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

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 $$$$\label{star} \big(\varphi \star \psi\big)(a) \; = \; \varphi(a) \star \psi(a) \qquad \quad \forall\,a\in G.$$$$

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?

• Do you mean $\phi: X\to G?$ If not, what role does $X$ play here? Commented May 18, 2022 at 0:59
• Does this answer your question? Show that the free abelian group is a group.
– user1007416
Commented May 18, 2022 at 4:49
• @fitzcarraldo That is essentially a duplicate. I wonder if this is a recent assignment. Commented May 18, 2022 at 11:04
• It might be worth trying to prove that this group is isomorphic to the direct product of $|X|$ copies of $G$, $\prod_{x\in X}G$. Commented May 18, 2022 at 18:12
• @user21820: I do not follow your "actually means"; never encountered such a claim before. Not sayiing it's false, just saying I've never seen it before. I don't consider the colon in functional notation a relational symbol: to me it's punctuation. Commented May 18, 2022 at 18:56

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.

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.

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

• This only makes sense if $X$ is finite without a bit of work to define $|X|$, normally Cartesian products of groups are only done in introductory lectures for finitely many multiplicands. Commented May 18, 2022 at 9:31