A group operation on $G^S$ My question is:

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 and operation on $G^S$ that will yield a
  group.

Can the operation on $G^S$ be the exact same as the operation that is defined for the group $G$? Why or why not? 
 A: Hint: Think about $G^S$ as a product of $S$ copies of $G$, how would you define the operation on the product? Now find a way to translate it back to the functions.
A: The operation of $G^S$ can't be the same as the operation on $G$, because the elements of $G^S$ are functions, and the operation on $G$ applies not to functions but to elements of $G$.
Perhaps an example will help.  Let's say that $G$ is the set $\{E,O\}$, with the following operation $\oplus$:
$$\begin{array}{c|cc}
\oplus & E & O \\
\hline
E&E&O \\
O&O&E
\end{array}$$
Let's also say that $S$ is the set $\{1,2,3,4,5\}$.
Then $G^S$ is the set of all functions that take some number between 1 and 5 and which give you back, for each number, either $E$ or $O$.
For example, one element of $G^S$ is a function I'll call $OOOOO$, which has $OOOOO(x) = O$ for each $x$ in $\{1,2,3,4,5\}$.
Another element of $G^S$ is a function I will call $OEOEO$ which has $OEOEO(x)$ equal to $O$ when $x$ is an odd number and to $E$ when $x$ is an even number.
In all, there are 32 different elements of $G^S$, each one a different function from $S$ to $G$.
Your job is to think of an operation, $\star$ that will make these  32 elements into a group. So for example you should be able to say what $OOOOO\star OEOEO$ is; it should be one of the 32 functions.
The $\star$ operation can't be $\oplus$, because that applies to $O$ and $E$, not to these 32 functions.  But there is a simple way to define $\star$ that is based on $\oplus$ in a natural way.
