General Morphism Property for Semigroups

Theorem
Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be semigroups.

Let $\phi: S \to T$ be a homomorphism.

Then:
 * $\forall s_k \in S: \phi \left({s_1 \circ s_2 \circ \cdots \circ s_n}\right) = \phi \left({s_1}\right) * \phi \left({s_2}\right) * \cdots * \phi \left({s_n}\right)$

Hence it follows that:
 * $\forall n \in \N^*: \forall s \in S: \phi \left({s^n}\right) = \left({\phi \left({s}\right)}\right)^n$

Proof
$\forall s_k \in S: \phi \left({s_1 \circ s_2 \circ \cdots \circ s_n}\right) = \phi \left({s_1}\right) * \phi \left({s_2}\right) * \cdots * \phi \left({s_n}\right)$ can be proved by induction.

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition $\phi \left({s_1 \circ s_2 \circ \cdots \circ s_n}\right) = \phi \left({s_1}\right) * \phi \left({s_2}\right) * \cdots * \phi \left({s_n}\right)$.

$P(1)$ is true, as this just says $\phi \left({s_1}\right) = \phi \left({s_1}\right)$.

Basis for the Induction
$P(2)$ is the case:
 * $\phi \left({s_1 \circ s_2}\right) = \phi \left({s_1}\right) * \phi \left({s_2}\right)$

This follows from the fact that $\phi$ is a homomorphism.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\phi \left({s_1 \circ s_2 \circ \cdots \circ s_k}\right) = \phi \left({s_1}\right) * \phi \left({s_2}\right) * \cdots * \phi \left({s_k}\right)$

Then we need to show:


 * $\phi \left({s_1 \circ s_2 \circ \cdots \circ s_k \circ s_{k+1}}\right) = \phi \left({s_1}\right) * \phi \left({s_2}\right) * \cdots * \phi \left({s_k}\right) * \phi \left({s_{k+1}}\right)$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\phi \left({s_1 \circ s_2 \circ \cdots \circ s_n}\right) = \phi \left({s_1}\right) * \phi \left({s_2}\right) * \cdots * \phi \left({s_n}\right)$

The result for $n \in \N^*$ follows directly from the above, by replacing each occurrence of $s_k$ with $s$.