I am going over the group action and unfortunately I am not sure if I understand it very well. So for this reason, I have a few questions that possibly could help me to understand this concept better. A group action is defined as: Group $G$ acts on a set $X$ if there is a homomorphism $\sigma:G \rightarrow S_X$. I am aware group action has other definitions but for me this one is easier to understand.

Now $S_X$ is the group of permutations of $X$. It is not crystal clear what is meant by this to me. Is $S_X$ the same thing as $S_{|X|}$? When we talk about $S_n$, I have a clear mental picture; $n$ is an integer and I think of all the permutations with $n$ elements. But in $S_X$, $X$ is not an integer but a set. So suppose $X=\{1,2,3\}$. Is $S_X$ the same thing as $S_3$ in this case?

The other part of my question is about what the definition implies: it says "... if there is a homomorphism". Is it possible that homomorphism does not exist? Is the homomorphism unique? Or is it possible to have more than one homomorphism? Once we specify a group and a set, how do we find such homomorphism(s)? Is there a requirement for the size of the group and the set to make this definition work, for instance $|G| \geq |X|$?

Please let me know if you would like me to clarify my question. Any help with making a better intuition is appreciated!

  • 3
    $\begingroup$ A permutation of a set $X$ is a bijection $X \to X$. So yes, $S_n$ can be defined as $S_{\{1,\dots,n\}}$. $\endgroup$
    – posilon
    Commented Mar 24, 2021 at 2:45
  • $\begingroup$ $X$ is here a finite set ,, and $S_X$ mean ,,the set of all permutting elements on $X$,,so if $|X|=n$ ,you can take $S_X$ as $S_n$, as $\{1,2,\cdot\cdot,n\}$ is in bijection with $X=\{x_1,x_2,\cdot\cdot,x_n\}$ $\endgroup$
    – A learner
    Commented Mar 24, 2021 at 2:52
  • $\begingroup$ I would define the group action stuff like: "Let $G$ be a group. We say that a set $X$ is a $G$-set if there exists a homomorphism $\rho : G \to S_X = \{\operatorname{bijections } X \to X\}$. In this case, we say that $G$ acts on X (via $\rho$) and that $\rho$ is the group action." $\endgroup$
    – azif00
    Commented Mar 24, 2021 at 2:53
  • $\begingroup$ Possible duplicate: math.stackexchange.com/questions/344790/… $\endgroup$
    – posilon
    Commented Mar 24, 2021 at 2:55
  • 3
    $\begingroup$ The group action is the homomorphism $\sigma$. The definition as you have quoted it is rather sloppy. A group $G$ can have many different actions on the same set $X$. Some examples to have in mind in addition to the permutation groups are any group acting on itself by left multiplication, right multiplication or conjugation and a group of $n\times n$-matrices acting on $\Bbb{R}^n$. $\endgroup$
    – Rob Arthan
    Commented Mar 24, 2021 at 2:56

4 Answers 4


For your first question, $S_X$ is defined as the group of bijections from $X$ to $X$. The group operation is function composition. If $X$ is finite, say with $n$ elements, then the groups $S_X$ and $S_n$ are obviously (noncanonically) isomorphic.

There is always at least one group homomorphism from $G$ to $S_X$, namely the one which sends everything in $G$ to the identity map in $S_X$. This is the "stupid group action" which doesn't do anything ($g.x = x$ for all $g \in G$ and $x \in X$).

There are typically many homomorphisms from $G$ to $S_X$. Therefore, there are typically many group actions of $G$ on $X$.

"Once we specify a group and a set, how do we find such homomorphisms?"

Good question. In many settings, the group action comes up naturally. It's not like people are taking random sets $X$ and $G$ and asking, "I wonder how many different actions of $G$ I can find on $X$." I mean maybe some people are doing this, but usually the group action is a convenient way to describe some existing phenomenon they are trying to study.

Example: Let $X = \{ z \in \mathbb C: \operatorname{Im}(z) > 0\}$ be the upper half plane, and let $G = \operatorname{SL}_2(\mathbb Z)$ be the group of integer matrices with determinant $\pm 1$. There is a natural group action of $G$ on $X$ by

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}.z = \frac{az+b}{cz+d}.$$

This came up (I believe) when people were working out problems like describing all holomorphic bijections from the Riemann sphere to itself, and it turns out that they all look like the above (where the matrix can be anything in $\operatorname{GL}_2(\mathbb R)$).

One example where people are taking certain groups $G$ and certain sets $X$, and asking what are the ways $G$ can act on $X$, is when $X$ is a vector space. The bijections from $X$ to itself coming from the elements of $G$ are required to also be linear transformations on $X$. Such group actions are called representations. Representation theory includes the problem of describing all representations of a given group on a given vector space, and this turns out to be an extremely difficult problem in general. Finding such group actions is no simple matter.

  • $\begingroup$ It's worth noting that there is a lot of information available about representations of a finite group on a finite-dimensional vector space over $\mathbb{C}$ (Maschke's theorem, character tables). $\endgroup$
    – jskattt797
    Commented May 2, 2021 at 4:34

In addition to the other excellent answers, I wanted to add another example of group actions. Let $S$ be a square inside a plane and let $G$ be the set of all rigid transformations $f$ of the plane (i.e. rotations, translations, reflections, or their combination), such that $f(S)=S$. Turn $G$ into a group by equipping it with composition. Now notice that $G$ acts in a natural way not only on $S$, but also on the set of vertices of $S$, as well as on the set of sides of $S$, the set of diagonals of $S$, and so on...

Of course you can do the same thing starting with any geometric object in any space with a properly selected set of transformations (bijections of the space that preserve/have some desired properties).


There may not be such a non-trivial group action, to answer your question, because for instance if the groups are finite, we need the orders of the groups to satisfy certain conditions. For instance there's no non-trivial homomorphism from $\mathbb Z_m\to\mathbb Z_n$, when $(m,n)=1$.

For example this prevents us from forming certain non-trivial semi-direct products, since we need a group action to do so. So, take the case when $p\nmid q-1, q\gt p\gt2$. Then we can't do anything but the standard direct product $\mathbb Z_q\times\mathbb Z_p\cong\mathbb Z_{pq}$.

Also we do have $|G|\ge|X|$ when the action is transitive. That's because in that case there is only one orbit. The orbit-stablizer theorem says the index of the stabilizer is the size of the orbit (of an element). We can easily say that each orbit has size less than or equal to that of $G$, since the orbit is $Gx$.

  • 2
    $\begingroup$ May I suggest \nmid to get $\nmid$; or some reason, MathJax really messes up \not|: $\not|$; $\LaTeX$ is a bit better, but MathJax really can’t handle it. (Alternatively, \not\mid also works; I know it leaves extra space around the bar, which I don’t like for the usual “divides” symbol, but for “does not divide” it really is better, IMHO...) $\endgroup$ Commented Mar 24, 2021 at 3:10
  • $\begingroup$ Thanks @ArturoMagidin $\endgroup$
    – user403337
    Commented Mar 24, 2021 at 3:11

As for the first part of your question. If $X$ is finite of size $n\in\Bbb N$, then there is a bijection $f:\{1,\dots,n\}\to X$; build the map $\varphi\colon S_X\to S_n$ by $\varphi(\sigma):=f^{-1}\sigma f$; note that:

  • $f^{-1}\sigma f\in S_n$ (good definition);
  • $\varphi(\sigma)=\varphi(\tau)\Rightarrow f^{-1}\sigma f=f^{-1}\tau f\Rightarrow \sigma=\tau$ (injectivity);
  • for every $\alpha\in S_n$, $\alpha=\varphi({f\alpha f^{-1}})$ (surjectivity);
  • $\varphi(\sigma\tau)=f^{-1}\sigma\tau f=f^{-1}\sigma(ff^{-1})\tau f=(f^{-1}\sigma f)(f^{-1}\tau f)=\varphi(\sigma)\varphi(\tau)$ (group homomorphism)

Therefore, $S_X\cong S_n$ (noncanonically, being the isomorphism dependent on the bijection $f$).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .