4
$\begingroup$

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

enter image description here

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
5
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $(g_1g_2) \cdot x)=g_1\cdot(g_2\cdot x)$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.