# Understanding the definition of the Group action.

Quoting from Wolfram definition of group action -

In general, a group action is when a group acts on a set, permuting its elements , so that the map from the group to the permutation group of the set is a homomorphism.

Is it always that when a group acts on a set it will always permute its elements ?. Like permutation word comes up then I can think of permutation group $S_{n}$, but like what happens when some other group acts on the set?

Also, a map from the group to the permutation group of the set is a homomorphism is like saying that there is a homomorphism between the group and the set of elements, hmm is it intuitive?

Also are there any stunning group action applications or visualizations?

• I know there are other equivalent definitions, but I prefer to define a group action of a group $G$ on a set $X$ as a homomorphism $\phi:G \to {\rm Sym}(X)$. – Derek Holt Mar 31 '18 at 8:29

## 3 Answers

If a group $G$ acts on any set $X$, the reason that the action of a group element $g \in G$ is a permutation --- i.e. a bijection --- is because it has an inverse, namely the action of $g^{-1}$. By definition of group action, the action of $g \in G$ followed by the action of $h \in G$ is equal to the action of $hg$; that's why we say the action is a homomorphism.

As a consequence, the action of $g$ on $X$, composed with the action of $g^{-1}$ on $X$ is equal to the action of $g^{-1} g = \text{Id}$ on $X$, which is the identity map. Similarly the action of $g^{-1}$ composed with the action of $g$ is equal to the action of $g g^{-1} = \text{Id}$. It follows that the action of $g$ is a permutation.

As for "stunning" examples, any time you have a highly symmetric object, i.e. the 2-sphere $S^2 = \{(x,y,z) \mid x^2 + y^2 + z^2 = 1\}$, the symmetries of $S^2$ form an interesting and often useful group. In this case, the symmetries of $S^2$ are the orthogonal group $O(3)$, the group of $3 \times 3$ matrices $M$ such that $M M^T = M^T M = \text{Id}_{3 \times 3}$. The usual vector multiplication formula $Mv$, restricted to unit vectors $v=\langle x,y,z\rangle$, defines the action of $O(3)$ on $S^2$. There are many, many other interesting actions like this in geometry.

"Permutation" is not a good word unless you are specifically focused on actions on finite sets. A group action of a group $G$ on a set $X$ is equivalent to a group homomorphism $G\to\mathsf{Bij}(X)$, where $\mathsf{Bij}(X)$ is the group of bijections, i.e. invertible functions, on $X$. When $X$ is a finite set, then this will be the group of permutations of the elements of $X$. The homomorphism $G\to\mathsf{Bij}(X)$ is not like a homomorphism from $G$ to $X$ as that doesn't even make sense: no group structure is defined or required for $X$. The homomorphism maps $g\in G$ to an invertible function $X\to X$ where the group structure is defined by function composition.

Historically, mathematicians usually cared about group actions. It was only later that the idea of a group became a stand-alone concept. Noether's Theorem completely revolutionized how we conceptualize physics and it is about (Lie) group actions. It produces results like the conservation of energy is due to symmetry under time translation. The Standard Model is a gauge theory and the gauge symmetries referred to are with respect to (Lie) group actions. This is closely related to Noether's theorem.

• I think using "automorphisms" in place of permutations is actually bad practice. If $X$ has some sort of algebraic structure then $\mathrm{Aut}(X)$ has a very specific meaning, and it is not required that a group action will specifically act as a group of automorphisms. On the other hand, there is no problem talking about permutations of an infinite set: they are the bijections, aka the invertible functions. – Morgan Rodgers Mar 31 '18 at 4:35
• @MorganRodgers If $X$ is something other than a plain set, then this gives a coherent notion of "group action" just in an enriched context. But that is a bit of an advanced point, so I'll reword it a bit. I still dislike "permutation" for bijections on infinite and especially uncountably infinite sets. – Derek Elkins Mar 31 '18 at 4:43

There's always a trivial action from any group to any set, which doesn't transitively permute everything. In general, we can think of a group action of $G$ on a set $X$ as a map $\phi:G\to S_X$, the symmetric group on elements of $X$. The group action, therefore, contains all the permutations of $X$ when the map $\phi$ is surjective. We don't want to say that this group action is a homomorphism between $G$ and $X$, because $X$ doesn't have any native structure. It's $S_X$, the permutation group, that has the structure that we want to deal with, and we already have a word for group homomorphisms.

In general, group actions are the way we study finite groups. A group acting on itself by conjugation gives us the class equation, which says that $|G|=|Z(G)|+\sum|\kappa_x|$, where $\kappa_x=\{y\in G\,|\,\exists g\in G \text{ s.t. } gxg^{-1}=y\}$. This approach also gives us the Sylow theorems, which are a sort of partial converse to Cauchy's theorem: for each prime $p$ with $p^k\big||G|$, there is a subgroup of $G$ of size $p^k$. In fact, one of the proofs for Cauchy's theorem also uses group actions and the orbit-stabilizer theorem.