Quotient Theorem for Group Homomorphisms/Examples/Integer Power on Circle Group
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
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 67 \ \delta$