Let $(G, \circ_G)$ be a group. Then, let

$$F(G) = \left(G^{(G^2)}, (f, g) \mapsto \big((x, y) \mapsto f(x, y) \circ_G g(x, y)\big)\right)$$

be the group of all functions from $G^2 \to G$, with the operation of applying $\circ_G$ pointwise. The identity is the function $(x, y) \mapsto e_G$, where $e_G$ is the identity element of $G$. $F(G)$ being a group follows from the already known group operation $\circ_G$.

Is there a name for this group?

  • $\begingroup$ Do you want the functions to be continuous? Edit: Oh wait, these aren't the circle and the sphere, are they? $\endgroup$ – Chris Culter May 15 '17 at 23:35
  • $\begingroup$ No. Since I begin with an arbitrary group $S$, I don't even think we can ask for continuity unless we ask for specific restrictions on which groups we begin with. (I am unsure what you mean by "circle and the sphere") $\endgroup$ – Enrico Borba May 15 '17 at 23:36
  • $\begingroup$ Usually $S^1$ is the unit circle, an important topological group, and $S^2$ is the unit sphere. Spaces of maps to or from $S^1$ are common enough that using the letter S here is short-circuiting my poor brain! $\endgroup$ – Chris Culter May 15 '17 at 23:40
  • $\begingroup$ Oh sorry, I'll change $S$ to $G$ then. $\endgroup$ – Enrico Borba May 15 '17 at 23:52

While I don't know the name of your group, it is simply a particular form of the direct product $\prod\limits_{\alpha \in A}\,G$ for an arbitrary set $A$. Note that $\prod\limits_{\alpha\in A}\,G$ can be viewed as the set of all functions $A\to G$ with the pointwise multiplication as the group operation. In your case, $A=G^2$.

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  • $\begingroup$ So according to your answer, $F(G) = G^{|G^2|}$? $\endgroup$ – Enrico Borba May 16 '17 at 0:08
  • $\begingroup$ I don't usually write that notation, because it seems ambiguous (there is also a notion of weak direct products of groups). But sure, with the usual definition, your group is basically $G^{G^2}\cong G^{\left|G^2\right|}$. $\endgroup$ – Batominovski May 16 '17 at 0:10
  • $\begingroup$ I'm having a hard time following your claim. You say $\prod_{\alpha \in A} G$ can be viewed as the set of all functions $A \to G$. However, if $|G| = 3$ and $|A| = 2$, we have $6$ such functions, but $\left|\prod_{\alpha \in A} G\right| = |G^2| = 9$ $\endgroup$ – Enrico Borba May 16 '17 at 0:13
  • $\begingroup$ There are $9$ functions, not $6$. $\endgroup$ – Batominovski May 16 '17 at 0:16
  • $\begingroup$ Well this is embarrassing, you are correct. I'm having a bit of a hard time seeing the isomorphism between this (possibly infinite) direct product, and $F(G)$, but I will take a closer look. Thank you $\endgroup$ – Enrico Borba May 16 '17 at 0:20

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