$( \ X\to G \ , \ \star \ )$ is a group if $(\varphi \star \psi)(a) = \varphi(a) \star \psi(a)$

Question

Let $X$ be a set and $G$ a group with the operation $\star$. Show that the set $$ \mathcal{X} = \Big\{ \varphi : X\to G \mid \text{$\varphi$ is a function} \Big\} $$ is a group with the operation \begin{equation}\label{star} \big(\varphi \star \psi\big)(a) \; = \; \varphi(a) \star \psi(a) \qquad \quad \forall\,a\in G. \end{equation}

So associative is pretty easy since $(G,\star)$ is a group: Let $\varphi,\tau,\phi\in\mathcal{X}$. Thus, \begin{align*} ((\varphi\star\tau)\star\phi)(g) &= (\varphi\star\tau)(g)\star\phi(g)\\ &= (\varphi(g)\star\tau(g))\star\phi(g)\\ &=\varphi(g)\star (\tau(g)\star\phi(g))&G \text{ group}\\ &= \varphi(g)\star(\tau\star\phi)(g)\\ &= (\varphi\star(\tau\star\phi))(g) \end{align*} And for identity, let $id:G\to G$ with the map $g\mapsto e$, with $e$ the identity on G, is a function and acts as an identity for $\mathcal{X}$: $$(\varphi\star id)(g) = \varphi(g)\star id(g) = \varphi(g)\star e = \varphi(g) = e\star \varphi(g) = id(g)\star\varphi(g) = (id\star\varphi)(g).$$

But I'm having trouble proving closure and inverses. Since we don't know if an element is bijective or not, then we can't construct an inverse. And for closure, how can I show that $\varphi(a) \star \psi(a)$ is still a function?

Answer 1

Hint. When the group operation is function composition a function must be a bijection to have an inverse. But that's not the group operation that matters here. To invert $f$ you have to find a function $g$ that inverts the values of $f$ as they occur.

If you stare at your last sentence you might see that you are actually describing the function that's the product of the two you started with.

Answer 2

You can actually work out what an inverse looks like quite explicitly. Take an arbitary $\varphi \in \mathcal X$. The question is, can we construct a function (let's denote it, completely coincidentally, as $\varphi^{-1}: X \to G$) which satisfies $$\varphi \star \varphi^{-1} \equiv \operatorname{id}.$$

I claim that you can define $\varphi^{-1}$ as $$\varphi^{-1}(a) = \varphi(a)^{-1} \quad \forall a \in X$$ Note that this is well-defined: $g \equiv \varphi(a) \in G$ so it has an inverse $g^{-1}$.

I will leave it as an exercise for you to prove this claim.

Answer 3

Your set is $G^X$ and it is a group isomorphic to the product of $|X|$ copies of $G$.