", you can only make sense (algebraically speaking) by talking about a binary relation $\le \subset S \times S $ satisfying the $3$ axioms of an order. In this case, you have yourself a pair $(S,\le)$, and you are talking about this pair and not the set itself. We don't define group actions on such objects (extremely technically, the pair $(S,\le)$ is a set and we can define a group action, but that would be stretching it!). We define group actions on raw sets, and don't care about what kind of order they might be endowed with.

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  • $\begingroup$ ok, so let's say that i have a group action with the group $\mathbb{Z} $ acting on $S = \{ A, B, C\}$ and that i map the element $1 \in \mathbb{Z}$ to a rotation to the right by 1 called $r^1$ because $r^1 \in Sym(S)$. what does $1$ acting on $S$ do? it cant necesarily be $ 1.S = \{ C, A, B\}$ since there is no default configuration in $S$ to tell me that the original configuration was $ A, B, C$ and therefore a rotation to the right would be $ C, A, B$ , that information is just not there on a set. $\endgroup$ Aug 20, 2017 at 20:53
  • $\begingroup$ Please don't throw in the word "action" randomly. I won't take an example from you unless you can actually prove that your declared "action" is actually an "action" in the mathematical sense. $\endgroup$
    – Cauchy
    Aug 21, 2017 at 2:39
  • $\begingroup$ But just to note, in your example, what on earth is "rotation to the right" suppose to mean? Can you give me this map explicitly? The image of $S$ under this map is not well-defined because this map itself is not well-defined at all. The source of your problem is that your so-called symmetries are not even well-defined functions on the set. $\endgroup$
    – Cauchy
    Aug 21, 2017 at 2:42
  • $\begingroup$ mhh, ok, let me give it a try. given $Sym(S)$ where $|S|=3$, i define "rotation to the right" as being the element $r^1 \in Sym(S)$ defined as a cycle of length $3$ described as $\begin{pmatrix} 1 & 2 & 3 & \\ 2 & 3 & 1 \end{pmatrix}$ in Cauchy's two-line notation. $\endgroup$ Aug 22, 2017 at 20:49

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