Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

5 Answers 5

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 $\mbox{Perm}{S}.$

share|improve this answer
All answers were helpful, but I think this answer was the easiest for me to understand. Thanks a lot. –  xiongtx 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.

share|improve this answer

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 $f$.

share|improve this answer
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 $\phi(g)(s)$.

share|improve this answer
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 $(g,s) \mapsto \phi_g(s)$.

share|improve this answer

protected by user26857 Apr 29 at 21:47

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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