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I'm trying to get more intuition on this definition:
Let $(G,\circ,e)$ be group. A group action is a mapping $G×X→X:(g,x) ↦g.x \,$ such that:
\begin{align*} ∀x∈X &: e.x=x \tag{1}\\ ∀g,h∈G,∀x∈X &: g_1.(g_2.x)=(g_1 \circ g_2).x \tag{2} & \end{align*}

Is there some kind of intuition behind this definitions ? From the examples I've seen it seems like you always get a bijective mapping $X →X:x↦g(x)$. Is this the idea ? That $G$ acts on $X$ if all elements of $g$ are bijective mappings from $X→X$ ?

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Yes, an action of $G$ on $X$ is just a homomorphism $G \to \mathrm{Sym}(X)$ and that's the idea behind the definition. –  Martin Brandenburg Mar 28 '13 at 15:43
    
@MartinBrandenburg Surely that will help OP improve his intuition... –  Alexander Gruber Mar 28 '13 at 17:56

1 Answer 1

up vote 4 down vote accepted

Let $\mathcal{F}$ be the collection of functions that map $X$ to itself. There is a group in $\mathcal{F}$, that is, the collection of invertible functions, $\mathcal{G}$ (The unit is the identity function, the binary operation is composition).

Now given a group $G$, we want to know whether there is a group homomorphism \begin{equation} G\xrightarrow{h}\mathcal{G}. \end{equation} If there is, then the image is a subgroup in $\mathcal{G}$ that behaves like $G$, so probably we can learn something about $\mathcal{G}$ or $X$ from $G$.

But now we have $h(e)$ is the identity function on $X$. Thus $h(e)(x)=x$ for all $x\in X$. Also, if $g_i\in G$, then $h(g_1g_2)(x)=h(g_1)\circ h(g_2)(x)$. These follow from the fact that $h$ is a homomorphism.

In particular, this $h$ defines a group action of $G$ on $X$. The other direction is just as natural.

So to conclude, by a group action we are just trying to find a subgroup of functions on $X$ that behaves like some group we already knew. Also note that the group of invertible functions are symmetries of $X$, we are looking at subgroups of symmetries of $X$, no wonder we can learn a lot about $X$ from the group actions.

In practice, we seldom just take $X$ as a set and look at all the invertible functions. We often impose more structures on $X$, and only look at the collection of functions that respect these structures. But the spirit remains.

For more examples and a much clearer exposition, look at this wonderful article by Gowers.

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