Endomorphism from Integers to Multiples

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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.

$\blacksquare$


Sources