Problem 5 - Algebra

Question

Für \(n \in \mathbb{N}\) sei \(E_{n}=\left\\{\mathrm{e}^{\frac{2 \pi k i}{n}} \mid k=0, \ldots, n-1\right\\}\) die Gruppe der \(n\)-ten Einheitswurzeln (mit dem üblichen Produkt der komplexen Zahlen). Begründen Sie, dass \(\varphi: \mathbb{Z} \rightarrow E_{n}, k \mapsto \varepsilon_{n}^{k}\) für \(\varepsilon_{n}=\mathrm{e}^{\frac{2 \pi i}{n}}\) ein Homomorphismus ist. Bestimmen Sie den Kern \(\operatorname{von} \varphi\).

Step-by-Step Solution

Answer

The homomorphism \(\varphi\) maps each integer \(k\) to \(\varepsilon_n^k\). Its kernel is \(n\mathbb{Z}\).

Step 1: Understand the mapping

The mapping \(\varphi: \mathbb{Z} \rightarrow E_n\) is defined by \(k \mapsto \varepsilon_n^k\), where \(\varepsilon_n = e^{\frac{2 \pi i}{n}}\). This suggests that each integer \(k\) is mapped to the \(k\)-th power of the \(n\)-th root of unity.

Step 2: Prove \(\varphi\) is a homomorphism

To show \(\varphi\) is a homomorphism, we need to demonstrate that \(\varphi(a + b) = \varphi(a) \varphi(b)\) for all \(a, b \in \mathbb{Z}\). We have \(\varphi(a + b) = \varepsilon_n^{a+b} = \varepsilon_n^a \cdot \varepsilon_n^b = \varphi(a) \varphi(b)\). Therefore, \(\varphi\) is a homomorphism.

Step 3: Define the kernel of \(\varphi\)

The kernel of a homomorphism is the set of all elements in the domain that are mapped to the identity element in the codomain. In this case, \(\varepsilon_n^k = 1\) implies \(e^{\frac{2\pi ki}{n}} = 1\). Therefore, \(\frac{2\pi k}{n} = 2\pi m\) for some integer \(m\). Solving for \(k\) gives \(k = mn\).

Step 4: State the kernel

The kernel of \(\varphi\) consists of all multiples of \(n\), i.e., \(\ker(\varphi) = n\mathbb{Z}\).

Key Concepts

Roots of Unity

Roots of unity are special complex numbers that arise when considering powers of complex numbers. They are defined as the solutions to the equation \[x^n = 1\]. This means that when you raise a number to the power of \(n\), you get 1. These numbers are spread out evenly along the unit circle in the complex plane. For example, the "3rd roots of unity" are the complex numbers which satisfy \(x^3 = 1\).
A key figure here is the primitive \(n\)-th root of unity, denoted by \(\varepsilon_n = e^{\frac{2 \pi i}{n}}\). It represents the exponential form of these roots. This primitive root can be used to generate all \(n\)-th roots of unity as powers from \(\varepsilon_n^0\) to \(\varepsilon_n^{n-1}\). For any integer \(k\), \(\varepsilon_n^k\) still remains on the unit circle. This makes \(\varepsilon_n\) ---a key component in several mathematical fields like Fourier transforms and algebraic equations.
In our exercise, the set of \(n\)-th roots of unity \(E_n\) forms a group under multiplication of complex numbers. This is crucial since operations within this set maintain its structure, an essential property for examining homomorphisms.

Kernel of a Homomorphism

The kernel of a homomorphism is a cornerstone concept in group theory. It helps identify the "structure-preserving" function's traits. In simple terms, a kernel includes all elements in the starting set (domain) that are changed into the special identity element in the ending set (codomain).
A homomorphism \(\varphi: G \rightarrow H\) in group theory is a function where the operation performed on two elements in group \(G\) reflects on their images in group \(H\). For multiplication, if \(\varphi(a \cdot b) = \varphi(a) \cdot \varphi(b)\), \(\varphi\) is a homomorphism.
In our specific case, the identity element here in the group of \(n\)-th roots of unity, \(E_n\), is 1---because multiplying any complex number by 1 leaves it unchanged. The kernel of the mapped homomorphism \(\varphi\) from earlier was found by setting \(\varepsilon_n^k = 1\). This directly means \(k = mn\) for integers \(m\), where \(n\) is constant. Thus, \(\ker(\varphi) = n\mathbb{Z}\), emphasizing that only multiples of \(n\) are mapped to identity in our case, holding significant importance in understanding the group's structure.

Complex Numbers

Complex numbers drastically extend our understanding of algebra beyond real numbers, by including imaginary units. A complex number is typically expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, satisfying \(i^2 = -1\).
They can also represent points in the 2D plane, with \(a\) as the x-coordinate and \(b\) as the y-coordinate (Complex Plane). This visualization helps in multiplying, adding, or performing other operations with them.
In polar form, complex numbers are depicted via magnitude (or modulus) and angle (or argument) relative to the positive x-axis. A complex number \(z = a + bi\) can also be presented as \(r(\cos \theta + i\sin \theta)\) or simply as \(re^{i\theta}\), providing an exponential representation crucial for understanding root behavior. This form is useful in finding roots of unity, as each "unit" complex number lies on the circle with radius 1 in this plane, simplifying multiplication and powers.
Complex numbers are core to numerous fields, from signal processing to quantum mechanics. Their rich structure, combined with operations such as the roots of unity, allow deep exploration into mathematical spaces and symmetries.