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I'm currently taking a class in abstract algebra, and the textbook we are using is Ted Shifrin's Abstract Algebra: A Geometric Approach. In the chapter on group actions and symmetry, he defines a group actions as follows

$$\phi: G \mapsto \operatorname{Perm}(S)$$

where $G$ is a group acting on set $S$. However, most internet sources I've come across define a group action as

$$\phi: G \times S \to S$$

There are a few sources that talk about how the two are equivalent, but the explanations are overly brief. Can someone help me reconcile these two definitions?

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In general, any map $X\times Y\rightarrow Z$ is an "adjoint" of some map $X\rightarrow Z^Y$, where $Z^Y$ is the set of all functions from $Y$ to $Z$. This is because if $f:X\times Y\rightarrow Z$, then you can define a $g:X\rightarrow Z^Y$ by defining $g(x)(y)=f(x,y)$. You'll find the definition of the action as $G\times S\rightarrow S$ has conditions that imply that the adjoint function $G:X\rightarrow S^S$ has $G(x)$ going to a permutation, with $G$ then being a group homomorphism. – Thomas Andrews Apr 3 '12 at 20:43
Wikipedia explains it. – lhf Apr 3 '12 at 20:45
You probably want to use $\to$ instead of $\mapsto$. Also, there should be some requirements on the map $G \times X \to X$. – Dylan Moreland Apr 3 '12 at 20:48
Considering the basic nature of the question, it might be more confusing than helpful to tell the OP that the notion of group action is "really just" a specialization of the notion of adjoint or of morphisms into endomorphism groups. – Robert Haraway Apr 3 '12 at 20:57
Also, I am surprised that everybody keeps referring to $\mbox{End}S$ as opposed to $\mbox{Perm}S;$ a group action, as I've always thought of it, must permute the elements of whatever it's acting upon. In particular, the functions resulting from a group action have to be invertible, and therefore be permutations. – Robert Haraway Apr 3 '12 at 21:04
up vote 5 down vote accepted

The first definition assumes that the map $\phi: G \to \mbox{Perm}S$ is a homomorphism of groups, i.e. that the map preserves multiplication. Now, already, elements of $\mbox{Perm}(S)$ are just bijections $p: S \to S.$ So for any $g \in G,$ $\phi(g): S \to S$ will be such a bijection. Perhaps it is good to emphasize that $\phi$ takes an element of $g$ to a function on $S.$ Sometimes we write this function as $\phi_g.$ This makes the notation nicer; it is nicer to write $\phi_g(s)$ for the image of an element $s \in S$ under the function $\phi(g) \in \mbox{Perm}(S)$ than it is to write $\phi(g)(s).$

In any case, the corresponding map $\psi: G \times S \to S$ is just $\psi(g,s) = \phi_g(s),$ i.e. the image of $s$ under $\phi(g).$ This map in fact yields a group action since $\phi_{g h} = \phi(g h) = \phi(g) \circ \phi(h)= \phi_g \circ \phi_h$ so $\psi(g h,s) = \phi_{g h}(s) = \phi_g(\phi_h(s)) = \psi(g,\psi(h,s)),$ which is what is required for a map to be a group action (cf. @Dylan Moreland's comment).

Conversely, if we were to start off with such a group action $\psi: G \times S \to S$ satisfying $\psi(g h, s) = \psi(g, \psi(h,s)), $ then we may define $\phi: G \to \mbox{Perm}(S)$ by how the image of an element of g under $\phi$ acts on elements of $s.$ That is, we define $\phi(g)$ to be that element $f:S\to S$ of $\mbox{Perm}{S}$ such that $f(s) = \psi(g,s).$ Then you may check that this yields a homomorphism of $G$ into $\mbox{Perm}{S}.$

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All answers were helpful, but I think this answer was the easiest for me to understand. Thanks a lot. – Tianxiang Xiong Apr 3 '12 at 22:20
You are very welcome! – Robert Haraway Apr 3 '12 at 23:37

Your author's definition is that of a permutation representation which is equivalent to a group action.

If $\varphi:G \to \mathrm{Perm}(S)$, then given $g \in G$, $\varphi(g) \in \mathrm{Perm}(S)$ (a permutation of $S$). Suppose $s\in S$, then one can define $g \cdot s = (\varphi(g))(s)$. It's not hard to show that this is a group action.

Conversely, if one has a group action, it is not hard to show that $s \mapsto g\cdot s$ is a permutation, call it $\varphi(g)$. Then one has a map $\varphi:G\to \mathrm{Perm}(S)$ which is a group homomorphism.

So you can look at a homomorphism from your group to a group of permutations on $S$ or you can consider your group acting on $S$. They are sort of two sides of the same coin.

This sort of thing appears in many contexts usually under the names representation and module.

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It’s an example of an even more general phenomenon. Suppose that $f:X\times Y\to Z$ is a function. For each $x\in X$ we can define a related function $f_x:Y\to Z$ by $f_x(y)=f(x,y)$. This correspondence gives us a map $$\varphi:X\to {^YZ}:x\mapsto f_x\;,$$ where ${^YZ}$ is the set of all functions from $Y$ to $Z$.

Clearly if we know $f$, we know $\varphi$: we just constructed $\varphi$ from $f$. But the reverse is true as well. Suppose that I have a function $$\varphi:X\to{^YZ}\;.$$ Then I can define from it a function $$f:X\times Y\to Z:\langle x,y\rangle\mapsto \big(\varphi(x)\big)(y)\;.$$ It’s not too hard to see that these constructions are inverse to each other: if I now apply the first construction to this $f$, I’ll recover the original $\varphi$.

In the case of the group action, the first definition is like the map $\varphi$ in my discussion, with $G$ for $X$ and $S$ for $Y$ and $Z$: it associates to each $g\in G$ a map in ${^SS}$. The second definition is like the map $f$.

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I disagree with your epilogue; a group action must go to the set of all invertible functions from $S$ to $S,$ which is just the set of permutations of $S.$ – Robert Haraway Apr 3 '12 at 21:07
@Ambrose: You’re right; I was thinking of semigroups. Fixed. – Brian M. Scott Apr 3 '12 at 21:13

The two are not quite the same, since permutations are bijective. But if we allow ourselves a map

$$ G\longrightarrow \operatorname{End}(S)$$

where $\operatorname{End}$ is just functions $S\rightarrow S$, then there is a map between the two definitions. Given a map $\sigma:G\times S\rightarrow S$ we get another map by sending $g\in G$ to the function $f_g$ which takes $s\in S$ to $f_g(s)=\sigma(g,s)$.

The other way around, if we have a map $\phi:G\rightarrow \operatorname{End}(S)$. Then we let $(g,s)\in G\times S$ map to $\phi(g)(s)$.

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Moreover, and this is important, the two constructions are inverses, in the sense that if you start from a homomorphism, construct the action from it, and then construct the homomorphism from the action, you get back the original homomorphism; and vice-versa. – lhf Apr 3 '12 at 20:52

In the first definition, $g \in G$ gets sent to some permutation of $S$, call it $\phi_g$. The map $\phi \colon G \to \text{Perm}(S)$ satisfies some properties (basically, that $\phi_g\circ\phi_h=\phi_{gh}$). On the other hand, the second definition gives a map $G \times S \to S$. What is this map? It's the map that sends $(g,s) \mapsto \phi_g(s)$.

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protected by user26857 Apr 29 '15 at 21:47

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