Binary Operation

Let $S$ be a set, a binary operation on $S$ is a function $\beta \x8b\x86:S\mathrm{\Gamma \x97}S\beta \x86\x92S$ where we denote $\beta \x8b\x86(a,b)$ as $a\beta \x8b\x86b$ using infix notation

Note the distinction between binary operation and binary relation, a binary operation is a function, which is a binary relation with some extra conditions.

Commutative Binary Operation

A binary operation on a set $S$ such that for any $a,b\beta \x88\x88S$, then $a\beta \x8b\x86b=b\beta \x8b\x86a$ is called commutative

Associative Binary Operation

A binary operation on a set $S$ such that for any $a,b,c\beta \x88\x88S$, then $(a\beta \x8b\x86b)\beta \x8b\x86c=a\beta \x8b\x86(b\beta \x8b\x86c)$ is called associative

Left Fold

Suppose that $\beta \x8b\x86:S\mathrm{\Gamma \x97}S\beta \x86\x92S$ is a binary operation, and $({x}_{1},{x}_{2},\mathrm{\beta \x80\xa6}{x}_{k})$ be a tuple in ${\mathbb{N}}_{0}$ for some $k\beta \x88\x88{\mathbb{N}}_{1}$ then $$foldl(\beta \x8b\x86,({x}_{1},{x}_{2},\mathrm{\beta \x80\xa6}{x}_{k})):=\{\begin{array}{cc}(foldl(\beta \x8b\x86,({x}_{1},{x}_{2},\mathrm{\beta \x80\xa6}{x}_{k\beta \x88\x921})))\beta \x8b\x86{x}_{k}\hfill & \text{\Beta if\Beta}k\beta \x89\u20af2\hfill \\ {x}_{1}\hfill & \text{\Beta if\Beta}k=1\hfill \end{array}$$

Left fold formalizes the idea that we want to look at objects of the form $((({x}_{0}\beta \x8b\x86{x}_{1})\beta \x8b\x86{x}_{2})\beta \x8b\x86\beta \x8b\u2015\beta \x8b\x86{x}_{k})$, we also have a right fold:

Right Fold

Suppose that $\beta \x8b\x86:S\mathrm{\Gamma \x97}S\beta \x86\x92S$ is a binary operation, and $({x}_{1},{x}_{2},\mathrm{\beta \x80\xa6}{x}_{k})$ be a tuple in ${\mathbb{N}}_{0}$ for some $k\beta \x88\x88{\mathbb{N}}_{1}$ then $$foldr(\beta \x8b\x86,({x}_{1},{x}_{2},\mathrm{\beta \x80\xa6}{x}_{k})):=\{\begin{array}{cc}{x}_{1}\beta \x8b\x86(foldr(\beta \x8b\x86,({x}_{2},\mathrm{\beta \x80\xa6}{x}_{k})))\hfill & \text{\Beta if\Beta}k\beta \x89\u20af2\hfill \\ {x}_{1}\hfill & \text{\Beta if\Beta}k=1\hfill \end{array}$$

Distributivity

Let $\beta \x8a\x95$ be a binary operation on a set $S$, and let $\beta \x8a\x97:X\mathrm{\Gamma \x97}S\beta \x86\x92S$ be a binary function for some set $X$ then we say that **$\beta \x8a\x97$ distributes into $\beta \x8a\x95$** whenever we have for any $a,b\beta \x88\x88S$ and $c\beta \x88\x88X$ $$c\beta \x8a\x97(a\beta \x8a\x95b)=(c\beta \x8a\x97a)\beta \x8a\x95(c\beta \x8a\x97b)$$

Bracket Placement Doesn't Matter for Associative Binary Operations

Suppose that $\beta \x8b\x86$ is an associative binary operation, then for any $n\beta \x88\x88{\mathrm{\beta \x84\u20ac}}^{\beta \x89\u20af3}$, then no matter how one places the brackets in the product ${x}_{1}\beta \x8b\x86{x}_{2}\beta \x8b\x86\mathrm{\beta \x80\xa6}\beta \x8b\x86{x}_{n}$ it always evaluates to the same value

Associativity Implies Folds are Equal

Suppose that $\beta \x8b\x86:S\mathrm{\Gamma \x97}S\beta \x86\x92S$ is associative, and $({x}_{1},\mathrm{\beta \x80\xa6}{x}_{k})$ is a tuple in ${\mathbb{N}}_{1}$, then $$foldl(\beta \x8b\x86,({x}_{1},\mathrm{\beta \x80\xa6}{x}_{k}))=foldr(\beta \x8b\x86,({x}_{1},\mathrm{\beta \x80\xa6},{x}_{k}))$$

Fold of a Composition Law

Suppose that $\beta \x8b\x86:S\mathrm{\Gamma \x97}S\beta \x86\x92S$ is a composition law, and $({x}_{1},\mathrm{\beta \x80\xa6}{x}_{k})$ a tuple in ${\mathbb{N}}_{0}$ for $k\beta \x88\x88{\mathbb{N}}_{1}$ $$fold(\beta \x8b\x86,({x}_{0},{x}_{1},\mathrm{\beta \x80\xa6},{x}_{k})):=foldl(\beta \x8b\x86,({x}_{0},{x}_{1},\mathrm{\beta \x80\xa6},{x}_{k}))$$

Sum Fold

Suppose we have $+:\mathbb{R}\mathrm{\Gamma \x97}\mathbb{R}\beta \x86\x92\mathbb{R}$ then and $R\beta \x8a\x86\mathbb{R}$ a finite tuple, then we define $$\beta \x88\x91R=fold(+,R)$$ in terms of the sum fold

Ellipses Notation

Suppose that $\beta \x8b\x86:S\mathrm{\Gamma \x97}S\beta \x86\x92S$ is a composition law, and $({x}_{1},\mathrm{\beta \x80\xa6}{x}_{k})$ a tuple in ${\mathbb{N}}_{0}$ for $k\beta \x88\x88{\mathbb{N}}_{1}$ $${x}_{1}\beta \x8b\x86{x}_{2}\beta \x8b\x86\beta \x8b\u2015\beta \x8b\x86{x}_{k}:=fold(\beta \x8b\x86,({x}_{0},{x}_{1},\mathrm{\beta \x80\xa6},{x}_{k}))$$

Since all bracketings yield the same value, our definition of fold to be the left fold was quite arbitrary. We can also now look at previous things under simpler lights, recall a definition of ${\beta \x88\x91}_{i=0}^{k}{a}_{i}$ that you're familiar with, this is simply just $fold(+,({a}_{0},\mathrm{\beta \x80\xa6},{a}_{k}))$

Composition Law

A composition law is an associative binary operation

Identity Element

Given a set $S$ and a binary relation $\beta \x8b\x86$, we say that an element $e\beta \x88\x88S$ is an identity element on $S$ when for every $s\beta \x88\x88S,e\beta \x8b\x86s=s\beta \x8b\x86e=s$

Binary Operation has Inverses For an Identity

Given a set $S$ and a binary operation $\beta \x8b\x86$ on $S$, that has an identity element $e$, then we say that **$\beta \x8b\x86$ has inverses for $e$** if for every $s\beta \x88\x88S$, there exists some $r\beta \x88\x88S$ such that $$s\beta \x8b\x86r=e$$

Group

We say that $(G,\beta \x8b\x86)$ where $G$ is a set and $\beta \x8b\x86$ is a composition law on $G$, when there exists an element $e\beta \x88\x88G$ satisfying the following properties:

- $e$ is an identity element on $G$
- inversion: for every $a\beta \x88\x88G$ we have some $b\beta \x88\x88G$ such that $a\beta \x8b\x86b=b\beta \x8b\x86a=e$

note: *informally this says that you can do nothing, and you can also undo anything*

Trivial Group

A trivial group is a group that has one element

The Integers are a Group

$\mathbb{Z},+$ form a group

Note that at this point in time we can write things like $a+b$ and that makes sense, if we were to write something like $xa+b$ this should not make sense because at this point we only have one operator, and not a multiplication operator yet, but we can think of $xa$ as "syntactic sugar" for $a+a+\mathrm{\beta \x80\xa6}+a$ (with $x$ repetitions) and it's alright.

One Point sets are Form Trivial Groups

Suppose that $T$ is a set such that $|T|=1$, and $\beta \x8b\x86$ is a constant composition law, then $(T,\beta \x8b\x86)$ forms a group

It should be clear by now that $\beta \x88\x85$ is not a group since it cannot have an identity element because there are no elements as candidate for this position. At this point in time we know that in a group there is at least one element $e$ with the above properties, we will now find out that there is exactly one such element.

identity element is unique in a group

Suppose $G$ is a group, then the identity element $e$ is unique

inverse element is unique

Suppose $G$ is a group, and $a\beta \x88\x88G$, then there is a unique element $b$ such that $a\beta \x8b\x86b=b\beta \x8b\x86a=e$

Inverse

Given a group $G$ and $a\beta \x88\x88G$, we denote the unique element $b$ such that $a\beta \x8b\x86b=b\beta \x8b\x86a=e$ as ${a}^{\beta \x88\x921}$ and call it the inverse of $a$

two inverses cancel

${\left({a}^{\beta \x88\x921}\right)}^{\beta \x88\x921}=a$

inverse of ab

Given a group $G$ and $a,b\beta \x88\x88G$, then ${(a\beta \x8b\x86b)}^{\beta \x88\x921}={b}^{\beta \x88\x921}\beta \x8b\x86{a}^{\beta \x88\x921}$

Cancellation

Suppose $G$ is a group and $a,b,c\beta \x88\x88G$, then $a\beta \x8b\x86c=b\beta \x8b\x86c\mathrm{\beta \x9f\u0389}a=b$ and $c\beta \x8b\x86a=c\beta \x8b\x86b\mathrm{\beta \x9f\u0389}a=b$, in other words, we can cancel common elements on right and left

Order of a group

The order of a group $G$ denoted by $\left|G\right|$ is the cardinality of $G$

Abelian

The Integers with Addition are Abelian

$\mathbb{Z},+$ where $+$ is defined in the usual sense form an abelian group

Cartesian Product of Abelian Groups is Abelian

Let $(G,\beta \x8b\x86)$ be an abelian group, then ${G}^{n}$ forms an abelian group with the operation $\beta \x8a\x95$ defined for $a,b\beta \x88\x88{G}^{n}$ as $$a\beta \x8a\x95b:=({a}_{1}\beta \x8b\x86{b}_{1},\mathrm{\beta \x80\xa6},{a}_{n}\beta \x8b\x86{b}_{n})$$

Can the above be generalized for any extra group properties?

Multiplicative Notation

Given a group $(G,\beta \x8b\x86)$ then since there is only one operation in question, instead of writing the operator between every two pairs of elements we may limit it to $ab$, to refer to $a\beta \x8b\x86b$ to reduce visual clutter

note: *since $\mathrm{{\rm B}\xb7}$ is also a simple thing to write, we may also use it in place of $\beta \x8b\x86$*

Power of an Element

Given a group $G$ and $a\beta \x88\x88G$, then we define ${a}^{n}:=\stackrel{\text{n\Beta times}}{\stackrel{\beta \x8f\x9e}{aa\mathrm{\beta \x80\xa6}a}}$ for any $n\beta \x88\x88{\mathrm{\beta \x84\u20ac}}^{>0}$ and ${a}^{0}:=e$, when $n\beta \x88\x88{\mathrm{\beta \x84\u20ac}}^{<0}$ then ${a}^{n}:={\left({a}^{\beta \x88\x921}\right)}^{n}$, thus we've defined ${a}^{n}$ for all $n\beta \x88\x88\mathrm{\beta \x84\u20ac}$

Power Properties

For all $a,b\beta \x88\x88G$, ${a}^{n}{a}^{m}={a}^{n+m}$ and ${\left({a}^{n}\right)}^{m}={a}^{nm}$ for any $m,n\beta \x88\x88\mathrm{\beta \x84\u20ac}$

Order of a Group Element

The order of an element $a\beta \x88\x88G$, denoted $\left|a\right|$, is the smallest $n\beta \x88\x88{\mathrm{\beta \x84\x95}}_{1}$ such that ${a}^{n}=e$

Suppose $G$ is a group and that $a\beta \x88\x88G$ if $a$ has finite order given by $n\beta \x88\x88{\mathbb{N}}_{0}$, then ${a}^{i}={a}^{j}$ iff $i\text{\Beta}\%\text{\Beta}n=j\text{\Beta}\%\text{\Beta}n$

If the power of an element with finite order yields the identity then the order divides the power

Let $G$ be a group and suppose $a\beta \x88\x88G$ has order $n$ then if ${a}^{k}=e$ then $n\beta \x88\pounds k$

Subgroup

Suppose that $(G,\beta \x8b\x86)$ is a group, then if $H\beta \x8a\x86G$ is a group under $\beta \x8b\x86$, then we say that $H$ is a subgroup of $G$ and write $H\mathrm{\beta \copyright \xbd}G$

Let $G$ be a group. Given a non-empty $H\beta \x8a\x86G$ if $a{b}^{\beta \x88\x921}\beta \x88\x88H$ for any $a,b\beta \x88\x88H$, then $H\mathrm{\beta \copyright \xbd}G$

Notice that the equation stated above is special, firstly given two elements in $a,b\beta \x88\x88H$ it combines $a$ and ${b}^{\beta \x88\x921}$, the special thing here is the ${b}^{\beta \x88\x921}$, originally we don't know if this element is a member of $H$ because in the proof we verify if indeed it is a group, and we don't know that inverses exists until we verify them, but we do know when it's combined with ${b}^{\beta \x88\x921}$ it produces an element of $H$.

Note that we need $H$ to be non-empty otherwise the empty set would satisfy the above, and we know that the empty set is not a group. It also helps us bootstrap the proof.

TODO

An element of a group has order $1$ if and only if $x=e$

TODO

Show that $\mathrm{\beta \x84\x9a},\mathrm{\beta \x84\x9d},\mathrm{\beta \x84\x82}$ are groups under multiplicaiton

Generator

The notation $\beta \x9f\xa8a\beta \x9f\copyright $ denotes the set $\{{a}^{n}:n\beta \x88\x88\mathrm{\beta \x84\u20ac}\}$. Then $a\beta \x88\x88G$ is said to be a generator of $g$ if and only if $G=\beta \x9f\xa8a\beta \x9f\copyright $

cyclic group

A group $G$ is considered a cyclic group if and only iff there exists some $a\beta \x88\x88G$ such that $G=\beta \x9f\xa8a\beta \x9f\copyright $

cyclic implies abelian

If a group $G$ is cyclic, then it is abelian

power equivalence

For a group $G$ with $a\beta \x88\x88G$, if $G$ has infinite order, then ${a}^{i}={a}^{j}$ if and only if $i=j$ otherwise if $G$ has finite order $n$ then ${a}^{i}={a}^{j}$ if and only if $i\%n=j\%n$

gcd generates the same

Suppose $G$ is a group. Let $a\beta \x88\x88G$ so that $\left|a\right|=n$, then $\beta \x9f\xa8{a}^{k}\beta \x9f\copyright =\beta \x9f\xa8{a}^{gcd(n,k)}\beta \x9f\copyright $ and $\left|{a}^{k}\right|=\frac{n}{gcd(n,k)}$

TODO

For a group $G$ and $a\beta \x88\x88G$, then $\left|a\right|$ divides $\left|G\right|$