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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?

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  • $\begingroup$ Do you mean $\phi: X\to G?$ If not, what role does $X$ play here? $\endgroup$ May 18 at 0:59
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    $\begingroup$ Does this answer your question? Show that the free abelian group is a group. $\endgroup$ May 18 at 4:49
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    $\begingroup$ @fitzcarraldo That is essentially a duplicate. I wonder if this is a recent assignment. $\endgroup$ May 18 at 11:04
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    $\begingroup$ It might be worth trying to prove that this group is isomorphic to the direct product of $|X|$ copies of $G$, $\prod_{x\in X}G$. $\endgroup$ May 18 at 18:12
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    $\begingroup$ @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. $\endgroup$ May 18 at 18:56

3 Answers 3

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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.

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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.

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

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    $\begingroup$ 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. $\endgroup$ May 18 at 9:31

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