Is there a name for this property of functions on groups?

Solution 1:

The set of such functions are in one-to-one correspondence with the set of all functions $G^{n-1}\to G.$

For example, if $H:G^{n-1}\to G$ we can define:

$$F(g_1,\dots,g_n)=g_nH(g_n^{-1}g_1,g_n^{-1}g_2,\dots,g_n^{-1}g_{n-1})$$

On the other hand, given an $F,$ we can get back $H$ by: $$H(g_1,\dots,g_{n-1})=F(g_1,\dots,g_{n-1},1).$$

So such functions $F$ don’t seem too interesting.

One might ask, more generally, if $G$ acts on a set $X,$ then $G$ acts on $X^n,$ and what can we say about functions $F:X^n\to X$ which is a map in the category of sets acted on by $G?$ In your case, $X=G.$

This might be more complicated. For example, if $G$ acts on $X$ $2$-transitively, and $n=2,$ then $F$ is entirely determined by one value of the form $F(x,x)$ and one value of the form $F(x,y), x\neq y.$ So there are at most $|X|^2$ such functions in that case.

In general, $X^n/G$ is very complicated, but when $X=G$ with the simple action, it is fairly simple.

Solution 2:

I would like to generalise Thomas Andrews's answer a bit more. Let $G$ act on two sets $X$ and $Y$, and consider functions $F:X\to Y$ such that $F(gx) = gF(x)$. In the original setting, we have $X=G^n$, $Y=G$. As Thomas Andrews has pointed out, $F$ is determined by its value on a representative of each orbit in $X/G$. This is of course because if $F(x_0)$ is known, then $F(gx_0)=gF(x_0)$ is forced for all $gx_0$ in the orbit of $x_0$.

In the original setting, Thomas Andrews showed (in slightly different words) that the functions $F$ are in in fact in a one-to-one correspondence with the set of all functions $G^n/G\to G$ (i.e. functions $X/G\to Y$). When do we have this nice property in general?

For each orbit in $a\in X/G$, choose a canonical representative $\psi(a)\in a$. Let's write $\hat x = \psi([x])$ for the representative of $x$'s orbit. Now we want to define $F$ by choosing a value at each orbit representative, and then setting $$ F(x) = F(g_x \hat x) := g_x F(\hat x). $$ This isn't well-defined in general, because there might be several choices for $g_x$. But if the action on $X$ is free, then $g_x$ is unique by assumption, and we're golden. Thus,

If $G$ acts on $Y$ and acts freely on $X$, then the functions $F:X\to Y$ such that $F(gx)=gF(x)$ are in a one-to-one correpondence withthe set of all functions $X/G \to Y$.

We recover the corresponding function $E:X/G \to Y$ by simply $$ E(a) = F(\psi(a)). $$ Note however that the concrete correspondence is not unique or natural, since it depends on our arbitrary choices of representatives when defining $\psi$.

Lastly, take the special case $G\le H$, where $G$ is a subgroup of $X=H$ and acts by left mulitplication. Then $h=g_h\hat h \implies g_h = h(\hat h)^{-1}$, so the action is free ($g_h$ is unique), and our result applies. This was the case in the original setting, if we identify $G$ with the diagonal subgroup of $G^n$. In this case $$ \widehat {(g_1, \ldots, g_n)} := (g_n^{-1}g_1, \ldots, g_n^{-1}g_{n-1}, 1). $$