# MATH3303

Last edited 17 Jun 2023

Here is a loose collection of notes from the Abstract Algebra component of MATH3303. The goal of these notes is to provide a quick reference to some of the ideas present in Abstract Algebra which will be relevant for MATH7133 and MATH4108.

## Revision of basic group theory

### Group definition

Recall a group $$\left(G, \cdot\right)$$ is a set $$G$$ equipped with a binary operation $$\cdot$$ satisfying the following requirements:

1. (Closure) For all $$a, b \in G$$, we have $$a \cdot b \in G$$

2. (Associativity) For all $$a, b, c \in G$$, we have $$\left( a \cdot b \right) \cdot c = a \cdot \left( b \cdot c \right)$$

3. (Identity element) There exists some $$e \in G$$ such that, for every $$g \in G$$, we have $$g \cdot e = g$$.

4. (Inverse element) For all $$g \in G$$, there exists some $$g^{-1} \in G$$ such that $$g \cdot g^{-1} = g^{-1} \cdot g = e$$ where $$e$$ is the identity element.

If a group is commutative (i.e. $$a \cdot b = b \cdot a$$ for all $$a, b \in G$$), we call it an Abelian group.

### Remark on group notation

Often, when referring to the group, we will just write $$G$$ and the binary operation is assumed from context. For example:

1. $\mathbb{Z} = \left( \mathbb{Z}, + \right)$
• Easy to see from context because multiplication would violate the inverse rule
2. $\mathbb{R} = \left( \mathbb{R}, + \right)$
• Easy to see from context because multiplication on $$\mathbb{R}$$ would imply the existance of a multiplicative inverse for $$0$$ which is not the case
3. $$\mathbb{R}^\times = \left( \mathbb{R}^\times, \cdot \right)$$, recall $$\mathbb{R}^\times := \mathbb{R}\setminus\left\{0 \right\}$$, sometimes written $$\mathbb{R}^*$$
• Using $$\mathbb{R}^\times$$ suggests we are using multiplication here to avoid worrying about an inverse for 0 (something we would not have to worry about if we were using addition)
4. $$\mathbb{Z}_n = \left(\mathbb{Z}_n, +_n \right)$$, i.e. addition modulo $$n$$
• Again, easy to see that multiplication would violate the inverse rule, therefore, addition is assumed. Furthermore, we are assuming addition modulo $$n$$ because of the context clue the set we are working with is $$\left\{ \overline{0}, \overline{1}, \cdots, \overline{n - 1} \right\}$$ which fits nicely (the overline notation represents the idea that the elements of $$\mathbb{Z}_n$$ are, themselves, sets i.e. congruence classes).
5. $$GL_n(\mathbb{C}) = \left( GL_n(\mathbb{C}), \cdot \right)$$, i.e. the general linear group of invertible $$n \times n$$ complex matrices with matrix multiplication.
• Working with the invertible matrices implies we are interested in multiplication as to not violate the inverse rule
6. $$SL_n(\mathbb{C}) = \left( SL_n(\mathbb{C}), \cdot \right)$$, i.e. the special linear group of $$n \times n$$ complex matrices of determinant $$1$$ equipped matrix multiplication.
• Working with matrices of non-zero determinant implies invertibility and hence implies multiplication (see above)
7. $$D_{2m} = \left( D_{2m}, \circ \right)$$, i.e. the Dihedral group of $$m$$ reflections and $$m$$ rotations, equipped with composition.
• Composition is a standard choice of operation for any two symmetries

### The Dihedral group

Consider the Dihedreal group of $$3$$ reflections and $$3$$ rotations. You can imagine these are the symmetries of an equilateral triangle. Call $$D_6$$’s elements:

• $$1$$: the rotation of $$0^\circ$$ (i.e., the identity)
• $$r_1$$: the counter-clockwise rotation of $$120^\circ$$
• $$r_2$$: the counter-clockwise rotation of $$240^\circ$$
• $$s_i$$: the reflection about the bisecting line from vertex $$0$$ to its opposite edge ($$i = 0, 1, 2$$) We can represent the composition operation with the Cayley table over at https://en.wikipedia.org/wiki/Dihedral_group#Group_structure.

### Group isomorphisms

An isomorphism between groups $$G$$ and $$H$$ is a homomorphism that is injective and surjective (bijective). It is denoted $$G \cong H$$.

• A homomorphism is a function $$f: G \to H$$ such that, for all $$a, b \in G$$, we have that $$f(a \cdot_G b) = f(a) \cdot_H f(b)$$. It is said that “$$f$$ is compatible with the group structure” or $$f$$ preserves algebraic structure”.

• An injective homomorphism has the additional property that $$f(a) = f(b) \Rightarrow a = b$$, i.e. each input maps to a unique output.

• A surjective homomorphism has the additional property that, for each $$b \in H$$, there exists an $$a \in G$$ such that $$f(a) = b$$, i.e. the image of $$f$$ completely covers the codomain $$H$$.

• Therefore, a group isomorphism is a function $$f: G \to H$$ having:

• $f(a \cdot_G b) = f(a) \cdot_H f(b)$
• $f(a) = f(b) \Rightarrow a = b$
• $\forall b \in H,\,\, \exists a \in G \,\mid f(a) = b$

### Kernel and image

The Kernel of a homomorphism $$f: G \to H$$ is the set of all elements in $$G$$ which are mapped to the identity of $$H$$, i.e. $$\text{ker}(f) := \left\{ g \in G \, \mid f(g) = 1_H \right\}$$. Most of the time, we don’t bother writing $$1_H$$ as it is clear that $$1$$ is in the image of $$f$$ and hence an element of $$H$$.

The Image of a homomorphism $$f: G \to H$$ is the set of all elements mapped to by $$f$$, i.e. 

Note:

1. If $$\text{ker}(f) = \left\{ 1 \right\}$$, $$f$$ is injective. The converse is also true. This means that the kernel can be viewed as a measure of the degree to which the homomorphism fails to be injective.
2. If $$\text{im}(f) = H$$, $$f$$ is surjective.

#### Proof of 1. (trivial kernel implies injectivity and vice-versa)

We want to show

• [ Forwards ]: $$\text{ker}(f) = \left\{ 1 \right\} \Rightarrow f$$ is injective; and
• [ Backwards ]: is injective $$\Rightarrow \text{ker}(f) = \left\{ 1 \right\}$$.

[ Forwards ]:

Let $$\text{ker}(f) := \left\{ 1 \right\}$$ and suppose $$f$$ is not injective, i.e. there exists some $$a \ne b \in G$$ such that $$f(a) = f(b)$$. Now consider $$1 = f(a) \cdot f(b)^{-1}$$. Then, due to the homomorphism property, $$1 = f(a \cdot b^{-1}) = f(1)$$. Again, due to homomorphism, only $$1$$ is mapped to $$1$$ under $$f$$, so $$a \cdot b^{-1} = 1 \Rightarrow a = b$$.

$$\Rightarrow \Leftarrow \square$$.

[ Backwards ]:

Let $$f$$ be injective and suppose $$\text{ker}(f) \ne \left\{ 1 \right\}$$. We know $$1$$ must be in the kernel for otherwise $$f$$ is not a homomorphism (consider $$f(a) = f(1 \cdot a) = f(1) \cdot f(a) = 1 \cdot f(a)$$). Then suppose $$1 \ne a \in \text{ker}(f)$$. Then $$f(a) = 1 = f(1)$$. But since $$f$$ is injective, $$a = 1$$.

$$\Rightarrow \Leftarrow \square$$.