Prove that $\circ$ defines a group action of $G$ on $L(X)$

Question

Let $G$ be a group, $X$ a set, and define $(\cdot)$ an action of $G$ on $X$ as $(\cdot): G\times X \rightarrow X$

Let $L(X)=\{f:X\rightarrow \mathbb{C}\}$

If $g\in G$ and $f\in L(X)$ we define $\circ$ such that $\circ: G\times L(X) \rightarrow L(X)$ is defined as:

$$(g\circ f)(x)=f(g^{-1}\cdot x)$$

My attempt

Note that for $g=e$ we have:

$$(e\circ f)(x)=f(e^{-1}\cdot x)=f(e\cdot x)=f(x)$$

Moreover, let $g_1\, \, g_2 \in G$, then

$$g_1\circ(g_2\circ f)(x)= g_1\circ f(g_2^{-1}\cdot x)=f(g_1^{-1}\cdot (g_2^{-1}\cdot x))=f((g_1^{-1}\cdot g_2^{-1})\cdot x)=f((g_2\cdot g_1)^{-1}\cdot x)=(g_2\circ g_1)\circ (f)(x)$$

But here, I'm stuck, because I'm supposed to get: $$g_1\circ(g_2\circ f)(x)=(g_1\circ g_2)\circ (f)(x)$$

Can someone help me? Thanks.

Answer 1

I shall rephrase the problem slightly and use $\star$ instead of $\circ$, since $\circ$ is liable to cause confusion, especially when writing it is a composition e.g. $g\circ f$.

Proposition: Let $(g,x)\mapsto g\cdot x:G\times X\to X$ be a group action of a group $G$ on a non-empty set $X$. Define $L(X)$ to be the set of complex-valued functions on $X$. Then $(g,\alpha)\mapsto g\star\alpha:G\times L(X)\to L(X)$ given by $$(g\star\alpha)(x):=\alpha(g^{-1}\cdot x)$$ defines a group action on $L(X)$.

---

Proof: You have shown correctly that $e\star\alpha=\alpha$ for all $\alpha\in L(X)$ (where $e$ is the identity of $G$).

Here is how I would do the second part:

$$(g_1\star (g_2\star\alpha))(x)=(g_2\star\alpha)(g_1^{-1}\cdot x)=\alpha(g_2^{-1}\cdot(g_1^{-1}\cdot x))=$$ $$\alpha((g_2^{-1}g_1^{-1})\cdot x)=\alpha((g_1g_2)^{-1}\cdot x)=((g_1g_2)\star\alpha)(x)$$ for all $x\in X$, so $g_1\star(g_2\star\alpha)=(g_1g_2)\star\alpha$ for all $g_1,g_2\in G$ and $\alpha\in L(X)$.

---

Your mistake appears to be the first equality in the final part:

$$g_1\circ (g_2\circ f)(x)=g_1\circ f(g_2^{-1}\cdot x).$$

Note that $g_1\circ (g_2\circ f)(x)$ would be better written as $(g_1\circ (g_2\circ f))(x)$ to avoid confusion, since the whole thing is being applied to $x$. It is important to tread carefully when dealing with equalities of functions; using more intuitive notation will reduce the likelihood of making errors :)

Answer 2

(Because of the usual meaning of $\circ$, it would be better to use a different symbol, like Oliver House does in his write-up; but I'll keep the symbol you use below.)

You are doing the computation wrong in the last part.

What you are trying to do is show that the functions $(g_1g_2)\circ f$ and $g_1\circ(g_2\circ f)$ are equal (you have an extra $\circ$ in the last term corresponding to my first expression, which makes no sense, since $\circ$ is not an operation on elements of $G$).

To that end, we verify that they take the same values. Now, the former is easy to compute: $$\bigl((g_1g_2)\circ f\bigr)(x) = f\bigl( (g_1g_2)^{-1}\cdot x\bigr) = f\bigl((g_2^{-1}g_1^{-1})\cdot x\bigr) = f\Bigl(g_2^{-1}\cdot(g_1^{-1}\cdot x)\Bigr),$$ with the last equality holding because $\cdot$ is an action.

With $g_1\circ(g_2\circ f)$ you have to be careful to do it in the correct order: $g_1$ is acting, so the computation is: $$\begin{align*} \Bigl(g_1\circ(g_2\circ f)\Bigr)(x) &= (g_2\circ f)(g_1^{-1}\cdot x)\tag{1}\\ &= f\Bigl(g_2^{-1}\cdot(g_1^{-1}\cdot x)\Bigr).\tag{2} \end{align*}$$ Why? Well, (1) is because you are computing $g_1\circ h$ at $x$, with $h=g_2\circ f$, and that's the definition. Similarly, (2) holds because you are computing $g_2\circ f$ at the point $g_1^{-1}\cdot x$, and that's the definition.

And now we see that you get the same thing.

Your error lay in trying to do $(g_2\circ f)(x)$ first, which is incorrect. You are acting on the functions, so you need to do it in the correct order.