Definition: Let $H$ and $K$ be any groups (not necessarily subgroups of another group). We call an (exponential) action of $H$ on $K$ sending $(h,k) \in H \times K$ to $k^h \in K$ an operation of $H$ on $K$ if, for each $h \in H$, the map sending $k$ to $k^h$ is an automorphism of $K$.
The Statement of the Problem:
Prove that given an operation of a group $H$ on a group $K$, the map from $H$ to $Aut(K)$ which sends each $h \in H$ to an automorphism sending $k$ to $k^h$ is a homomorphism.
Where I Am:
So, I don't think that I fully understand what an "operation" is (as it is, I'm somewhat shaky with the notion of group "actions," as well, which probably doesn't help). I'm trying to setup the problem to figure out what it is I need to do. Clearly, I need to prove that some map, which I'll denote
$$ \phi: H \to \operatorname{Aut}(K) $$
is an homomorphism.
The thing is, I'm having trouble even defining this map. Here's what I think that it is:
Let $\pi \in\operatorname{Aut}(K)$. Then $\phi$ is given by
$$h \mapsto \{\pi \in\operatorname{Aut}(K) : \pi(k) = k^h \}.$$
Well, ok. But, then what does $\phi (h)$ even look like? Certainly, I would need to know that to even begin to show that it's an homomorphism. Any help here would be appreciated.
You are on the right track; you want to show that the map $$\varphi:\ H\ \longrightarrow\ \operatorname{Aut}(K):\ h\ \longmapsto\ (k\ \longmapsto\ k^h),$$ is a homomorphism. You are given that the 'exponential' map $$\varepsilon:\ H\times K\ \longrightarrow\ K:\ (h,k)\ \longmapsto\ k^h,$$ defines an action of $H$ on $K$. What does this mean for $\varepsilon$? Look up your definition of group action. This tells you everything you need to know about $\varphi$ to show that it is a homomorphism.