Quotient Theorem for Group Homomorphisms/Examples/Integer Power on Circle Group

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Example of Use of Quotient Theorem for Group Homomorphisms

Let $K$ denote the circle group.


Let $\phi: K \to K$ be the homomorphism defined as:

$\forall z \in K: \map \phi z = z^n$

for some $n \in \Z_{>0}$.


Then $\phi$ can be decomposed into the form:

$\phi = \alpha \beta \gamma$

in the following way:


$\alpha: K \to K$ is defined as:
$\forall z \in K: \map \alpha z = z$
that is, $\alpha$ is the identity mapping


$\beta: S \to K$ is defined as:
$\forall z \in S: \map \phi z = z^n$
where $S$ denotes the set defined as:
$S := \set {z \in \C: z = e^{2 \pi i x}, 0 \le x < \dfrac 1 n}$


$\gamma: K \to S$ is defined as:
$\forall z \in K: \map \gamma z = z \bmod \dfrac 1 n$
where $\bmod$ denotes the modulo operation.


Proof

It is first demonstrated that $\phi$ is a homomorphism:

\(\ds \map \phi {w \times z}\) \(=\) \(\ds \paren {w \times z}^n\) Definition of $\phi$
\(\ds \) \(=\) \(\ds w^n \times z^n\)
\(\ds \) \(=\) \(\ds \map \phi w \times \map \phi z\)


We have that $1$ is the identity element of $K$, and to confirm:

$\map \phi 1 = 1^n = 1$


Now we can establish what the kernel of $\phi$ is:

\(\ds \map \phi z\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds z^n\) \(=\) \(\ds 1\) Definition of $\phi$
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds \set {z \in \C: z^n = 1}\)

Thus $\map \ker \phi$ is the set of complex $n$th roots of unity:

$\map \ker \phi = U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$

where $\N_n = \set {0, 1, 2, \ldots, n - 1}$.


Next we establish what the image of $\phi$ is.

Let $w \in \Img \phi$ such that:

$w = e^{x i}$

for some $x \in \R$.

\(\ds w\) \(\in\) \(\ds \Img \phi\)
\(\ds \leadsto \ \ \) \(\ds \exists z \in K: \, \) \(\ds w\) \(=\) \(\ds z^n\)
\(\ds \leadsto \ \ \) \(\ds e^{x i}\) \(=\) \(\ds z^n\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds e^{x i / n}\)

Thus every element of $K$ has a preimage under $\phi$.

Hence:

$\Img \phi = K$

Thus, from the Quotient Theorem for Group Homomorphisms, $\phi$ can be decomposed into:

$\phi = \alpha \beta \gamma$

where:

$\alpha: K \to K$, which is the identity mapping
$\beta: K / \map \ker \phi \to K$, which is an isomorphism
$\gamma: K \to K / \map \ker \phi$, which is an epimorphism.


The result follows.

$\blacksquare$


Sources

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