Group Actions

We're finally going to get back to some group theory today. Group actions give us ways for potentially unrelated things to interact with each other. If $G$ is a group and $S$ is any arbitrary set then a group action gives a way to combine an element of the group and an element of a set - that is if $g \in G$ and $s \in S$ then group actions define $g \cdot s$. The question, now, is what sort of thing should $g \cdot s$ equal? More specifically, should $g \cdot s$ equal an element of $S$ or an element of $G$? The way it works out is that it makes much more sense to make this operation map to an element of $S$.

Before I write out the definition (which is right below) I want to talk about some of the language used in the definition. The definition is going to talk about a binary function $G \times S \to S$ denoted by $\left ( g , s \right ) \mapsto g \cdot s$. We've seen lots of binary operations before, we just haven't called them that - the term "binary" tells us that the function needs two elements. Then the notation $G \times S$ means that the two elements we need are an element of $G$ and an element of $S$ (in that order) and the notation $\to S$ means that these two elements are mapped to an element of $S$. The last part of the notation just tells us how two write down this operation. If $g \in G$ and $s \in S$ then this binary function maps $G \times S$ to an element of $S$ that we denote by $g \cdot s$.

I think we are now ready to define group actions.

Definition:Group Action

If $G$ is a group and $S$ is a set then a group action of $G$ on $S$ is a binary function $G \times S \to S$ denoted by $\left ( g , s \right ) \mapsto g \cdot s$ that satisfies the following two axioms:

• $\left ( g h \right ) \cdot s = g \cdot \left ( h \cdot s \right ) \; \forall g,h \in G \; \text{and} \; \forall s \in S$
• $e \cdot s = s \; \forall s \in S$

If there exists a group action of $G$ on $S$ then we say that $G$ acts on $S$ and we call $S$ a $G$-set.

The definition of a group action tells us the rules that the operation must follow, but it doesn't tell us what this operation does or how exactly it maps the elements. This is okay, though, because when you start with a group, $G$, and a set, $S$, and you want to define a group action of $G$ on $S$, you need to specify and define the particular operation that you claim creates a group action and then show that it follows the rules in the definition. The best way to see how these group actions really work is through an example, and shown below is maybe the most "natural" example of a group action.

Example:S_n acts on Ω_n

Choose $n \in \mathbb{N}$ and consider the group $S_n$ and the set $\Omega_n$. Define the binary operation $S_n \times \Omega_n \to \Omega_n$ by $\pi \cdot k = \pi(k)$ for $\pi \in S_n$ and $k \in \Omega_n$. Let $\epsilon$ be the identity element of $S_n$. Now, observe the following:

• Choose $\pi,\tau \in S_n$ and $k \in \Omega_n$. Then
  \[ \left ( \pi \tau \right ) \cdot k = \left ( \pi \circ \tau \right )\left (k\right ) = \pi \left (\tau (k) \right ) = \pi \left (\tau \cdot k \right ) = \pi \cdot \left ( \tau \cdot k \right ) \]
  so that we have $\left ( \pi \tau \right ) \cdot k = \pi \cdot \left ( \tau \cdot k \right )$ and the first condition is satisfied.
• Choose $k \in \Omega_n$. Then $\epsilon \cdot k = \epsilon \left ( k \right ) = k$ so that $\epsilon \cdot k = k$ and the second condition is satisfied.
Thus this operation defines an action of $S_n$ on $\Omega_n$.

You can see how in the example with the symmetric groups, the action is extremely natural and, in fact, the definition of the symmetric group is basically built to support the action of $S_n$ on $\Omega_n$. Below gives another extremely natural example of a group action, but it requires some linear algebra, so if you haven't had any experience with matrix multiplication feel free to skip it.

Example:GL(n,R) acts on Rⁿ

Choose $n \in \mathbb{N}$, let $GL \left ( n,\mathbb{R} \right )$ be the general linear group over $\mathbb{R}$, and consider $\mathbb{R}^n$, the standard $n$-dimensional vector space over $\mathbb{R}$. Define the binary operation $GL \left ( n,\mathbb{R} \right ) \times \mathbb{R}^n \to \mathbb{R}^n$ by $M \cdot \mathbf{v} = M \mathbf{v}$ for $M \in GL \left ( n,\mathbb{R} \right )$ and $\mathbf{v} \in \mathbb{R}^n$ where $M \mathbf{v}$ represents standard matrix multiplication. Recall that the identity matrix, $I_n$, is the identity element of $GL \left ( n,\mathbb{R} \right )$ and observe the following:

• Choose $M,N \in GL \left ( n,\mathbb{R} \right )$ and $\mathbf{x} \in \mathbb{R}^n$. By definition, matrix multiplication is associative, so
  \[ \left ( M N \right ) \cdot \mathbf{x} = \left ( M N \right ) \mathbf{x} = M \left ( N \mathbf{x} \right ) = M \left ( N \cdot \mathbf{x} \right ) = M \cdot \left ( N \cdot \mathbf{x} \right ) \]
  and we have that $\left ( M N \right ) \cdot \mathbf{x} = M \cdot \left ( N \cdot \mathbf{x} \right )$ so the first condition is satisfied.
• Choose $\mathbf{x} \in \mathbb{R}^n$. Then we have $I_n \cdot \mathbf{x} = I_n \mathbf{x} = \mathbf{x}$ so that $I_n \cdot \mathbf{x} = \mathbf{x}$ and the second condition is satisfied.

Thus this operation defines an action of $GL \left ( n,\mathbb{R} \right )$ on $\mathbb{R}^n$.

The definition of a group action says that if $G$ acts on $S$ then $G$ must be a group but $S$ can be any arbitrary set. In the last example, the set was $\mathbb{R}^n$ which is a group (under vector addition) which is perfectly fine and actually happens a lot. In fact, we use actions of groups on themselves quite a bit - there are two important ones that we'll explore soon.

Group actions crop up a lot and get us a lot of fun properties that we'll keep exploring in future posts.

References

Previous Related Post: More About Homomorphisms

Text Reference: Gallain Chapter 29

Wikipedia: Group Action

Wolfram: Group Action

Planet Math: Group Action