Homomorphism: How do we get the equality? Let $Z = (\mathbb{Z},+)$ the additive group of integers and $G = (M,\star )$ an arbitrary group. 
I want to show that for all $a \in G$ the map $\phi_a : Z \rightarrow G$ defined by $\phi_a(k) = a^k$ is an homomorphism from $Z$ to $G$. 
Let $m,n\in Z$. Then we have that $\phi_a(m+n)=a^{m+n}$ and $\phi_a(m)\star \phi_a(n)=a^m\star a^n$. 
Do we consider $\star$ as a multiplication and so $a^{m+n}=a^m\star a^n$, or how do we get the equality? 

Yes we see that as the multiplication between elements of group.
Since $a^m$ and $ a^n$ are elements of your group, $a^m \star a^n = a^{m+n}$ is given as the associative property of the group operation. 
