Exponential notation for group homomorphisms I recently saw in The Theory of Finite Groups by Kurzweil and Stellmacher a different notation for writing homomorphisms.
Consider a group homomorphism $f: G \rightarrow H$. Usually we write $f(g)$ to indicate the element which $f$ maps $g$ to (although I've also seen it written $(g)f$. This textbook writes the homomorphism as $g^f$.
I was curious to see if there were any explanations of why this notation is used and/or how common it is. I do realize that the following are true:
$$
(gh)^f=g^fh^f
$$
$$
(g^{-1})^f = (g^f)^{-1}
$$
and so we see that this exponential notation for homomorphisms displays many of the rules that we would expect exponents to have. However, this just doesn't seem worth it to me because


*

*For students first learning group theory, it's probably more important to think of group homomorphisms as a kind of function.

*Learning the basic properties of homomorphisms isn't so difficult and so thinking that homomorphisms share many properties with exponents isn't terribly useful, and possibly a source of confusion.


However, in order for the notation to appear in a textbook, there must be some kind of reason for it. So, thoughts? 
 A: As noted in the comment, the notation is common for right actions of a group, and the group of automorphisms acts on the group.
Recall that given a group $G$ and a set $X$, a right action of $G$ on $X$ is a function $X\times G\to X$, denoted with infix notation as $x\cdot g$, such that (i) For all $x\in X$, $x\cdot e = x$; and (ii) For all $x\in X$, $g,h\in G$, $x\cdot(gh) = (x\dot g)\cdot h$.
One can use exponential notation: $x^g = x\cdot g$. Then you have $x^e = x$, and $(x^g)^h = x^{gh}$.
This notation is also common when the action of $G$ on itself (or on a normal subgroup) by conjugation, defining $x^g = g^{-1}xg$. 
Now let $K$ be a group, and let $G$ be the group of all automorphisms of $K$. We can let $K$ act on $G$ by letting $x\cdot f = f(x)$ for all $x\in X$, $f\in K$. Writing composition in $K$ from right to left, so that $fg$ means "$f$ first, $g$ second", we have that this is a right action.
(Using functions on the right is common in ring theory, since then a module homomorphism would satisfy a homogeneity rule that reads a "$(am)f = a(mf)$"; for a while during the 60s, some group theorists adopted this notational convention by putting morphisms on the right; e.g., Hanna Neumann's Varieties of Groups has functions on the right. It also has the advantage that $f\circ g$ means "$f$ composed with $g$", i.e., $f$ first, $g$ second, instead of meaning the usual $g$ composed with $f$.) 
Now, in fact, you can even realize this action of $\mathrm{Aut}(G)$ on $G$ as conjugation: the holomorph of a group $G$ is the semidirect product $G\rtimes \mathrm{Aut}(G)$, where the action is by $x\cdot f = x^f$. 
