The following is a homework problem. The conclusions are extremely intuitive and easy to see, but I am having proving this. Could someone please help?

Show that, given a group action $G \times X \to X$, for every $g \in G$, the map $a_g(x) = g \cdot x$ is a permutation of $X$, and that the map $G \to \text{Perm}(X), g \mapsto a_g$ is a homomorphism. Conversely, given a homomorphism $G \to \text{Perm}(X)$, show this defines a group action. Show that these are inverse operations: going from an action to a homomorphism and back gives the original action. Hence, homomorphisms $G \to \text{Perm}(X)$ are the same as group actions.


Let's show (for every $g \in G$) the map $a_g$ is bijective from $X \to X$. There are some different ways we can do this, and what follows is merely one way.

So suppose for an arbitrary $g \in G$ we have that for $x,y \in X$ that $a_g(x) = a_g(y)$, that is to say that $g\cdot x = g\cdot y$.

Using the particular map $a_{g^{-1}}$, and the fact that the images of the same function are equal for equal domain elements, we have:

$a_{g^{-1}}(a_g(x)) = a_{g^{-1}}(a_g(y))$ (since $a_g(x) = a_g(y)$) and "unravelling the definitions":

$a_{g^{-1}}(g\cdot x) = a_{g^{-1}}(g \cdot y)$

$g^{-1}\cdot(g\cdot x) = g^{-1}\cdot(g\cdot y)$, and using the rules of group actions:

$x = e\cdot x = (g^{-1}g)\cdot x = g^{-1}\cdot(g\cdot x) = g^{-1}\cdot(g\cdot y) = (g^{-1}g)\cdot y = e\cdot y = y$

Since $g$ was arbitrary, this shows that each map $a_g$ is injective. Now we need to show that each such $a_g$ is also surjective.

So let $y$ be any element of $X$. We seek to always be able to find (for any such $y$) some $x \in X$ with $a_g(x) = y$. I claim $x = a_{g^{-1}}(y)$ will serve:

$a_g(x) = a_g(a_{g^{-1}}(y)) = a_g(g^{-1}\cdot y) = g\cdot(g^{-1}\cdot y) = (gg^{-1})\cdot y = e\cdot y = y$ , as desired.

So we have a bijection, that is $a_g \in \text{Perm}(X)$. The next question is: is the map $g \mapsto a_g$ a homomorphism of $G$ into $\text{Perm}(X)$? That is, does the equality:

$a_{gh} = a_g \circ a_h$ hold? If the two sides of this equation take the same value for every $x \in X$, we can conclude they define the same function (this is a standard mathematical trick). But this merely means:

$a_{gh}(x) = a_g(a_h(x))$ for all $x \in X$, which we can show because we have an action:

$a_{gh}(x) = (gh)\cdot x = g\cdot(h\cdot x) = g\cdot(a_h(x)) = a_g(a_h(x))$

Now, if we start with a homomorphism $\phi:G \to \text{Perm}(X)$, the question becomes: how do we squeeze a group action from it? Since for any $g \in G,\ \phi(g)$ is a bijection on $X$, we might try to set as our action, for any $x \in X$:

$g\cdot x \stackrel{\text{def}}{=} \phi(g)(x)$ Is this actually a bona-fide group action? We need to verify the group action rules:

  1. $e\cdot x = x$, for all $x \in X$
  2. $g \cdot(h \cdot x) = (gh)\cdot x$ for all $g,h \in G$ and $x \in X$.

Note that since $\phi$ is given to be a homomorphism, it must map $e \in G$ to the identity of $\text{Perm}(X)$, which is the identity FUNCTION on $X$, and showing the first rule of group actions is an easy consequence of this. So we turn to examining rule 2:

$g\cdot(h\cdot x) = g\cdot(\phi)(h)(x)) = \phi(g)[\phi(h)(x)] = (\phi(g)\circ\phi(h))(x)$

$= \phi(gh)(x)$ (since $\phi$ is a homomorphism)

$=(gh)\cdot x$,as we hoped. Note that if our homomorphism $\phi$ is given by: $\phi(g) = a_g$ (the homomorphism we found in the first part of this post), we then recover our original action:

$g\cdot x = a_g(x) = \phi(g)(x) = g\cdot' x$ (I put the prime in there to indicate we might possibly have a "different" action, hypothetically).

A word about the "why" of these two equivalent views. The set $X$ might represent some collection of objects (like the vertices, or edges of a polygon) and $G$ might be a group of reversible transformations on the set $X$ (like rotations, for example). The "concrete action" (the notation $g\cdot x$) emphasizes what happens to the set. The "abstract action" (viewing it as a homomorphism into the full symmetric group on $X$) emphasizes the mappings, that is, highlights the group character of $G$.

One of the nice features of the abstract view, is that it generalizes nicely; for example replacing $X$ with an abelian group $A$, and $\text{Perm}(X)$ with the ring of abelian group endomorphisms (homomorphisms $A \to A$) $\text{End}_{\Bbb Z}(A)$, replacing $G$ with a field $F$ and insisting we have a ring-homomorphism: $\phi: F \to \text{End}_{\Bbb Z}(A)$, we obtain the notion of a "field acting (compatibly) on an abelian group", or what is more commonly called a vector space. You will no doubt encounter similar constructions like this in the future.


Some hints:

To show $a_g$ is a permutation. Try to show that it has a two sided inverse by using $a_{g^{-1}}$ and the property of composition of group action: $a_{gh}(x)=a_g(a_h(x))$.

To show $\phi(g)=a_g$ is a homomorphism you need to establish that $\phi(gh)=\phi(g) \circ \phi(h)$. Observe that $\phi(g)=a_g$ is a permutation so if you can show that $\phi(gh)(x)=\phi(g)\circ \phi(h) (x)$ for all $x \in X$ (i.e they agree as functions on all possible inputs). Then you will be done. Again composition property will help you.

Once you do the above you will get the idea to tackle rest. But feel free to ask in case you need some more help.

  • $\begingroup$ I'm having trouble on this: "Show that these are inverse operations: going from an action to a homomorphism and back gives the original action." Obviously, this is true, but I don't know how to show or prove it. $\endgroup$
    – clay
    Feb 17 '15 at 1:25

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