Is there any set in which exponents/powers can be used as a group operator. I am a student studying math who recently took an abstract algebra class that focused on group theory. Are there any sets such as maybe complex numbers that fulfill all of the group axioms using exponentiation?
 A: This heavily depends on what "exponentiation" means. The definition of the standard exponential map over reals, complex numbers or matrices heavily depends on their internal structure (Banach algebras). I don't really see how you can define an exponential map over other structures in an analogous way.
Note that typical exponential maps have this nasty property: $a^x=0$ has no solution regardless of $a$. And so it cannot induce a group operation, because group operations are always surjective.
This gets even worse: for some $a$ the $a^x$ is not even well defined, e.g. $-1^x$ fails to be well defined for $x\not\in\mathbb{Z}$. Even inside complex numbers $-1^{1/2}$ has two solutions, while say $i^i$ infinitely many. Of course you can apply an arbitrary choice here, but then you end up with an infinite number of exponential maps, with bad properties (none of them are continuous) and still none of them induces a group operation.
And finally the standard exponentiation is not associative, e.g. $3^{3^3}\neq (3^3)^3$.
The only set I can think of that is a group under the exponential map is the trivial $\{1\}$ set. Except for that, I suppose this cannot be done in a reasonable way.
A: Another answer already points out that you are unlikely to find an example where proper exponentiation is the group law, except for trivial situations like $G = \{1\}$. You may find the following example compelling none-the-less: it sort of looks like exponentiation, and understanding the precise sense in which it is not exponentiation is worthwhile. The set $\mathbb R_+ = \{x \in \mathbb R \mid x > 0\}$ is a group under the operation $x \circ y = x^{\log(y)}$.
