A group is a set with a defined product. This product is associative. The group is closed under this product. We have an identity element and each element has an inverse. Formally,

Definition 1.10.1

A group is a (nonempty) set $G$ with a product satisfying the following properties:

The group is closed under the operation.
The product is associative. For all $a, b, c \in G$, we have $(ab)c = a(bc)$.
There is an identity element $e \in G$ with property that for all $a \in G$, $ea = ae = a$.
For each element $a \in G$, there is an element $a^{-1}\ \in G$ satisfying $aa^{-1} = a^{-1}a = e$ (The inverse).

What are examples of the groups?

$\mathbf{Z}$ with the addition operation.
The set of invertible $n \times n$ matrices with entries in $\mathbf{R}$ with matrix multiplication as the product.
Any vector space is a group if you forget about the scalar multiplication.
For any set $T$, define the set of all bijections $g: T \rightarrow T$. The set of all bijections is the symmetric group of $T$. The operation here is the composition of these maps.

The Symmetric Group

A little more on the symmetric group example above. A permutation is a bijection from a set to itself. A symmetric group is the set of all permutations on a set with the composition as the group operation.

This is a group because the identity element is the identity transformation and it is a valid permutation so it’s in the set. The inverse transformation is the inverse element. We are guaranteed to have an inverse because any bijection has an inverse and the inverse of a permutation is another permutation. The composition of permutations is another permutation so the group is closed under composition. Finally, the product is associative.

The symmetric group on a set $A$ is denoted by $S_A$. The symmetric group on the set $\{1,2,3,...,n\}$ is denoted by $S_n$ where $n$ is the number of elements.

The simplest symmetric group is $S_1 = \{e\}$ because a group can’t be empty. It must contain at least one element.

What about $S_2$? The set is $\{1,2\}$. We have an identity map $e$. We also have the map that takes 1 to 2 and 2 to 1. Let this permutation be $\tau$. So $S_2 = \{e, \tau\}$. Composing $\tau \circ \tau = e$. So $\tau^{-1} = \tau$. Side note here: When a permutation only changes two elements, it is called a transposition.

Uniqueness of the Identity and the Inverses

In the next two propositions, we’ll prove that the identity element in a group is unique and similarly the inverses are unique.

Proposition 2.1.1 (Uniqueness of the identity)

Let $G$ be a group and suppose $e$ and $e'$ are both identity elements in $G$; that is, for all $g \in G$, $eg = ge = e'g = ge' = g$. Then $e = e'$.

Proof
Since $e$ is an identity element, then $e'e = e'$ and since $e'$ is an identity element $ee' = e$. Therefore, $e = e'$. $\ \blacksquare$

Proposition 2.1.2 (Uniqueness of inverses)

Let $G$ be a group and let $h, g \in G$. If $hg = e$, then $h = g^{-1}$. Likewise, if $gh = e$, then $h = g^{-1}$.

Proof
Suppose $hg = e$, then

$$ \begin{align*} h = h(gg^{-1}) = (hg)g^{-1}= eg^{-1} = g^{-1}. \end{align*} $$

Suppose now that $gh = e$, then

$$ \begin{align*} g = g(hh^{-1}) = (gh)h^{-1} = eh^{-1} = h^{-1}. \ \blacksquare \end{align*} $$

Corollary 2.1.3

Let $g$ be an element of a group $G$. We have $g = (g^{-1})^{-1}$

Proof
We know $gg^{-1} = e$. By (2.1.2) $g = (g^{-1})^{-1}$. $\ \blacksquare$

Proposition 2.1.4

Let $G$ be a group and let $a,b \in G$. Then We have $(ab)^{-1} = b^{-1}a^{-1}$

Proof
Notice that $(ab)(b^{-1}a^{-1}) = a(b((b^{-1}a^{-1})) = a((bb^{-1})a^{-1}) = a(ea^{-1}) = aa^{-1} = e$. Therefore, $(ab)^{-1} = b^{-1}a^{-1}$.

The Left and Right Multiplication Maps

Proposition 2.1.5

Let $G$ be a group and $a \in G$.
The map $L_a: G \rightarrow G$ defined by $L_a(x) = ax$ is a bijection. Similarly,
the map $R_a: G \rightarrow G$ defined by $R_a(x) = xa$ is a bijection.

Proof
To show that the map $L_a$ is a bijection, we need to show that $L_a$ is both one to one and onto or equivalently that $L_a$ has an inverse. We claim that the left multiplication by $a^{-1}$ or $L_{a^{-1}}$ is the inverse of $L_a$. To see that,

$$ \begin{align*} L_{a^{-1}}(L_a(x)) = a^{-1}(ax) = (a^{-1}a)x = ex = x. \end{align*} $$

Similarly,

$$ \begin{align*} L_a(L_{a^{-1}}(x)) = a(a^{-1}x) = (aa^{-1})x = ex = x. \end{align*} $$

So $L_{a^{-1}}$ and $L_a$ are inverse maps so both are bijective. $\ \blacksquare$

Corollary 2.1.6

Let $G$ be a group and let $a$ and $b$ be elements of $G$. The equation $ax = b$ has a unique solution $x$ in $G$. and likewise the equation $xa = b$ has a unique solution in $G$.

Proof
For $ax = b$ to have a solution. The map $L_a$ needs to be onto or surjective. For the solution to be unique, the map needs to be one to one or injective. Similarly for $xa = b$ to have a solution, we want $R_a$ to be a bijective. Since we proved earlier that $L_a$ and $R_a$ are bijections, then both equations have unique solutions. $\ \blacksquare$

Corollary 2.1.7 (Cancellation)

Suppose $a, x, y$ are elements of a group $G$. If $ax = ay$, then $x = y$. Similarly, if $xa = ya$, then $x = y$.

Proof
Suppose $ax = ay$. We know that $L_a = ax$ is one to one. So for any elements $x, y \in G$, $ax = ay$ must imply that $x = y$ by definition of a one to one or injective map. A similar arguments shows that if $xa = ya$ must imply that $x = y$ by the injectivity of $R_a$. $\ \blacksquare$

Corollary 2.1.8

If $G$ is a finite group, each row and each column of the multiplication table of $G$ contains each element of $G$ exactly once.

Proof
A row in the multiplication table can be represented by a left multiplication map $G \rightarrow G$ if you fix the element multiplied on the left. We know the left multiplication map is a bijection. Therefore every element/result must be unique and each element of $G$ must show up in the row. Similarly, each column can be represented by a right multiplication map. The map is a bijection and so each element must be unique and shown exactly once. (TODO clean up this proof)

Isomorphic Groups

Definition 2.1.10

The order of a group is its size or cardinality. We will denote the order of a group $G$ by $|G|$.

[TODO: Examples]

Definition 2.1.13

We say two groups $G$ and $H$ are isomorphic if there is a bijection $\phi: G \rightarrow H$ such that for all $g_1, g_2 \in G, \phi(g_1g_2) = \phi(g_1)\phi(g_2)$. The map $\phi$ is called an isomorphism.

Definition 2.1.14

A group $G$ is called abelian (or commutative) if for all elements $a, b \in G$, the products in the two orders are equal $ab = ba$.

Proposition 2.1.17

Up to isomorphism, $\mathbf{Z_1}$ is the unique group of order 1.
Up to isomorphism, $\mathbf{Z_2}$ is the unique group of order 2.
Up to isomorphism, $\mathbf{Z_3}$ is the unique group of order 3.
Up to isomorphism, there are exactly two groups of order 4, namely $\mathbf{Z_4}$, and the group of rational symmetries of the rectangular card.
Up to isomorphism, $\mathbf{Z_5}$ is the unique group of order 5.
All groups of order no more than 5 are abelian.
There are at least two non-isomorphic groups of order 6, one abelian and one non-abelian.

Up to isomorphism just means that that two objects will be considered the same if they are isomorphic. We just want to classify groups into distinct non-isomorphic groups. For example, for statement (c), This means groups of order 3 are all isomorphic to $\mathbf{Z_3}$ while statement (d) implies that we have two distinct groups of order 4.

Proof
TODO

Again two groups are isomorphic means that the multiplication tables match up. In fact, not only that but the identity elements and inverses of elements match up as well. The following propositions state these facts.

Proposition 2.1.18

If $\phi : G \rightarrow H$ is an isomorphism, then $\phi(e_G) = e_H$, and for each $g \in G$, $\phi(g^{-1}) = \phi(g)^{-1}$

Proof
Since $\phi$ is an isomorphism, then we know that for each $h \in H$, there is a $g \in G$ such that $\phi(g) = h$. Therefore,

$$ \begin{align*} \phi(e_G)h &= \phi(e_G)\phi(g) \\ &= \phi(e_Gg) \quad \text{$\phi$ is an isomorphism}\\ &= \phi(g) \\ &= h \\ \end{align*} $$

So $\phi(e_G)$ is an identity element. But since the identity element is unique, then $\phi(e_G) = \phi(e_H)$.

To show that $\phi(g^{-1}) = \phi(g)^{-1}$, see that

$$ \begin{align*} \phi(g^{-1})\phi(g) &= \phi(g^{-1}g) \quad \text{$\phi$ is an isomorphism} \\ &= \phi(e_G) \\ &= e_H \quad \text{by the previous result} \end{align*} $$

So $\phi(g)$ is the inverse of $\phi(g^{-1})$ or in other words $\phi(g)^{-1} = \phi(g^{-1})$ as we wanted to show. $\ \blacksquare$

The General Associative Law

We know by definition that the product is associative so for all $a, b, c \in G$, we have $(ab)c = a(bc)$. What about the product of 4 or more elements? is it associative? For example, there are five ways to group four elements

$$ \begin{align*} a(b(cd)), a((bc)d), (ab)(cd), (a(bc))d, ((ab)c)d \end{align*} $$

$a(b(cd))$ and $((ab)c)d$ follow from the definition. For the rest, see that

$$ \begin{align*} a(bcd) = a(b(cd)) = (ab)(cd) = ((ab)c)d = (abc)d \end{align*} $$

In general this works for any number of elements. This is formally presented in the next proposition.

Proposition 2.1.19 (General associative law)

Let $M$ be a set with an associative operation, $M \times M \rightarrow M$, denoted by juxtaposition. For every $n \geq 1$, there is a unique product $M^n \rightarrow M$, $$ \begin{align*} (a_1, a_2,...,a_n) \rightarrow a_1a_2...a_n, \end{align*} $$ such that

The product of one element is that element $(a) = a$.
The product of two elements agrees with the given operation $(ab) = ab$.
For all $n \geq 2$, for all $a_1,...,a_n \in M$, and for all $1 \leq k \leq n-1$, $$ \begin{align*} a_1a_2...a_n = (a_1...a_k)(a_{k-1}...a_n) \end{align*} $$

Proof
By induction on $n$.
Base Case: For $n \leq 3$, property $(c)$ holds by definition.
Inductive Case: Suppose this is true for all $1 \leq r \leq n$ where a unique product of $r$ elements satisfies the properties $(a)-(c)$ above. Suppose now that we have $n$ elements. Fix elements $a_1, ...,a_n \in M$. By the inductive hypothesis, the $n-1$ products

$$ \begin{align*} p_k = (a_1...a_k)(a_{k+1}...a_n), \end{align*} $$

are defined since we have at most $n-1$ elements. … [TODO]

[2.1] Group Theory: First Results

The Symmetric Group

Uniqueness of the Identity and the Inverses

The Left and Right Multiplication Maps

Isomorphic Groups

The General Associative Law

References