Checking out Wikipedia and Quora (and some papers) was a bit difficult for understanding Group Action. The closest to understanding was this part:

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I understand the definition of homomorphism (a structure-preserving map between two algebraic structures of the same type). And from what I understand the symmetric group is a group of functions (bijective functions) mapping elements of a set $X$ onto itself.

However I get confused at this line:

...an action of $G$ on $X$ may be formally defined as a group homomorphism $\phi$ from $G$ to the symmetric group of $X$.

That seems to be the definition of group action, but I am having trouble understanding how to apply that.

What that says to me is it is a homomorphism (function) $f \colon X \rightarrow S$ where $S$ is a set of bijective functions on $X$. That itself seems weird, I thought a group action would result in a number or something related to $\{1, 2, 3\}$ (defined next).

Continuing, if we have a set $X = \{1, 2, 3\}$ and a group $(*,\{4, 5, 6\})$, then the symmetric group of $X$ would be something like $S = \{1 \rightarrow 2, 2 \rightarrow 3, \dots \}$, and $\phi$ would be something like $f : \{4, 5, 6\} \rightarrow S$. That doesn't make sense to me and I feel like I'm interpreting something wrong. I don't see how mapping to a set of functions $S$ will result in an action on the set.

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    $\begingroup$ The notation $1\to 2$ does not represent a function on the set $\{1,2,3\}$ as far as I'm aware. Do you mean the $2$-cycle $(12)$? Also, the text you highlighted clearly states the action is a homomorphism $G\to{\rm Sym}(X)$, and right after that you say you interpret this as a function $X\to{\rm Sym}(X)$. You can't understand the text if you replace what it says with something wrong! $\endgroup$ – anon Apr 11 '18 at 5:28
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    $\begingroup$ Let $G$ be the symmetry group of a 3D cube. Any such symmetry can be interpreted as a permutation of the vertex set $V$, the edge set $E$, or the face set $F$ (among others, like axes, space diagonals, and inscribed tetrahedra). That means we end up with homomorphisms $G\to{\rm Sym}(V)$, $G\to{\rm Sym}(E)$, $G\to{\rm Sym}(F)$. Thus, instead of thinking of the group elements intrinsically as permutations, we think of them as taking up the role of certain permutations in certain contexts. $\endgroup$ – anon Apr 11 '18 at 5:31

Unfortunately, you are interpreting something wrong… Let's try to make more sense out of it.

What that says to me is it is a homomorphism (function) $f\colon X\to S$ where $S$ is a set of bijective functions on $X$.

No, it doesn't say that. There are three different objects here, but you keep confusing them with other (in different ways throughout your post). Those three objects are:

  • a group $G$,
  • a set $X$,
  • and the symmetric group of all bijective functions on $X$, which we may denote $S=\Sym(X)$.

Now, the definition of a group action (of a group $G$ on a set $X$) says that is it a group homomorphism $f\colon G\to S$, i.e. $f\colon G\to\Sym(X)$. Note that the domain of the group action $f$ is $G$, not $X$ as you said.

… if we have … a group $(∗,\{4,5,6\})$ …

This is a bit hard to understand. I guess you're saying that we have a group $G=(∗,\{4,5,6\})$. Okay, so $\{4,5,6\}$ is the underlying set of this group, but it's not quite clear how the operation $*$ works. Since this is not one of the common standard notations for a concrete group (like $\mathbb{Z}$ or $V_4$), you should define it. Of course, since you're having a general discussion here and don't use this operation explicitly, I can't say that there's anything wrong with it per se. It is possible to define a group operation $*$ on the set $\{4,5,6\}$. But it would be a bit unusual considering the choice of the symbols for the elements of the group (for example, one of them — either $4$ or $5$ or $6$ — will have to be the identity element). So the only reason I am pointing this out is because I suspect it is also a sign of one of your confusions.

… the symmetric group of $X$ would be something like $S=\{1\to2,2\to3,\ldots\}$ …

Sorry, but this doesn't make much sense. We can say that an individual element of the symmetric group looks more or less like that. For example, if $X=\{1,2,3\}$, then one of the elements of $S=\Sym(X)$ is the bijective mapping $\sigma_1=\{1\mapsto2,2\mapsto3,3\mapsto1\}$, or rather $\sigma_1=\{(1,2),(2,3),(3,1)\}$ in ordered pairs notation. Another examples of an elements of $S=\Sym(X)$ is the bijective mapping $\sigma_2=\{1\mapsto2,2\mapsto1,3\mapsto3\}=\{(1,2),(2,1),(3,3)\}$. But then $S=\Sym(X)$ is the set of all such bijective mappings. In this example (hold your breath): $$\begin{align} S&=\{\sigma_1,\sigma_2,\sigma_3,\sigma_4,\sigma_5,\sigma_6\}, \quad \text{where}\\ \sigma_1&=\{1\mapsto2,2\mapsto3,3\mapsto1\}=\{(1,2),(2,3),(3,1)\},\\ \sigma_2&=\{1\mapsto2,2\mapsto1,3\mapsto3\}=\{(1,2),(2,1),(3,3)\},\\ \sigma_3&=\{1\mapsto3,2\mapsto2,3\mapsto1\}=\{(1,3),(2,2),(3,1)\},\\ \sigma_4&=\{1\mapsto1,2\mapsto3,3\mapsto2\}=\{(1,1),(2,3),(3,2)\},\\ \sigma_5&=\{1\mapsto3,2\mapsto1,3\mapsto2\}=\{(1,3),(2,1),(3,2)\},\\ \sigma_6&=\{1\mapsto1,2\mapsto2,3\mapsto3\}=\{(1,1),(2,2),(3,3)\}, \end{align}$$ listed in no particular order.

… and $\varphi$ would be something like $f\colon\{4,5,6\}\to S$.

Yes, under three conditions:

  • you understand correctly what $S$ is;
  • you clearly define a group structure, i.e. the multiplication operation $*$ on the group $G=\{4,5,6\}$;
  • and you make sure that $f\colon G\to S$ is group homomorphism between the group $G=\{4,5,6\}$ with the operation $*$ and the group $S=\Sym(X)$ with composition as the group operation.
  • $\begingroup$ I still don't understand. I think I'm confused by "group homomorphism". "The image of a homomorphism is the set of elements that occur as images of elements. It turns out that the image of any homomorphism is a subgroup of the group on the right." groupprops.subwiki.org/wiki/Homomorphism_of_groups From that I can only get as far as: The image $h(g)$ is the image of every element of the $Sym(X)$ with the group operation on the left. $\endgroup$ – Lance Pollard Apr 11 '18 at 6:22
  • $\begingroup$ I'm confused as to where $x$ goes (from wikipedia). $\phi(g)(x)$ it seems. What exactly happens going from $Sym(X) \to \phi(g)$, there is tricky stuff going on if that ends up being a permutation of $X$. Something like $Sym(X) \to (f_1, \dotsc, f_n) \to f_i(X)\ or\ f_i(x) \to A \subseteq X$ where $f_i \dots$ are the bijective functions in $Sym(X)$. $\endgroup$ – Lance Pollard Apr 11 '18 at 6:26
  • $\begingroup$ Then wikipedia has ${\displaystyle \varphi \colon G\times X\to X\colon (g,x)\mapsto \varphi (g,x)}$ which says $\varphi(g, x)$, so still don't see how that plugs in. $\endgroup$ – Lance Pollard Apr 11 '18 at 6:33
  • $\begingroup$ This is what confuses me: $σ_1=\{(1,2),(2,3),(3,1)\}$ if the element of S is a bijective mapping $f : X \to X$, then instead of ordered pairs it would be a function $σ_1=\{f_1,f_2,f_3\}$. But hmm. I guess I am leaking in the programming definition of function rather than the math one (of relation between input/output). So $f_1 : X \to X$ is equal to $\{(1,2),(2,3),(3,1)\}$. So $φ(g,x)$ is a permutation because $X \to X$ is a function/relation. $\endgroup$ – Lance Pollard Apr 11 '18 at 7:49

Let $\newcommand{\Map}{\textrm{Map}}\newcommand{\Sym}{\textrm{Sym}}$ us look first at sets $A$, $B$ and $C$. Write $\Map(A,B)$ for the set of all mappings (functions) from $A$ to $B$. There is a natural correspondence (in the sense of category theory) $$\Map(A\times B,C)\leftrightarrow\Map(A,\Map(B,C)).$$ Computer scientists call this currying after Haskell Curry. It says that maps $f:A\times B\to C$ correspond to maps $F:A\to\Map(B,C)$. How does this work? If $f:A\times B\to C$ and $a\in A$, define $F_a:B\to C$ by $F_a(b)=f(a,b)$. Then $F_a\in\Map(B,C)$ and $F:a\mapsto F_a$ is a map from $A$ to $\Map(B,C)$.

Now let's get to group actions. If $G$ is a group and $X$ a set, then a group action is an element of $\Map(G\times X,X)$ or equivalently $\Map(G,\Map(X,X))$ satisfying certain conditions. If we consider it as a map $F:G\to\Map(X,X)$ then its image must lie in $\Sym(X)\subseteq\Map(X,X)$, the collection of bijections from $X$ to $X$. Moreover $F$ must be a group homomorphism. Then $F$ corresponds to $f:G\times X\to X$ by currying. The condition that $F$ be a group homomorphism translates to $f(e,x)=x$ and $f(g_1g_2,x)=f(g_1,f(g_2,x))$. If we write $f$ in infix notation as $f(g,x)=g\cdot x$ then these become $e\cdot x=x$ and $(g_1g_2) \cdot x)=g_1\cdot(g_2\cdot x)$.

  • $\begingroup$ Will have to let this sink in. I understand currying and can follow along up to "How does this work", a little confused by the $F_a\dots$ and remainder of that paragraph. "then its image must lie in $Sym(X)⊆Map(X,X)$, the collection of bijections from $X$ to $X$" ok then that part I was partially understanding yay. "Then $F$ corresponds to $f:G\times X\rightarrow X$ by currying". Would be interested to know how you knew this required currying to follow in this form. (why wikipedia doesn't have that part). $\endgroup$ – Lance Pollard Apr 11 '18 at 5:24
  • $\begingroup$ "its image must lie in Sym$(X)$" is part of the condition for $F$ to be a group action. $\endgroup$ – Lord Shark the Unknown Apr 11 '18 at 5:26
  • $\begingroup$ "the action of its image $\phi(g) \subset Sym(X)$ on the point $x$." (wikipedia) I think I'm getting it. This is essentially $\phi(g)$ gives a function from $f : X \rightarrow X$, then $\phi(g)(x) \mapsto a \in X$. Not sure. Or no, $\phi(g)$ is a permutation of $X$, which is built into the definition of $\phi$. So $g \mapsto permutation_X$. $\endgroup$ – Lance Pollard Apr 11 '18 at 5:40

A less abstract definition of the action of a group G on a set S is as a mapping from $G\times S$ to $S.$ We write $gs$ for the image of $(g,s)$ under this mapping. There are two conditions, $$es=s, \forall s\in S$$ $$g_1(g_2s)= (g_1g_2)s, \forall g_1,g_2\in G, \forall s\in S$$ Here $e$ is the identity element of $g$.

It's east to see that for any $g\in G$ we have that $s\mapsto gs$ is a permutation of $S$ (just left-multiply by $g^{-1})$.

With that hint, you ought to be able to show the equivalence of this definition to the one you found on the Web.

  • $\begingroup$ Thank you. Still new to this, so the image is basically the output of the function say $f : (g, s) \rightarrow gs$. But $s \mapsto gs$ is a permutation of $S$ I read as being a random combination of elements of $S$. So if $S = \{1,2,3\}$, $s \mapsto gs$ might be like $s \mapsto \{3,1,2\}$. I'm not sure I understand what $gs$ is doing. $\endgroup$ – Lance Pollard Apr 11 '18 at 4:02
  • $\begingroup$ There are a lot of examples of group actions, and what $gs$ means depends n the particular example. Often the set set $S$ has some algebraic structure itself, but not always. Think of the action of $\mathbf R^*$ the multiplicative group of non-zero reals on $\mathbf R^n$ by scalar multiplication. $\endgroup$ – saulspatz Apr 11 '18 at 4:38
  • $\begingroup$ @LancePollard The notation $s\mapsto\{3,1,2\}$ doesn't mean anything here as far as I can tell. Perhaps you mean the function $f(s)=gs$ could for instance look like $f(1)=3,f(2)=1,f(2)=2$, in which case sure that could happen. $\endgroup$ – anon Apr 11 '18 at 5:34

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