General Morphism Property for Semigroups

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

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

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

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

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

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


 * $$P(1)$$ 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 $$\phi \left({s_1 \circ s_2}\right) = \phi \left({s_1}\right) \ast \phi \left({s_2}\right)$$.

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:

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

Then we need to show:

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

Induction Step
This is our induction step:

$$ $$

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

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


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