# Concrete examples of group actions.

First, a little motivation:

I have read the section on Group Actions in Dummit & Foote, the wikipedia page, and (countably many) other references. And seemingly without exception, they only offer rote and/or abstract examples, such as:

1. Let $ga = a$ for al $g \in G, a \in A$
2. The symmetric group $S_N$ acting on $A$ by $\sigma \cdot a = \sigma(a)$
3. Something about regular n-gons and $D_{2n}$
4. $g \cdot a = ga$...

I don't mean to undermine the importance of these examples, but I'm left with no hands-on experience with these things. Exercise $\S$ 1.7.8(b) in D&F says:
"Describe explicitly how the elements $(1 \ 2)$ and $(1 \ 2 \ 3)$ act on the six 2-element subsets of $\left \{1, 2, 3, 4 \right \}$.

How does a three cycle permute two elements?

Furthermore, what are some concrete examples of (computational exercises of) group actions?
Thanks.

• Pick any set $X$ and consider the symmetric group $S(X)$. Then any group homomorphism $G\to S(X)$ gives an action of $G$ on $X$ and, in fact, all actions of $G$ are of this form. Make a list of 25 sets and find homomorphisms! – Mariano Suárez-Álvarez Oct 4 '11 at 2:07
• @Mariano: Can you give me an example of such a homomorphism that isn't trivial or "canonical"? Say, on $X =$ {1, 2, 3, 4}?? – Altar Ego Oct 4 '11 at 2:18
• Regarding your question on how a three cycle permutes two elements : It does not. But permutations act on the set of 2-element subsets (as well as the set of $k$-element subsets for any $k$). Here, $\sigma.\{a,b\}$ is just the subset $\{\sigma(a), \sigma(b)\}$ (this is still a 2-element subset because $\sigma$ is bijective). – Joel Cohen Oct 4 '11 at 2:48
• How much linear algebra do you know? The set of invertible linear transformations on $\mathbb{R}^n$ form a group called the general linear group of degree $n$, written GL_n($\mathbb{R}$). This group acts on $\mathbb{R}^n$ as linear transformations! – Adam Saltz Oct 4 '11 at 3:01
• Also, you might be interested in the following paper. The author claims it is an "elementary" proof that subgroups of free groups are free based on group actions. However, it uses tensor products and I maintain that anything which uses tensor is not elementary...but anyway, it is a short paper which uses group actions to prove a big result, so it should be of interest to you. arxiv.org/abs/1006.3833 – user1729 Oct 4 '11 at 9:12

## 3 Answers

You ask, in a comment, for a non-obvious action on $X=\{1,2,3,4\}$. Let me give you, instead, a non-trivial action of $S_5$ on $X=\{1,2,3,4,5,6\}$: it is given by an homomorphism $\phi:S_5\to S_6$ such that \begin{align} (1,2)&\longmapsto(1,2)(3,4)(5,6) \\ (1,2,3,4,5)&\longmapsto(1,2,3,4,5) \end{align} You should check that this homomorphism is injective. In fact, you should find all ways in which $S_5$ can act on this $X$.

If you change the numbers $5$ and $6$ and look for examples, you'll have lots of fun.

• This is EXACTLY what I was looking for - a way to get my hands dirty, so to speak! Many thanks. – Altar Ego Oct 4 '11 at 2:30
• Since I've never done this before, I'm having a bit of trouble at the very start. The definition I have is a map $G \times X \to X$, written $g \cdot a$ (with certain properties...). In your example, what is $X$? I was expecting the domain and codomain to be the same ... ?? – Altar Ego Oct 4 '11 at 2:40
• If you have an homomorphism $\phi:G\to S(X)$, you can construct an action in your style by constructing the map $(g,x)\in G\times X\mapsto \phi(g)(x)\in X$. – Mariano Suárez-Álvarez Oct 4 '11 at 2:41

The dihedral group $D_8$, which is generated by elements $a$ and $b$ with the relations $a^2=1$, $b^2=1$, $(ab)^4=1$, acts on two coins resting side-by-side on a table, as follows: $a$ flips the coin on the left (so if it was heads, it becomes tails, and vice versa), and $b$ swaps the two coins (so the one that was on the left is now on the right, and vice versa).

• Thanks! I'll have to go put some money on the table :) – Altar Ego Oct 4 '11 at 5:25

Consider a cube in 3d space. There is a set of rotations of space that map the cube back onto itself - for example, rotation by 90 degree around an axis that cuts through two opposite faces in their middle. This set is in fact a group.

How does this group act on the set of 8 vertices? Name the vertices in some way and try to list the various permutations. Which group elements leave a vertex fixed? Which group elements leave two vertices fixed? Which vertices can those be? What if we consider the action on the edges of the cube instead of vertices? The faces? The main diagonals? What if we replace the cube by some other platonic solid such as an octahedron?

I hope this is less rote and abstract than the examples you saw so far.

It's hard to give examples without knowing a little about what you know and what you're comfortable with. If you know something about graphs, you can draw several small graphs and consider their automorphism groups - the different ways you can map their vertices to themselves while respecting the edge relationships. Or if you know some linear algebra and finite fields, consider a vector space of dimension 3 over the field $\mathbb{F}_2$ with two elements. How many subspaces of dimension 1 are there? What group naturally acts on those subspaces? How does the same group act on subspaces of dimension 2?

You also asked "How does a three cycle permute two elements?" Well, the 3-cycle $(1\, 2\, 3)$ maps 1 to 2, 2 to 3 and 3 to 1. Therefore, it maps the set $\{1,2\}$ to the set $\{2,3\}$, and the set $\{1,4\}$ to the set $\{2,4\}$. Can you take it from here?

• I don't even know enough to give you an informed statement of my background (maybe my comments on the question and other answer will shed some light on this), but I think the 3D example is within my grasp. Thanks for the answer, and for the clarification in the last paragraph. – Altar Ego Oct 4 '11 at 2:43