binary operations and groups

binary operation
Let $$S$$ be a set, a binary operation on $$S$$ is a function $$\star : S \times S \to S$$ where we denote $$\star \left ( a , b \right )$$ as $$a \star b$$ using infix notation
Commutative Binary Operation
A binary operation on a set $$S$$ such that for any $$a , b \in S$$, then $$a \star b = b \star a$$ is called commutative
Associative Binary Operation
A binary operation on a set $$S$$ such that for any $$a , b , c \in S$$, then $$\left ( a \star b \right ) \star c = a \star \left ( b \star c \right )$$ is called associative
Distributivity
Let $$\oplus$$ be a binary operation on a set $$S$$, and let $$\otimes : X \times S \to S$$ be a binary function for some set $$X$$ then we say that $$\otimes$$ distributes into $$\oplus$$ whenever we have for any $$a, b \in S$$ and $$c \in X$$ $c \otimes \left( a \oplus b \right) = \left( c \otimes a \right) \oplus \left( c \otimes b \right)$
Bracket Placement Doesn't Matter for Associative Binary Operations
Suppose that $$\star$$ is an associative binary operation, then for any $$n \in \mathbb{Z}^{\ge 3}$$, then no matter how one places the brackets in the product $$x_{1} \star x_{2} \star \ldots \star x_{n}$$ it always evaluates to the same value
TODO
Composition Law
A composition law is an associative binary operation
Identity Element
Given a set $$S$$ and a binary relation $$\star$$, we say that an element $$e \in S$$ is an identity element on $$S$$ when for every $$s \in S, e \star s = s \star e = s$$
Binary Operation has Inverses For an Identity
Given a set $$S$$ and a binary operation $$\star$$ on $$S$$, that has an identity element $$e$$, then we say that $$\star$$ has inverses for $$e$$ if for every $$s \in S$$, there exists some $$r \in S$$ such that $s \star r = e$
Group
We say that $$\left( G, \star \right)$$ where $$G$$ is a set and $$\star$$ is a composition law on $$G$$, when there exists an element $$e \in G$$ satisfying the following properties:
• $$e$$ is an identity element on $$G$$
• inversion: for every $$a \in G$$ we have some $$b \in G$$ such that $$a \star b = b \star a = e$$

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 $$x a + b$$ 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 $$x a$$ as "syntactic sugar" for $$a + a + \ldots + a$$ (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 $$\star$$ is a constant composition law, then $$\left( T, \star \right)$$ forms a group

Let $$t$$ be the only element of $$T$$, we claim this is the identity, which we can verify by checking $$t \star t = t$$, because $$\star$$ is constant, thus it is an identity element.

Now we can also show that we have inversion as well, because for every element in $$T$$ (which is just $$t$$) we have an element (which is also $$t$$) such that $$t \star t = t$$, where the $$t$$ on the right hand side is the identity.

It should be clear by now that $$\emptyset$$ 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
Suppose that $$f$$ and $$g$$ are both identity elements, then by the identity property of a group, we know that $$f{\star} g{=} g{\star} f{=} f$$ since $$g$$ is an identity element, but also $$g{\star} f{=} f{\star} g{=} g$$ since $$f$$ is the identity, as well. Relating the equalities yields that $$f{=} g$$ and therefore the identity element is unique.
inverse element is unique
Suppose $$G$$ is a group, and $$a \in G$$, then there is a unique element $$b$$ such that $$a \star b = b \star a = e$$
Suppose that $$b$$ and $$c$$ are both inverses to $$a$$, thus $$a \star b = e = c \star a$$, then $$c = c \star e = c \star \left ( a \star b \right ) = \left ( c \star a \right ) \star b = e \star b = b$$ so then $$c = b$$
Inverse
Given a group $$G$$ and $$a \in G$$, we denote the unique element $$b$$ such that $$a \star b = b \star a = e$$ as $$a^{- 1}$$ and call it the inverse of $$a$$
two inverses cancel
$$\left ( a^{- 1} \right )^{- 1} = a$$
To show this is true we want to show that $$a^{- 1} \star a = a \star a^{- 1} = e$$, though the definition of $$a^{- 1}$$ is such that $$a \star a^{- 1} = a^{- 1} \star a = e$$ and since the equality symbol is a reflexive relation, this proves what we needed to show
inverse of ab
Given a group $$G$$ and $$a , b \in G$$, then $$\left ( a \star b \right )^{- 1} = b^{- 1} \star a^{- 1}$$
For ease let $$c = \left ( a \star b \right )^{- 1}$$, so that $$\left ( a \star b \right ) \star c = c \star \left ( a \star b \right ) = e$$ since we know $$\left ( a \star b \right ) \star c = a \star \left ( b \star c \right ) = e$$ , and thus $$a^{- 1} \star \left ( a \star \left ( b \star c \right ) \right ) = a^{- 1} \star e$$, thus $$\left ( a^{- 1} \star a \right ) \star \left ( b \star c \right ) = a^{- 1} e$$, which simplifies to $$e \star \left ( b \star c \right ) = a^{- 1}$$, so we can see that $$a^{- 1} = b \star c$$, as needed
Cancellation
Suppose $$G$$ is a group and $$a , b , c \in G$$, then $$a \star c = b \star c \implies a = b$$ and $$c \star a = c \star b \implies a = b$$, in other words, we can cancel common elements on right and left
Suppose that $$a \star c = b \star c$$, let $$c ^ { -1 }$$ be the inverse of $$c$$ , then we have the following $\begin{gather} a \star c = b \star c &\iff \\ \left( a \star c \right) \star c ^ { -1 } = \left( b \star c \right) \star c ^ { -1 } &\iff \\ a \star \left( c \star c ^ { -1 } \right) = b \star \left( c \star c ^ { -1 } \right) &\iff \\ a \star e = b \star e &\iff \\ a = b \end{gather}$ Where we've used associativity followed by the definition of inverse and then the identity element. The proof for right cancellation can be obtained by placing a mirror on the left of the above chain of equalities and reversing individual elements, so that if you see ɘ you interpret it as e.
Order of a group
The order of a group $$G$$ denoted by $$\left | G \right |$$ is the cardinality of $$G$$
Abelian
Given a group, if it's composition law is commutative, then we say that the group is 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 $$\left( G, \star \right)$$ be an abelian group, then $$G ^ n$$ forms an abelian group with the operation $$\oplus$$ defined for $$a, b \in G ^ n$$ as $a \oplus b := \left( a _ 1 \star b _ 1, \ldots , a _ n \star b _ n \right)$

Can the above be generalized for any extra group properties?

Multiplicative Notation

Given a group $$\left ( G , \star \right )$$ 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 $$a b$$, to refer to $$a \star b$$ to reduce visual clutter

note: since $$\cdot$$ is also a simple thing to write, we may also use it in place of $$\star$$

Power of an Element
Given a group $$G$$ and $$a \in G$$, then we define $$a^{n} := \overbrace{a a \ldots a}^{\text{n times}}$$ for any $$n \in \mathbb{Z}^{\gt 0}$$ and $$a^{0} := e$$, when $$n \in \mathbb{Z}^{\lt 0}$$ then $$a^{n} := \left ( a^{- 1} \right )^{n}$$, thus we've defined $$a^{n}$$ for all $$n \in \mathbb{Z}$$
Power Properties
For all $$a , b \in G$$, $$a^{n} a^{m} = a^{n + m}$$ and $$\left ( a^{n} \right )^{m} = a^{n m}$$ for any $$m , n \in \mathbb{Z}$$
TODO
Order of a Group Element
The order of an element $$a \in G$$, denoted $$\left | a \right |$$, is the smallest $$n \in \mathbb{N}_{1}$$ such that $$a^{n} = e$$
Suppose $$G$$ is a group and that $$a \in G$$ if $$a$$ has finite order given by $$n \in \mathbb{ N } _ 0$$, then $$a ^ i = a ^ j$$ iff $$i ~\%~ n = j ~\%~ n$$
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 $$a \in G$$ has order $$n$$ then if $$a ^ k = e$$ then $$n \mid k$$
We know that $$a ^ k = e = a ^ 0$$ therefore $$k ~\%~ n = 0 ~\%~ n = 0$$ therefore $$n \mid k$$ as needed.
Subgroup
Suppose that $$\left( G, \star \right)$$ is a group, then if $$H \subseteq G$$ is a group under $$\star$$, then we say that $$H$$ is a subgroup of $$G$$ and write $$H \leqslant G$$
Let $$G$$ be a group. Given a non-empty $$H \subseteq G$$ if $$a b ^ { -1 } \in H$$ for any $$a, b \in H$$, then $$H \leqslant G$$

Since $$H$$ is non-empty then there is some element $$h \in H$$, thus we know that $$h h ^ { -1 } \in H$$, since $$h \in H$$, since $$H \subseteq G$$, then we know that $$h \in G$$, and therefore in $$G$$ we have an inverse $$h ^ { -1 }$$ such that $$h h ^ { -1 } = e$$, where $$e \in G$$ but by assumption this is also an element of $$H$$, thus $$e \in H$$.

So now we know that $$e \in H$$, how can we be sure this is the identity in $$H$$? Perhaps it is not, but given any $$s \in H$$, since $$H \subseteq G$$, we know $$s \in G$$ then using the fact that $$G$$ is a group we see $$e \star s = s \star e$$, since $$e$$ is also in $$H$$ this confirms that it is $$H$$'s identity.

Now we'll show that $$H$$ has inverses, so let $$a \in H$$, at the same time we know we have $$e \in H$$, thus $$e a ^ { -1 } \in H$$ and thus $$a ^ { -1 } \in H$$ as needed.

At this point it may seem like we've verified all the requirements for $$H$$ to be a group, but there's something a little subtle, when we were working with $$\star$$ we know that it was composition law meaning that this operation is closed for elements in $$G$$, so we also need to verify that $$\star$$ is closed in $$H$$, that is to say that $$\star$$ is a composition law in $$H$$, so let $$x, y \in H$$, then from the previous paragraph we know that $$y ^ { -1 }$$ is a member of $$H$$, thus $$x \left( y ^ { -1 } \right) ^ { -1 } \in H$$

Notice that the equation stated above is special, firstly given two elements in $$a, b \in H$$ it combines $$a$$ and $$b ^ { -1 }$$, the special thing here is the $$b ^ { -1 }$$, 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 $$b ^ { -1 }$$ 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 $$x = e$$

Suppose that $$a \in G$$, then $$\left | a \right | = 1$$ if and only if $$a^{1} = e$$ note that $$a^{1} = a$$, so $$a = e$$ as needed

For the reverse direction, follow the above proof backwards, noting that each connection is a bi-implication

TODO
Show that $$\mathbb{Q} , \mathbb{R} , \mathbb{C}$$ are groups under multiplicaiton
TODO
Generator
The notation $$\left \langle a \right \rangle$$ denotes the set $$\left \lbrace a^{n} : n \in \mathbb{Z} \right \rbrace$$. Then $$a \in G$$ is said to be a generator of $$g$$ if and only if $$G = \left \langle a \right \rangle$$
For a positive integer n, all groups of order n are cyclic iff GCD(n, phi(n)) = 1
cyclic group
A group $$G$$ is considered a cyclic group if and only iff there exists some $$a \in G$$ such that $$G = \left \langle a \right \rangle$$
cyclic implies abelian
If a group $$G$$ is cyclic, then it is abelian
TODO
power equivalence
For a group $$G$$ with $$a \in G$$, 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 $$i \% n = j \% n$$
TODO
gcd generates the same
Suppose $$G$$ is a group. Let $$a \in G$$ so that $$\left | a \right | = n$$, then $$\left \langle a^{k} \right \rangle = \left \langle a^{\gcd{\left ( n , k \right )}} \right \rangle$$ and $$\left | a^{k} \right | = \frac{n}{\gcd{\left ( n , k \right )}}$$
TODO
TODO
For a group $$G$$ and $$a \in G$$, then $$\left | a \right |$$ divides $$\left | G \right |$$
TODO