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?

  • 2
    $\begingroup$ Commutativity isn't a requirement for a group, so if you can show closure, associativity, and the existence of an identity you're finished. $\endgroup$ – walkar Feb 24 '15 at 17:50
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    $\begingroup$ In fact this group is isomorphic to the direct product $\prod_X G$. $\endgroup$ – Crostul Feb 24 '15 at 17:53
  • $\begingroup$ Either of those comments would be perfectly fine as answers... $\endgroup$ – Jim Feb 24 '15 at 18:19

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.

  • $\begingroup$ It was precesily this result with the set of Homomorphisms, thank you ;) $\endgroup$ – Sebastien Feb 24 '15 at 19:04

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).


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