MATH3303
Last edited 24 Dec 2022
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:

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

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

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

(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:

\[\mathbb{Z} = \left( \mathbb{Z}, + \right)\]
 Easy to see from context because multiplication would violate the inverse rule

\[\mathbb{R} = \left( \mathbb{R}, + \right)\]
 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
 \(\mathbb{R}^\times = \left( \mathbb{R}^\times, \cdot \right)\), 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)
 \(\mathbb{Z}_n = \left(\mathbb{Z}_n, +_n \right)\), 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).
 \(GL_n(\mathbb{C}) = \left( GL_n(\mathbb{C}), \cdot \right)\), 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
 \(SL_n(\mathbb{C}) = \left( SL_n(\mathbb{C}), \cdot \right)\), i.e. the special linear group of \(n \times n\) complex matrices of determinant \(1\) equipped matrix multiplication.
 Working with matrices of nonzero determinant implies invertibility and hence implies multiplication (see above)
 \(D_{2m} = \left( D_{2m}, \circ \right)\), 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 counterclockwise rotation of \(120^\circ\)
 \(r_2\): the counterclockwise 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:
 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.
 If \(\text{im}(f) = H\), \(f\) is surjective.
Proof of 1. (trivial kernel implies injectivity and viceversa)
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, \(a = 1\).
\(\Rightarrow \Leftarrow \square\).