0

Up until now I thought a group action of a group $G$ on a set $X$ was a homomorphism fron the group to the symmetric group on $X$. But today I took a class on geometric group theory and several things seem to imply that the professor thinks it‘s rather an antihomomorphism. ( Beware : I think of the geoup operation in the symmetric group of „$fg$ is first $g$ then $f$“)

Do the conventions differ regarding which research area you are in?

Example (everything quoted from the lecture) Let $H\leq G$ be a subgroup of $G$ and let $X$ be the set of right cosets of $H$. Then $\rho\colon G \to Sym(X), g \mapsto \rho(g)$, where $\rho(g(Hx))=Hxg$ is a transitive right-action. BEWARE : $\lambda’\colon G\to Sym(Y)$, where $Y$ is the set of left cosets of $H$ and $\lambda’(g)(xH)=gxH$ is not a homomorphism and so no group action.

As I understand it it is just the other way around.

  • No, the definition does not depend on the research area. For the same question see here. – Dietrich Burde Apr 12 '18 at 08:23
  • A group action is a group action, whether you work in geometric group theory or not. There is no definition of a group action specific to GGT. Can you give an example of your issue? – user1729 Apr 12 '18 at 08:26
  • 2
    @DietrichBurde But it might be common to use right actions in some field, in which case one does get an antihomomorphism. – Tobias Kildetoft Apr 12 '18 at 08:29
  • Just to add to Tobias Kildetoft's comment, you can look ok [Wikipedia]{https://en.wikipedia.org/wiki/Group_action} for the definition of a \emph{left} group action, for which the notation agrees with usual functional notation (functions "act on the left of elements": $x\mapsto f(x)$). Right group actions can be though as "functions acting on the right": $x\mapsto xf$ (this agrees with [Reverse Polish Notation]{https://en.wikipedia.org/wiki/Reverse_Polish_notation}. Then right actions correspond to anti-homomorphism of the group to the symmetric group of the set that is being acted upon. – Luiz Cordeiro Apr 12 '18 at 08:37
  • 1
    You only get an anti-homomorphism if you mix right actions and left actions, which is not usually necessary. I generally prefer to use right actions and to compose permutations from left to right. If it is really necessary to use both right and left actions (this can happen when working with bimodules for example) then you can still avoid anti-homomorphisms by mapping $f$ (right action) to $f^{-1}$ (left action). – Derek Holt Apr 12 '18 at 09:17
  • I will edit my wuestion and provide an example. –  Apr 12 '18 at 09:37

1 Answers1

2

When it comes to group actions, the conventions seem to depend more on geography or personal preference than on applications, although this is anecdotal.

Personally, I always think of a group as acting on the left, since each element of the group represents a function $X \to X$, and typically we use the notation $g(x)$ for the function $g$ acting on the element $x$. Since I'm used to composing functions right to left, I read $fg$ as "$g$ then $f$", so that $fg(x)$ is exactly $f(g(x))$. It follows that the convention I assume when multiplying in the symmetric group is the reverse of yours, and that group actions are homomorphisms into the symmetric group for me.

My Italian professor apparently uses the same convention you do for the symmetric group, and thinks of symmetric group elements as acting on the right, but still thinks of other functions as acting and composing on the left, which lead to some confusion about the order that operations should appear when group actions were interwoven with functions.

Ultimately any convention is fine as long as you stick to it. Whether a group action corresponds to a homomorphism or an antihomomorphism depends entirely on the conventions chosen and on whether you choose the axiom $fg(x)=f(g(x))$ or $fg(x)=g(f(x))$ for your group actions.

Finally, don't forget that every left group action is also a right group action by inverting: We can define $(x)f = f^{-1}(x)$ to get $(x)fg = (fg)^{-1}(x) = g^{-1}f^{-1}(x) = ((x)f)g$. This is the reason the choice of convention isn't especially important in practice.

Morgan Rogers
  • 1,221
  • Yeah, seems to me my professor multiplies the elements of the symmetric group from left to right ...

    Thanks for your answer!

     –  Apr 12 '18 at 09:53