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Let $\cdot$ be an action of a group $G$ on a set $X$. For $g\in G$, define the map $\phi(g):X\to X$ by $\phi(g)(x)=g\cdot x$. I'm asked to show that $\phi(g)\in {\rm Sym}(X)$ and that $\phi :G\to{\rm Sym}(X)$ is a homomorphism.

I've shown that $\phi(g)\in{\rm Sym}(X)$, but I don't understand how to show $\phi$ is a homomorphism? I know it follows from the definition of an action, but what even is the function $\phi: G\to Sym(X)$ explicity?

The solution says that $\phi$ being a homomorphism follows immediately from the fact that $\phi (gh)(x)=g\cdot (h\cdot x)$, but aren't $\phi$ and $\phi (g)$ different functions?

Shaun
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Loobear23
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1 Answers1

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The group operation in $\mathrm{Sym}(X)$ is composition of functions, and $\phi:G\to\mathrm{Sym}(X)$ sends a group element to the function mapping $x\mapsto g\cdot x$. So you just have to check that the composition checks out.

Let $g,h\in G$. Consider the function $\phi(g)\circ \phi(h)$. This maps $x\mapsto \phi(g)(h\cdot x)=g\cdot (h\cdot x)$. The key observation is that since this is an action, $g\cdot (h\cdot x)=(gh)\cdot x$, so $\phi(g)\circ \phi(h)=\phi(gh)$.

Matt Samuel
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