I'm reading Basic Algebra by Knapp and on Page 161 of Chapter 4 Section 6 he's talking about how a group action on a set can define other group actions. I can see how this makes sense in theory, but could anyone give any examples of how this is done? He gives an example:

For example, from an action G on X, we can construct an action on the space of >all complex-valued functions on X. The definition is $(gf)(x) = f(g^-1 x)$...

and then he goes on to show it satisfies the first of the group action axioms. Like I said, I feel like it's natural that some group actions induce other group actions, but how does one usually arrive at these definitions? It seems like he kind of pulled it out of nowhere.

P.S. I've only had basic Algebra training (Algebra I-II in University) and I'm starting representation theory in the fall, so take that as you will!

Thank you!



A group action of $G$ on a set $X$ is a group homomorphism $G\to \mathrm{Sym}(X)$. So if you could take another set $Y$ and define a homomorphism $\mathrm{Sym}(X)\to\mathrm{Sym}(Y)$, then you would have defined a group action on $Y$ by $G\to \mathrm{Sym}(X)\to \mathrm{Sym}(Y)$.

For example, if $Y$ is the set of all complex-valued functions on $X$, then you can define $\varphi\colon\mathrm{Sym}(X)\to\mathrm{Sym}(Y)$ by setting $\sigma\mapsto (f\mapsto f\circ \sigma^{-1})$. The $\ ^{-1}$ is there to ensure that this is really a group homomorphism, so that $\sigma\tau$ takes $f$ to $\varphi(\sigma)(\varphi(\tau)(f))$. We see that $\varphi$ gives a group action of $G$ on $Y$ by restricting our attention to the symmetries of $X$ coming from the group action of $G$ on $X$. This gives back the definition in your text.

  • $\begingroup$ Could you explain the first sentence of your section paragraph a bit more? $\endgroup$ – hijasonno May 25 '16 at 23:15
  • $\begingroup$ the accepted answer to this question is a good explanation: math.stackexchange.com/questions/127736/… $\endgroup$ – Orlando Marigliano May 26 '16 at 16:28
  • $\begingroup$ note that he uses the notation $\mathrm{Perm}$ instead of $\mathrm{Sym}$ $\endgroup$ – Orlando Marigliano May 26 '16 at 16:29

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