# Representation of group actions

## Main Question or Discussion Point

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?

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fresh_42
Mentor
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##.

##\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
Mentor
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