2
$\begingroup$

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?

$\endgroup$
  • 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
  • 3
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It was precesily this result with the set of Homomorphisms, thank you ;) $\endgroup$ – Sebastien Feb 24 '15 at 19:04
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.