Difference between revisions of "Group"
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| + | {{Requires references|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== | ==Definition== | ||
| − | A group is a set {{M|G}} and an operation <math>*:G\times G\rightarrow G</math>, denoted <math>(G,*:G\times G\rightarrow G)</math> but [[Mathematicians are lazy|mathematicians are lazy]] so we just write <math>(G,*)</math> | + | A group is a set {{M|G}} and an operation <math>*:G\times G\rightarrow G</math>, denoted <math>(G,*:G\times G\rightarrow G)</math> but [[Mathematicians are lazy|mathematicians are lazy]] so we just write <math>(G,*)</math> (or sometimes {{M|(G,*,e_G)}} where {{M|e_G}} is the identity element of the group), often just "Let {{M|G}} be a group" with the implicit operation of "[[juxtaposition]]", meaning {{M|ab}} denotes the group's operation applied to the elements {{M|a\in G}} and {{M|b\in G}}. |
Such that the following axioms hold: | Such that the following axioms hold: | ||
| − | |||
{| class="wikitable" border="1" | {| class="wikitable" border="1" | ||
|- | |- | ||
| Line 16: | Line 16: | ||
|- | |- | ||
| <math>\forall g\in G\exists x\in G[xg=gx=e]</math> | | <math>\forall g\in G\exists x\in G[xg=gx=e]</math> | ||
| − | | All elements of {{M|G}} have an [[Inverse element|inverse element]] under {{M|*}}, that is | + | | All elements of {{M|G}} have an [[Inverse element|inverse element]] under {{M|*}}, that is to say for each item there is an item such that {{M|*}}-ing the item with {{M|g\in G}} we end up with the identity. |
|- | |- | ||
| − | !colspan="2"|For an "Abelian" or "commutative" group | + | !colspan="2"|For an [[Abelian group|"Abelian" or "commutative" group]] |
|- | |- | ||
| <math>\forall g\in G\forall h\in G[gh=hg]</math> | | <math>\forall g\in G\forall h\in G[gh=hg]</math> | ||
| Order of the operation does not matter - it is [[Commutative|commutative]] | | Order of the operation does not matter - it is [[Commutative|commutative]] | ||
|} | |} | ||
| + | |||
| + | ==Trivial group== | ||
| + | The trivial group {{M|G}} is the group of just one element, naturally all these groups are basically the same group, they are "isomorphic groups" (which is a bijective [[Homomorphism|group homomorphism]]) | ||
==Notations== | ==Notations== | ||
| Line 43: | Line 46: | ||
One need not write "(because {{M|G}} is Abelian)" after steps in proofs, additive implies Abelian so for example if I am writing: | One need not write "(because {{M|G}} is Abelian)" after steps in proofs, additive implies Abelian so for example if I am writing: | ||
| − | + | * <math>...\implies A+B=C+A</math> but <math>C+A=A+C</math> so we may use the cancellation laws.... | |
| − | + | ||
| − | + | ||
it is clear that I am using the property of commutativity | it is clear that I am using the property of commutativity | ||
| − | + | * <math>...\implies ab=ca</math> but <math>ca=ac</math> so we may use the cancellation laws.... | |
| − | + | ||
This looks like <math>ca=ac</math> may have come from a lemma or previous part, so writing: | This looks like <math>ca=ac</math> may have come from a lemma or previous part, so writing: | ||
| − | + | * <math>...\implies ab=ca</math> but <math>ca=ac</math> (as {{M|G}} is Abelian) so we may use the cancellation laws.... | |
| − | + | ||
Perfect | Perfect | ||
==Important theorems== | ==Important theorems== | ||
| Line 132: | Line 131: | ||
# <math>ba=ca\implies b=c</math> | # <math>ba=ca\implies b=c</math> | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
| − | Proof | + | Proof that <math>ab=ac\implies b=c</math> and <math>ba=ca\implies b=c</math> |
{{Begin Proof}} | {{Begin Proof}} | ||
| − | {{ | + | '''Proof that: ''' <math>ab=ac\implies b=c</math> |
| + | |||
| + | |||
| + | Suppose <math>ab=ac</math> | ||
| + | |||
| + | Pre-multiplying by {{M|a^{-1} }} we get: | ||
| + | |||
| + | <math>a^{-1}(ab)=a^{-1}(ac)</math>, by associativity | ||
| + | |||
| + | <math>\iff (a^{-1}a)b=(a^{-1}a)c</math> | ||
| + | |||
| + | <math>\implies b=c</math> as required | ||
| + | |||
| + | |||
| + | '''Proof that: ''' <math>ba=ca\implies b=c</math> | ||
| + | |||
| + | |||
| + | Essentially the same, just post-multiply instead. | ||
{{End Proof}} | {{End Proof}} | ||
{{End Theorem}} | {{End Theorem}} | ||
| − | {{Definition|Abstract Algebra}} | + | ==If <math>ab=e</math> then <math>b=a^{-1}</math>== |
| − | {{Theorem|Abstract Algebra}} | + | This is the final theorem on this page, and it is easy to show. |
| + | |||
| + | If <math>ab=e=aa^{-1}\implies ab=aa^{-1}\implies b=a^{-1}</math> | ||
| + | |||
| + | ===<math>(a^{-1})^{-1}=a</math>=== | ||
| + | |||
| + | As <math>a^{-1}a=e</math> by definition of inverse, we see from the theorem <math>a=(a^{-1})^{-1}</math> | ||
| + | |||
| + | |||
| + | ==See also== | ||
| + | * [[Generated subgroup]] | ||
| + | * [[Cyclic subgroup]] | ||
| + | * [[Commutator]] | ||
| + | * [[Coset]] | ||
| + | * [[Subgroup]] | ||
| + | * [[Normal subgroup]] | ||
| + | * [[Commutator subgroup]] | ||
| + | |||
| + | {{Definition|Abstract Algebra|Group Theory}} | ||
| + | {{Theorem Of|Abstract Algebra|Group Theory}} | ||
| + | [[Category:First-year friendly]] | ||
| + | [[Category:Exemplary pages]] | ||
Latest revision as of 07:17, 16 October 2016
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Contents
Definition
A group is a set [ilmath]G[/ilmath] and an operation [math]*:G\times G\rightarrow G[/math], denoted [math](G,*:G\times G\rightarrow G)[/math] but mathematicians are lazy so we just write [math](G,*)[/math] (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 |
|---|---|
| [math]\forall a,b,c\in G:[(a*b)*c=a*(b*c)][/math] | [ilmath]*[/ilmath] is associative, because of this we may write [math]a*b*c[/math] unambiguously. |
| [math]\exists e\in G\forall g\in G[e*g=g*e=g][/math] | [ilmath]*[/ilmath] has an identity element |
| [math]\forall g\in G\exists x\in G[xg=gx=e][/math] | 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 | |
| [math]\forall g\in G\forall h\in G[gh=hg][/math] | 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.
Additive
Seriously, additive notation unofficially [math]\implies[/math] 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, [math]nx=x+x+...+x[/math] 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 [math]x^n=xxx...x[/math] 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:
- [math]...\implies A+B=C+A[/math] but [math]C+A=A+C[/math] so we may use the cancellation laws....
it is clear that I am using the property of commutativity
- [math]...\implies ab=ca[/math] but [math]ca=ac[/math] so we may use the cancellation laws....
This looks like [math]ca=ac[/math] may have come from a lemma or previous part, so writing:
- [math]...\implies ab=ca[/math] but [math]ca=ac[/math] (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 [math]e\ne e'[/math] are both identities, reach a contradiction)
Assume there are two identity elements, [ilmath]e[/ilmath] and [ilmath]e'[/ilmath] with [math]e\ne e'[/math].
That is both:
- [math]\forall g\in G[e*g=g*e=g][/math]
- [math]\forall g\in G[e'*g=g*e'=g][/math]
But then [math]ee'=e[/math] and also [math]ee'=e'[/math] thus we see [math]e'=e[/math] 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 [math]a+0=0+a=a[/math] |
| [ilmath](G,*)[/ilmath] - multiplicative notation [ilmath]ab[/ilmath] | We denote the identity [ilmath]1[/ilmath], so [math]1a=a*1=a[/math] |
| [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 [math]AI=IA=A[/math] |
Inverse is unique
Proof that the inverse is unique. (Suppose that [math]x[/math] 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: [math]\forall g\in G\exists x\in G[xg=gx=e][/math])
Then we have both:
- [math]gx=xg=e[/math]
- [math]gx'=x'g=e[/math]
Take the product [math]xgx'[/math]
By associativity
- [math]xgx'=(xg)x' = ex' = x'[/math]
- [math]xgx'=x(gx') = xe = x[/math]
Thus [math]x=x'[/math] 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 [math]x+(-x)=(-x)+x=0[/math] |
| Note that [math]a+(-x)[/math] is often written as [math]a-x[/math] - 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 [math]x^{-1}x=xx^{-1}=1[/math] |
| [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.
- [math]ab=ac\implies b=c[/math]
- [math]ba=ca\implies b=c[/math]
Proof that [math]ab=ac\implies b=c[/math] and [math]ba=ca\implies b=c[/math]
Proof that: [math]ab=ac\implies b=c[/math]
Suppose [math]ab=ac[/math]
Pre-multiplying by [ilmath]a^{-1} [/ilmath] we get:
[math]a^{-1}(ab)=a^{-1}(ac)[/math], by associativity
[math]\iff (a^{-1}a)b=(a^{-1}a)c[/math]
[math]\implies b=c[/math] as required
Proof that: [math]ba=ca\implies b=c[/math]
Essentially the same, just post-multiply instead.
If [math]ab=e[/math] then [math]b=a^{-1}[/math]
This is the final theorem on this page, and it is easy to show.
If [math]ab=e=aa^{-1}\implies ab=aa^{-1}\implies b=a^{-1}[/math]
[math](a^{-1})^{-1}=a[/math]
As [math]a^{-1}a=e[/math] by definition of inverse, we see from the theorem [math]a=(a^{-1})^{-1}[/math]
See also
- Pages requiring references
- Pages requiring references of unknown grade
- Todo
- Definitions
- Abstract Algebra Definitions
- Abstract Algebra
- Group Theory Definitions
- Group Theory
- Theorems
- Theorems, lemmas and corollaries
- Abstract Algebra Theorems
- Abstract Algebra Theorems, lemmas and corollaries
- Group Theory Theorems
- Group Theory Theorems, lemmas and corollaries
- First-year friendly
- Exemplary pages
