# Group

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Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
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The bulk of this page was written when this was a 'note project' and was taken from books (even though I was already really familiar with it) and is accurate and trustworthy, however references are on the to-do list and will be easy to find

## Definition

A group is a set [ilmath]G[/ilmath] and an operation $*:G\times G\rightarrow G$, denoted $(G,*:G\times G\rightarrow G)$ but mathematicians are lazy so we just write $(G,*)$ (or sometimes [ilmath](G,*,e_G)[/ilmath] where [ilmath]e_G[/ilmath] is the identity element of the group), often just "Let [ilmath]G[/ilmath] be a group" with the implicit operation of "juxtaposition", meaning [ilmath]ab[/ilmath] denotes the group's operation applied to the elements [ilmath]a\in G[/ilmath] and [ilmath]b\in G[/ilmath].

Such that the following axioms hold:

Formal Words
$\forall a,b,c\in G:[(a*b)*c=a*(b*c)]$ [ilmath]*[/ilmath] is associative, because of this we may write $a*b*c$ unambiguously.
$\exists e\in G\forall g\in G[e*g=g*e=g]$ [ilmath]*[/ilmath] has an identity element
$\forall g\in G\exists x\in G[xg=gx=e]$ All elements of [ilmath]G[/ilmath] have an inverse element under [ilmath]*[/ilmath], that is to say for each item there is an item such that [ilmath]*[/ilmath]-ing the item with [ilmath]g\in G[/ilmath] we end up with the identity.
For an "Abelian" or "commutative" group
$\forall g\in G\forall h\in G[gh=hg]$ Order of the operation does not matter - it is commutative

## Trivial group

The trivial group [ilmath]G[/ilmath] is the group of just one element, naturally all these groups are basically the same group, they are "isomorphic groups" (which is a bijective group homomorphism)

## Notations

Usually with groups we use "multiplicative notation", if the group is Abelian we use additive. This is (probably) motivated from linear algebra. Addition of matrices is commutative, just like with numbers however multiplication is not (always) commutative, so we do not.

Seriously, additive notation unofficially $\implies$ the group is Abelian - we use [ilmath]0[/ilmath] for the identity and [ilmath](-x)[/ilmath] for the inverse, [ilmath]y-x[/ilmath] is simply a short hand for [ilmath]y+(-x)[/ilmath]

Doing things multiple times is denoted as one would expect, $nx=x+x+...+x$ Always be explicit that n is not in the group

TODO: Relate to group action

To do this write something like "Let [ilmath]G[/ilmath] be an Abelian group with the operation [ilmath]+:G\times G\rightarrow G[/ilmath] given by (definition of addition)"

If the operation is obvious then "Let [ilmath]G[/ilmath] be the set of (whatever) and let [ilmath](G,+)[/ilmath] be a group"

### Multiplicative

Multiplicative groups may be Abelian, but it really ought to be explicit We use [ilmath]1[/ilmath] to denote the identity element and [ilmath]x^{-1} [/ilmath] to denote the inverse. Note that $x^n=xxx...x$ x multipled n times. Make it clear that n is not a member of the group!

TODO: Link with group action

## Convention notes

One need not write "(because [ilmath]G[/ilmath] is Abelian)" after steps in proofs, additive implies Abelian so for example if I am writing:

• $...\implies A+B=C+A$ but $C+A=A+C$ so we may use the cancellation laws....

it is clear that I am using the property of commutativity

• $...\implies ab=ca$ but $ca=ac$ so we may use the cancellation laws....

This looks like $ca=ac$ may have come from a lemma or previous part, so writing:

• $...\implies ab=ca$ but $ca=ac$ (as [ilmath]G[/ilmath] is Abelian) so we may use the cancellation laws....

Perfect

## Important theorems

### Identity is unique

Proof that the identity is unique. (Method: assume [ilmath]e[/ilmath] and [ilmath]e'[/ilmath] with $e\ne e'$ are both identities, reach a contradiction)

Assume there are two identity elements, [ilmath]e[/ilmath] and [ilmath]e'[/ilmath] with $e\ne e'$.

That is both:

• $\forall g\in G[e*g=g*e=g]$
• $\forall g\in G[e'*g=g*e'=g]$

But then $ee'=e$ and also $ee'=e'$ thus we see $e'=e$ contradicting that they were different.

Now we know the identity is unique, so we can give it a symbol:

Group Identity element
[ilmath](G,+)[/ilmath] - additive notation [ilmath]a+b[/ilmath] We denote the identity [ilmath]0[/ilmath], so $a+0=0+a=a$
[ilmath](G,*)[/ilmath] - multiplicative notation [ilmath]ab[/ilmath] We denote the identity [ilmath]1[/ilmath], so $1a=a*1=a$
[ilmath]\text{GL}(n,F)[/ilmath] - the General linear group

(All [ilmath]n\times n[/ilmath] matrices of non-zero determinant)

We denote the identity by [ilmath]Id,I,I_n[/ilmath] or sometimes [ilmath]Id_n[/ilmath]
that is $AI=IA=A$

### Inverse is unique

Proof that the inverse is unique. (Suppose that $x$ and [ilmath]x'[/ilmath] are both inverses with [ilmath]x\ne x'[/ilmath] and reach a contradiction

Suppose that for any [ilmath]g\in G[/ilmath] we have [ilmath]x[/ilmath] and [ilmath]x'[/ilmath] being inverses (recall the inverse statement: $\forall g\in G\exists x\in G[xg=gx=e]$)

Then we have both:

• $gx=xg=e$
• $gx'=x'g=e$

Take the product $xgx'$

By associativity

• $xgx'=(xg)x' = ex' = x'$
• $xgx'=x(gx') = xe = x$

Thus $x=x'$ contradicting that they were distinct.

We may now denote the inverse of an element uniquely.

Here [ilmath]x[/ilmath] is some arbitrary member of [ilmath]G[/ilmath]

Group Inverse element
[ilmath](G,+)[/ilmath] - additive notation [ilmath]a+b[/ilmath] We denote the inverse by [ilmath]-x[/ilmath], so $x+(-x)=(-x)+x=0$
Note that $a+(-x)$ is often written as $a-x$ - this is a shorthand, no "subtraction" is defined
[ilmath](G,*)[/ilmath] - multiplicative notation [ilmath]ab[/ilmath] We denote the inverse of [ilmath]x[/ilmath] by [ilmath]x^{-1} [/ilmath], so $x^{-1}x=xx^{-1}=1$
[ilmath]\text{GL}(n,F)[/ilmath] - the General linear group

(All [ilmath]n\times n[/ilmath] matrices of non-zero determinant)

We denote the inverse of [ilmath]X\in GL(n,F)[/ilmath] by [ilmath]X^{-1} [/ilmath]

### Cancellation laws

These are extremely important.

1. $ab=ac\implies b=c$
2. $ba=ca\implies b=c$

Proof that $ab=ac\implies b=c$ and $ba=ca\implies b=c$

Proof that: $ab=ac\implies b=c$

Suppose $ab=ac$

Pre-multiplying by [ilmath]a^{-1} [/ilmath] we get:

$a^{-1}(ab)=a^{-1}(ac)$, by associativity

$\iff (a^{-1}a)b=(a^{-1}a)c$

$\implies b=c$ as required

Proof that: $ba=ca\implies b=c$

Essentially the same, just post-multiply instead.

## If $ab=e$ then $b=a^{-1}$

This is the final theorem on this page, and it is easy to show.

If $ab=e=aa^{-1}\implies ab=aa^{-1}\implies b=a^{-1}$

### $(a^{-1})^{-1}=a$

As $a^{-1}a=e$ by definition of inverse, we see from the theorem $a=(a^{-1})^{-1}$