Homomorphism: How do we get the equality?

Question

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?

Answer 1

The question as phrased is leaving it to you to figure out that the notation $a^n$ means $\overbrace{a \star \ldots \star a}^\mbox{$n$ $a$s}$ . I.e. $a^n$ is defined by:

$$ \begin{align} a^0 &= e \\ a^{n+1} &= a^n \star a \end{align} $$

where $e$ is the identity element of the group $G$. Now you can prove by induction that $a^m \star a^n = a^{m+n}$.

Answer 2

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.