Example of a group action of group $D_6$ on set $A = \{1,2,3\}$

Question

I am an undergraduate currently taking Abstract Algebra. We have learned about groups and examples of groups (dihedral, symmetric, general linear, etc.). We are just starting to learn about subgroups in more detail. I am reading through Dummit and Foote's Abstract Algebra 3ed, and I will refer to their section 1.7 on Group Actions.

I am confused because an arbitrary group action is a map $\eta: G \times A \rightarrow A$ (written $\eta(g, a) = g \cdot a$) that satisfies: $$g_2 \cdot (g_1 \cdot a) = (g_1g_2) \cdot a,$$ $$1 \cdot a = a.$$

I understand that $\cdot$ is not the group operation of $G$. However, what can it be? How can we guarantee that an operation defined by us can consistently combine elements $g \in G$ and $a \in A$ to produce another element of $A$? Or, is this just a natural challenge of coming up with group actions?

In particular, let's say that I want to define a group action that describes how elements of $D_6$ permute the vertices of a regular triangle. I will represent the vertices of the regular triangle by a set $A = \{1,2,3\}$. Is it possible to write the group action I am describing with just the elements of $D_6$?

Or, must we observe that $D_6$ is isomorphic to $S_3$ and define our action using $S_3$ (because $S_3$ specifically contains all bijections from the set A to A and so elements of $S_3$ can naturally "affect" elements of $A$). Additionally, what is this process of representing an initial group by another group in order to more naturally define a group action?

Answer 1

Another thing is that the two conditions for a group action, "compatibility" and "identity", as they are often referred to, guarantee a homomorphism from the group into the symmetric group on as many points or vertices as elements in the $G$-set (the set being acted upon).

So in this case, we have $h:D_6\to S_3$. Thus the number of actions is precisely the number of such homomorphisms.

Since $D_6$ is $S_3$, we are considering all endomorphisms of $S_3$. Now, $S_3=\langle (12),(123)\rangle $. There's $4$ choices where to send $(12)$ and $3$ choices for $(123)$. That gives $4\cdot3=12$ total (don't forget the ones where one or both generators go to the identity).

Also, you can recover the action from the endomorphism: $$\sigma \cdot x=h(\sigma) (x)$$.

Answer 2

These are great questions! First of all your suspicion that it’s a “challenge of coming up with group actions” is partially correct. An arbitrary map $\eta: G \times A \to A$ is not likely to be group action. Those properties must be satisfied for all $g \in G$ and all $a \in A$, so usually the action “comes from” somewhere else, often a geometric or combinatorial structure.

To understand this better, you need to know that there’s an alternative way to define a group action: a representation. Here’s how.

Given a set $A$, we can look at the set of all symmetries (i.e. bijections) $A \to A$. But this set naturally has the structure of a group! If you compose two such symmetries, the resulting map is again a symmetry, and composition is associative. Moreover, there is always the identity symmetry and any bijection naturally has an inverse. Great! In fact, historically, this is the definition of a group. The modern definition abstracts these properties from those that are always satisfied by symmetries of a set.

So how do we turn a group action on a set into a representation in the symmetries of that set? And vice versa? Let’s set notation $\operatorname{Sym}(A) = \{ f: A \to A \mid f \text{ is a bijection} \}$, for the group of symmetries of $A$ described above. A representation of the group $G$ is a map $\rho: G \to \operatorname{Sym}(A)$. So for any $g \in G$, we need to understand $\rho(g): A \to A$. It’s typical to write this symmetry as $\rho_g = \rho(g)$. Let’s define $$ \rho_g(a) = g \cdot a. $$ What do the properties of the action tell us about this representation?

First, the identity property gives us $$ \rho_1(a) = 1 \cdot a = a $$ for all $a \in A$. This means that $\rho_1$ is the identity in $\operatorname{Sym}(A)$. In other words, $\rho$ satisfies $$ \rho(1_G) = 1_{\operatorname{Sym}(A)}. $$ What about the “associative” property? $$ \rho_g \bigl( \rho_h(a) \bigr) = \rho_g ( h \cdot a ) = g \cdot ( h \cdot a ) = (gh) \cdot a = \rho_{gh}(a) $$ for all $a \in A$. In other words, $\rho_g \circ \rho_h = \rho_{gh}$ as symmetries.

These two properties, taken together, define $\rho$ as a group homomorphism! In other words, via the group action, the representation exhibits a shadow of the structure of the group in the symmetries of the set $A$.

But we can reverse the correspondence. Given a representation, that is a group homomorphism $\rho:G \to \operatorname{Sym}(A)$, we can define an action by the same equation, but with information flowing in the opposite direction: $$ g \cdot a = \rho_g(a) $$ for all $g \in G$ and all $a \in A$. The homomorphism properties of the map $g \mapsto \rho_g$ immediately give us the properties that define a group action. These two notions are tautologically the same thing!

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To your example, where you define the symmetric group $S_3$ as group of bijections of the set $A = \{1, 2, 3\}$ under composition, the homomorphism $\rho: D_3 \to S_3$ (which happens to be an isomorphism) is the representation of $D_3$ on $A$ that corresponds to the action you desire!

$^*$I prefer to denote by $D_n$ the group of symmetries of the regular $n$-gon of order $2n$. Some authors prefer to call this group $D_{2n}$ to reflect its order instead.