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

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 $$n \in \N^*$$, 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 $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

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

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