I How is Conjugacy a Group Action?

[Mr Davis 97]

I am told that $\varphi_g (x) = g x g^{-1}$ is a group action of G on itself, called conjugacy. However, I am a little confused. I thought that a group action was defined as a binary operation $\phi : G \times X \rightarrow X$, where $G$ is a group and $X$ is any set. However, this $\varphi_g$ is just a normal function $\varphi_g : G \rightarrow G$. If this is not a binary operation, how is it a group action?

 

[fresh_42]

I am told that $\varphi_g (x) = g x g^{-1}$ is a group action of G on itself, called conjugacy. However, I am a little confused. I thought that a group action was defined as a binary operation $\phi : G \times X \rightarrow X$, where $G$ is a group and $X$ is any set. However, this $\varphi_g$ is just a normal function $\varphi_g : G \rightarrow G$. If this is not a binary operation, how is it a group action?

$\phi = \varphi : G \times G \rightarrow G$ defined by $\phi(g,x) = \varphi (g,x) = \varphi_g(x) = gxg^{-1}$ with $X=G$.

 

[Mr Davis 97]

$\phi = \varphi : G \times G \rightarrow G$ defined by $\phi(g,x) = \varphi (g,x) = \varphi_g(x) = gxg^{-1}$ with $X=G$.

So is $\varphi_g$ the group action or is $\varphi$?

 

[fresh_42]

So is $\varphi_g$ the group action or is $\varphi$?

Formally, $(\phi,G,X=G)$ is, because it assigns to every group element $g$ a conjugation $g \mapsto (x \mapsto gxg^{-1})$ on $X=G$, that is $\phi = \varphi = (\phi=\varphi,G,X)$ operates via conjugation $(\varphi ,G,X) \ni (\varphi,g)=\varphi_g = (x \mapsto gxg^{-1})$ on $X=G$.

It is more important to know what a group action is, which is another word for "operates on". And another expression is "G is represented (here by conjugation) on $X$ (here =$G$)". So it's really easy to get confused when learning these terminology. If you actually want to define it rigorously then you will have to take the entire triple $(\textrm{form of action, group, set upon the action takes place}) = (\phi, G, X) = (\varphi , G, G)$.

Your first question is based on the confusion, that you might not have considered that $g \mapsto \varphi_g$ is already a mapping which gives you the missing argument in $\phi$.

 

[mathwonk]

You might like to check that a group action by G on a set S is also equivalent to a homomorphism G-->Bij(S) where Bij(S) is the group of bijections of the set S with itself. This shows you that a group action is a way of representing an abstract group inside a concrete group of permutations.