Is $\mathscr F(X,G)$ a group when $G$ is a group? Let $(G,\cdot)$ be a group. Define the law $\cdot$ on $\mathscr F(X,G)$ with the usual: $(f\cdot g )(x) = f(x)\cdot g(x)$.
I would like to say that $(\mathscr F(X,G), \cdot)$ is also a group without additionnal conditions (closure, associativity, neutral element and inverse seem easy to prove), but I've remembered that there is a trap (something to do with the commutativity).
Is my memory good? If not, do you see the result which has confused me?
 A: As said in the comments, your construction defines a group that is (isomorphic to) the direct product $\prod_{x\in X}G$ so there is no trap in this case.
However, in a relatively similar case commutativity is necessary: if $H$ is a group, you'd like to make the set of homomorphisms $\hom (H,G)$ into a group by defining $(f\cdot g)(x)=f(x)\cdot g(x)$. But then, $f\cdot g$ is a homomorphism iff for all $x,y\in H$ we have $(f\cdot g)(xy)=f(x)f(y)g(x)g(y)$ agree with $(f\cdot g)(x)(f\cdot g)(y)=f(x)g(x)f(y)g(y)$. Commutativity of $G$ is sufficient to guarantee this, and without that assumption $f\cdot g$ might not be a homomorphism. 
Relating the two constructions, the situation is that $\hom(H,G)$ is naturally a subset of the group $\mathscr F(H,G)$, but it might not be closed under multiplication unless $G$ is commutative.
A: Yes, the set of all maps from a set $X$ to a group $G$ with pointwise operation is again a group. Moreover, if the group $G$ is commutative this group is also commutative. 
It might be that you vaguely recalled that for the set of all maps from a set $X$ to a field $F$ one does not get a field again (when considering pointwise operations), as there one gets non-trivial zero-divisors. 
Thus, for example $\mathcal{F}( \mathbb{N}, \mathbb{R})$, in other words real valued sequences, form an additive group, yet not a field (only a commutative ring).  
