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Let $X$ be a $G-$set , then there exists a group homomorphism

$\lambda : G \rightarrow T(X)$ where

$T(X) = \{ f : X \rightarrow X $ such that $f$ is a bijection $\}$

forms a group w.r.t operation of composition of maps. And conversely , if $\lambda : G \rightarrow T(X) $ is a group homomorphism , then it defines a group action $G$ on $X$.

Is there a simpler proof to this..

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    What's the proof you know? – 2017-01-26
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    by proving $\lambda$ to be bijective and group preserving operation implying homomorphism .and conversely if $\lambda$ is a group homorphism then by defining a group action and proving that its well defined . – 2017-01-26
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    See, these proofs are in some sense canonical, in that when you see the question, it follows by just proceeding from definition and trivial leaps of logic. In that sense, the proof you have is of utmost simplicity and triviality, and is the simplest proof I can think of for this question. – 2017-01-26
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    Some people use the homomorphism $\lambda$ as the definition of a group action, in which case the proof is even easier! – 2017-01-26

1 Answers 1

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Usually this is framed in terms of showing that there are two (equivalent) definitions of a group action.

Definition 1: A group action of a group $G$ on a set $X$ is a map: $\alpha:G \times X \to X$ such that:

a) For any $g,h \in G$, and $x \in X$, we have: $\alpha(g,(\alpha(h,x)) = \alpha(gh,x)$.

b) For any $x \in X$, $\alpha(e_G,x) = x$.

(If we write $\alpha(g,x) = g\cdot x$, this becomes the notationally simpler:

a) $g\cdot(h \cdot x)) = (gh)\cdot x$

b) $e_G \cdot x = x$).

Definition 2: A group action of a group $G$ on a set $X$ is a group homomorphism:

$\phi: G \to \text{Sym}(X)$ ($\text{Sym}(X)$ is your $T(X)$).

If we start with definition 1), we can define $\phi$ by:

$\phi(g) = \alpha(g,-)$, where $\alpha(g,-)$ is the mapping that takes $x \mapsto \alpha(g,x)$.

It may not be obvious that $\alpha(g,-) \in \text{Sym}(X)$, although it is clearly a mapping $X \to X$. We have to show it is bijective, as well.

So suppose that $\alpha(g,x) = \alpha(g,y)$. Then, of course:

$x = \alpha(e_G,x) = \alpha(g^{-1}g,x) = \alpha(g^{-1},\alpha(g,x)) = \alpha(g^{-1},\alpha(g,y))$

(since $\alpha(g^{-1},-)$ is likewise a function $X \to X$, and functions take equal values on equal arguments)

$=\alpha(g^{-1}g,y) = \alpha(e_G,y) = y$. This shows $\alpha(g,-)$ is injective.

Now suppose $y \in X$ is arbitrary. We will exhibit $x \in X$, such that $\alpha(g,x) = y$.

Let $x = \alpha(g^{-1},y)$. Then $\alpha(g,x) = \alpha(g,\alpha(g^{-1},y)) = \alpha(gg^{-1},y) = \alpha(e_G,y) = y$. This shows $\alpha(g,-)$ is surjective, and thus bijective.

So $\phi: g \mapsto \alpha(g,-)$ is indeed a mapping from $G \to \text{Sym}(X)$, and all that is left to satisfy definition 2) is to show $\phi$ is a group homomorphism (the group operation in $\text{Sym}(X)$ being understood to be functional compostion). That is, we must show:

$\phi(gh) = \phi(g) \circ \phi(h)$.

Now the left hand side is $\phi(gh) = \alpha(gh,-)$, while the right hand side is:

$\phi(g) \circ \phi(h) = \alpha(g,-) \circ \alpha(h,-)$. If these are equal for every $x \in X$, they are the same function. So we compute:

$\alpha(gh,x) = \alpha(g,\alpha(h,x))$ (by definition 1))

$= [\alpha(g,-)](\alpha(h,x)) = [\alpha(g,-)](\alpha(h,-)(x)) = (\alpha(g,-) \circ \alpha(h,-))(x)$.

On the other hand, if we start from definition 2), with a group homomorphism:

$\phi: G \to \text{Sym}(X)$, we can define $\alpha: G \times X \to X$ as follows:

$\alpha(g,x) = \phi(g)(x)$. Now we must verify this satisfies definition 1):

$\alpha(g,\alpha(h,x)) = \phi(g)(\alpha(h,x)) = \phi(g)[\phi(h)(x)]$

$= [\phi(g) \circ \phi(h)](x) = \phi(gh)(x)$ (since $\phi$ is a homomorphism)

$= \alpha(gh,x)$, which show a) holds.

b) is almost trivial: since $\phi$ is a group homomorphism, it must map identity to identity, that is, it maps $e_G \mapsto \text{id}_X$. Hence:

$\alpha(e_G,x) = \phi(e_G)(x) = \text{id}_X(x) = x$.

But wait! There's more! Suppose we call the action we obtain from going from definition 2) to definiton 1) $\beta$, and we start with an action $\alpha$ that gives us the homomorphism $\phi$ (going from definition 1) to 2)), that is:

$\phi(g) = \alpha(g,-)$, and $\beta(g,x) = \phi(g)(x)$. I claim $\beta = \alpha$.

For any $g \in G, x \in X:\ \beta(g,x) = \phi(g)(x) = \alpha(g,-)(x) = \alpha(g,x)$.

And if we start with $\phi$, derive an action $\alpha$, and then derive a homomorphsim $\psi$, I claim $\phi = \psi$:

For $\psi(g)(x) = \alpha(g,x)$ which is, in turn, defined as $\phi(g)(x)$. Thus for every $x \in X$, we have $\psi(g)(x) = \phi(g)(x)$, that is $\psi(g)$ and $\phi(g)$ define the same mapping $X \to X$. So, for any $g \in G$, we have $\psi(g) = \phi(g)$, that is $\psi = \phi$.

This shows that the maps: $\alpha \mapsto \phi$ and $\phi \mapsto \alpha$ are inverses of each other, so we have a one-to-one correspondence between actions via definition 1), and actions via definition 2).