Finding an operation on $G^S$ that yields a group Problem: Assume $S$ is a nonempty set and $G$ is a group. Let $G^S$ denote the set of all mappings from $S$ to $G$. Find an operation on $G^S$ that will yield a group.
Update (full attempted proof--is it correct?): Let $*$ be the operation on the given group $G$ and let $\bigstar$ be the operation on $G^S$. For all $\alpha,\beta\in G^S$, define $(\alpha\,\bigstar\,\beta)(x)=\alpha(x)*\beta(x)$. Now we must prove $G^S$ is a group.


*

*Closure: This is inherited from $G$.

*Associativity: This is also inherited from $G$.

*Existence of an identity element: This is also inherited from $G$; more explicitly, $e_G(x)=1=e_{G^S}(x)$ because $(\alpha\,\bigstar\,e_G)(x)=\alpha(x)*e_G(x)=\alpha(x)$ and, in the other direction, $(e_G\,\bigstar\,\alpha)(x)=e_G(x)*\alpha(x)=\alpha(x)$.

*Existence of inverse elements: If $\eta(x)\in G$, then $\eta^{-1}(x)\in G$ also. Thus, for $\alpha\in G^S$, we have $(\alpha\,\bigstar\,\alpha^{-1})(x)=\alpha(x)*\alpha^{-1}(x)=e_G(x)=e_{G^S}(x)$ and, in the other direction, $(\alpha^{-1}\,\bigstar\,\alpha)(x)=\alpha^{-1}(x)*\alpha(x)=e_G(x)=e_{G^S}(x)$.


This concludes the proof that the operation $\bigstar$ on $G^S$ yields a group. $\blacksquare$

$\color{red}{\mathbf{\text{Question:}}}$ Is this above proof correct? Also, by the way the operation $\bigstar$ works, it seems like $G^S$ would be a subgroup of $G$ or am I wrong?

Original work shown for problem:
Book solution: For $\alpha,\beta\in G^S$, define $(\alpha\beta)(x)=\alpha(x)\beta(x)$ for each $x\in S$.
My questions: I thought about this problem for a good while but got nowhere and finally decided to look at the answer, but it does not make a ton of sense to me. How does the given operation yield a group? And what really is the operation being considered? Is it composition? That is, $(\alpha\beta)(x)\equiv(\alpha\circ\beta)(x)$? If it is composition, then I know all compositions are associative, and thus I do not need to prove that as part of proving that $G^S$ is a group. But what about closure? Existence of an identity element? Existence of inverse elements?
For the existence of an identity element, I thought about $e_G(x)=1$ and having $(\eta e_G)(x)=\eta(x)e_G(x)=\eta(x)$ and $(e_G\eta)(x)=e_G(x)\eta(x)=\eta(x)$, but this doesn't really seem right or is it? 
Lastly, the problem gives the condition that $S\neq\varnothing$ (why is that important?), and no operation is specified concerning $G$--does that matter? I guess not, but it seems interesting to me that no operation is specified for $G$ when we are trying to find an operation for $G^S$ that involves $G$.
 A: The group operation is not defined by composition, unless you mean the composition $S \to S\times S \to G\times G \to G$, where the first map is the diagonal $s \to (s,s)$, the second map is the product of the two given maps $\alpha, \beta$, and the third map is the group operation on $G$.  When the author writes $(\alpha\beta)(x) = \alpha(x)\beta(x)$, the latter product is defined using the operation of $G$, just to clarify.
This definition is using the operation of $G$ pointwise, just like how addition and multiplication are defined for real-valued functions, ie $(f+g)(x) = f(x) + g(x)$.  The identity element is the constant function $e_G(x) = 1$ for every $x$, and you correctly give the proof that this works.  The inverse of a function $\alpha: S \to G$ is given by taking inverses pointwise, namely $\alpha^{-1}(x) = (\alpha(x))^{-1}$.  Associativity follows from the associativity of $G$.
As regards to the empty set, if $S = \emptyset$ then there is a unique empty function $S \to G$, and I guess the authors would rather rule out this case than give the trivial product on this set.
Hopefully this makes things more clear.
A: It definitely looks much better!
I'm a little confused about the distinction between $e_G$ and $e_{G^S}$; they both seem like the identity of $G^S$, and you seem to be using $1$ for the identity of $G$. Maybe make it more explicit that you're defining $e_{G^S}: S \to G$ so that $e_{G^S}(x) = 1_G$; it's the map sending everything in $S$ to the identity of $G$. So my only suggestion would be to clean up the notation (how exactly are you writing the identity of $G$? Of $G^S$?) and make it clear exactly what the identity of $G^S$ does.
For the inverses, essentially the same thing: Given $\alpha: S \to G$, you just start using $\alpha^{-1}$ without explicitly stating that, if $\alpha(x) = g$, then $\alpha^{-1}(x) = (\alpha(x))^{-1} = g^{-1}$. That is, make it clear that we invert $\alpha:S \to G$ pointwise, by inverting its output (using the inverse in $G$).
For your other question, very rarely will it be the case that $G^S \cong G$. I think it's more likely that $G^S$ is isomorphic to a direct product of several copies of $G$ (how many copies?).
