# Endomorphism from Integers to Multiples

## Theorem

Let $\struct {\Z, +}$ be the additive group of integers.

Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be a mapping.

Then $\phi$ is a group endomorphism if and only if:

$\exists k \in \Z: \forall n \in \Z: \map \phi n = k n$

## Proof

### Necessary Condition

Let $\phi: \struct {\Z, +} \to \struct {\Z, +}$ be an endomorphism.

Let $k = \map \phi 1$.

We have that $n = 1 + \cdots \paren n \cdots + 1$ for any positive integer $n$.

Thus:

 $\displaystyle \map \phi n$ $=$ $\displaystyle \map \phi 1 + \cdots \paren n \cdots + \map \phi 1$ $\displaystyle$ $=$ $\displaystyle k + \cdots \paren n \cdots + k$ $\displaystyle$ $=$ $\displaystyle k n$

Also:

 $\displaystyle \map \phi 1$ $=$ $\displaystyle \map \phi {1 + 0}$ $\displaystyle$ $=$ $\displaystyle \map \phi 1 + \map \phi 0$ $\displaystyle \leadsto \ \$ $\displaystyle \map \phi 0$ $=$ $\displaystyle 0$ $\displaystyle$ $=$ $\displaystyle k \cdot 0$ $\displaystyle \leadsto \ \$ $\displaystyle \map \phi {-1}$ $=$ $\displaystyle -k$ $\displaystyle \leadsto \ \$ $\displaystyle \map \phi {-n}$ $=$ $\displaystyle -k n$ for all positive integers $n$

Thus:

$\forall n \in \Z: \map \phi n = k n$

$\Box$

### Sufficient Condition

Let $k \in \Z$ such that:

$\forall n \in \Z: \map \phi n = k n$

Then:

 $\displaystyle \forall n, m \in \Z: \ \$ $\displaystyle \map \phi {n + m}$ $=$ $\displaystyle k \paren {n + m}$ $\displaystyle$ $=$ $\displaystyle k n + k m$ Integer Multiplication Distributes over Addition $\displaystyle$ $=$ $\displaystyle \map \phi n + \map \phi m$

Thus $\phi: \struct {\Z, +} \to \struct {\Z, +}$ is a group homomorphism from $\Z$ to $\Z$.

Hence by definition $\phi$ is a group endomorphism.

$\blacksquare$