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 $n = \overbrace {1 + \cdots + 1}^n$ for any positive integer $n$.

Thus:

\(\ds \map \phi n\) \(=\) \(\ds \map \phi {\overbrace {1 + \cdots + 1}^n}\)
\(\ds \) \(=\) \(\ds \overbrace {\map \phi 1 + \cdots + \map \phi 1}^n\)
\(\ds \) \(=\) \(\ds \overbrace {k + \cdots + k}^n\)
\(\ds \) \(=\) \(\ds k n\)

Also:

\(\ds \map \phi 1\) \(=\) \(\ds \map \phi {1 + 0}\)
\(\ds \) \(=\) \(\ds \map \phi 1 + \map \phi 0\)
\(\ds \leadsto \ \ \) \(\ds \map \phi 0\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds k \cdot 0\)
\(\ds \leadsto \ \ \) \(\ds \map \phi {-1}\) \(=\) \(\ds -k\)
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N_{>0}: \, \) \(\ds \map \phi {-n}\) \(=\) \(\ds -k n\) similar with above

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:

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

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