MA136 - Groups, Rings, Fields
Problem with ℝ/ℚ
- if you take a=π and b=e then a-bar + b-bar = (π+e)-bar - can this be simplified further to e.g. 0-bar? - it is iff π+e is rational
Prove that V2 of Lagrange's theorem implies V1
- let G be a finite group and g∈G and g has order n - the cyclic subgroup generated by g is <g> which also has order n - by V2, #<g>|#G so n|#G QED
Define commutativity
- let S be a set and ∘ be a binary operation on S - we say ∘ is commutative if a∘b=b∘a ∀a,b∈S
G is a group satisfying a²=1 ∀a∈G. Show that G is abelian
- let a, b∈G - as G is a group ab, ba∈G - 1=(ab)²=(ab)(ab)=a(ba)b - so ab=a²(ba)b² = ba QED
Find unit group of M₂(ℤ)
- suppose A∈M₂(ℤ) is a unit then there is a B∈M₂(ℤ) st AB=BA=I - so det(A)det(B)=1 so det(A)=±1 and det(B)=±1 - so M₂(ℤ)^={A∈M₂(ℤ) : det(A)=±1} - define GL₂(ℤ)={A∈M₂(ℤ) : det(A)=±1} then M₂(ℤ)^=GL₂(ℤ)
Why is dot product not a binary operation on ℝⁿ?
- you take two elements from ℝⁿ - but the output is an element of ℝ
Define transposition
Cycle of length 2
Explain whether or not subtraction is a binary operation on ℕ
It's not because subtraction can lead to negative numbers which aren't in ℕ
Describe cycle notation for permutations
Let a₁, a₂, ..., aₙ be distinct elements of the set {1, 2, ..., n} (a₁, a₂, ..., aₙ) denotes the element of Sₙ that takes a₁ to a₂, a₂ to a₃..., aₙ to a₁ The permutation is called a cycle of length m
Define nth symmetric group
Sₙ=Sym({1, 2, ..., n}) is the nth symmetric group
Show that ℝ⁺ has exactly two cosets in ℝ^
The only possibilities are 0ℝ⁺ = {0} and nℝ⁺=ℝ^ for any real n. Easy to show
Elements of S₂
((1, 2), (1, 2)) and ((1, 2), (2, 1))
Is the transposition (1,2,3)(4,5) odd or even?
(1,2,3) is even and (4,5) is odd so the product is odd
How can every permutation be written as a product of transpositions?
(a₁, a₂, ..., aₘ) = (a₁, aₘ)(a₁, aₘ₋₁)...(a₁, a₂)
4th alternating polynomial
(x₁-x₂)(x₁-x₃)(x₁-x₄)(x₂-x₃)(x₂-x₄)(x₃-x₄)
Unit group of ℤ/mℤ
(ℤ/mℤ)^ = {[a]∈ℤ/mℤ : 0≤a≤m-1, hcf(a,m)=1}
Prove V={(x, x): x∈ℝ} is a subgroup (under addition). Is V'={(x, -x): x∈ℝ} a subgroup?
- (0,0)∈V because 0=0 and 0∈ℝ so identity element is in V - let (a, a)∈V and (b, b)∈V then (a+b, a+b)∈V because a+b∈ℝ so V is closed under addition - let (a, a)∈V then it's inverse is (-a, -a)∈V so inverse exists 3 criterion are satisfied so V is a subgroup - similarly V' is a subgroup
Cosets of 𝕊 in ℂ^
- (e^iθ)𝕊 multiplies rotates the circle anticlockwise about O by angle θ so you just get the same set 𝕊 - so (re^iθ)𝕊 = r𝕊, r=/=0, is a circle centred origin radius r
How to represent a permutation as a matrix?
- 2 rows and n columns - first row is numbers 1 to n - second row indicate where each element of first row maps to
What is group of symmetries of a square
- D₄ is the set of symmetries of a square - ρ₀ ρ₁ ρ₂ ρ₃ are rotations where ρᵢ represents a rotation of 90i° anticlockwise - σ₀ σ₁ σ₂ σ₃ are reflections about different axes: top right to bottom left, top midpoint to bottom midpoint, top left to bottom right, left midpoint to right midpoint - if a and b are two symmetries then define a∘b to be the symmetry obtained by doing b first and then a - then (D₄, ∘) is a group
Prove that Sₙ has order n!
- Sₙ is set of bijections from {1, 2, ..., n} to itself - as the set {1, 2, ..., n} is finite, by pigeohole principle all the injections from it to itself are bijections - to count these, f(1) can be any one of 1, 2, ..., n - when f(1) is chosen, there are n-1 choices for f(2). - etc and there is 1 choice for f(n) - so number of injections is n(n-1)...(1)=n! QED
Prove that (ℤ/mℤ, +) is a group
- [0] is obviously the identity element - the sum [a]+[b] = [a+b] is a congruence class modulo m so ℤ/mℤ is closed under addition - for associativity ([a]+[b])+[c] = [a+b] + [c] = [(a+b)+c] = [a+(b+c)] because addition in ℤ is associative = [a] + [b+c] = [a]+([b]+[c]) - [a] + [-a] = [0] so every element has an inverse QED
Define ring
- a ring is a triple (R, +, •) where + and • are binary operations on R such that - (R, +) is an abelian group with identity element 0 - it's closed under multiplication • - associativity of multiplication holds - it has multiplicative identity 1 - distributivity a•(b+c)=a•b+a•c and (b+c)•a=b•a+c•a hold
Describe composition table
- a table where for every possible a∘b where a and b are in a group, the a represents the row and b the column - the value of a∘b is written at the intersection
Proof of Euler's theorem
- as hcf(a,m)=1, [a]∈(ℤ/mℤ)^ - order of (ℤ/mℤ)^ is by definition φ(m) - [a]^φ(m)=[1] by Lagrange's theorem - so a^φ(m)≡1 (mod m)
Show that ℚ is not cyclic
- assume it is cyclic and is generated by a rational a/b - the rational a/2b is in the group - but this is not an integer multiple of a/b so contradiction
Why is multiplication not well-defined in general in quotient groups? What quotient groups is it well-defined for?
- because we're trying to define the product of [a] and [b] in terms of their representatives a and b but each class has many representatives. - for a multiplication definition to make sense it must be independent of the the choice of these representatives - multiplication is well-defined on ℤ/mℤ
Proof Fermat's little theorem
- by definition of congruence we a≡b (mod p) for some b=0, 1, ..., p-1 - as p∤a we know that b≠0 so [b] is in the unit group of ℤ/pℤ which is (ℤ/pℤ)^={[1], [2], [p-1]} - this group has order p-1 - by corollary to Lagrange's theorem [b]^(p-1)=[1] - so b^(p-1)≡1 (mod p) and since a≡b (mod p) we get a^(p-1)≡1 (mod p) QED
Why is division not a binary operation on ℝ? Solution? Slight problem with solution?//
- division is not a binary operation on ℝ because if pick the real numbers 1 and 0, 1/0 is undefined and therefore isn't in ℝ - solution is to define ℝ* = {a∈ℝ : a≠0} which is closed under division - however, this no isn't closed under addition (i.e. addition is no longer a binary operation) because a-a=0 which isn't in the set
ℝ[x] is closed under division. True or false + 2 reasons?
- false for same reason as ℝ isn't closed under division - also, even if you take e.g. x and x+1 from ℝ[x], the result of x/(x+1) isn't a polynomial
Properties of the group D₄
- finite - non-abelian - no additational structure
How to show ℚ[sqrt(2)] is a field?
- first show it's a subring of ℝ, which shows that it is a commutative ring - then α∈ℚ[sqrt(2)]\{0} be a non-zero element in ℚ[sqrt(2)] - write α=a+bi and WTS α is a unit i.e. that there is a β∈ℚ[sqrt(2)] st αβ=βα=1 - easy to find this by finding 1/α = a/(a²-2b²)-bsqrt(2)/(a²-2b²) - there won't be problems with division by 0 because you can show that the denominator is never 0 - therefore 1/α∈ℚ[sqrt(2)] so α is a unit so ℚ[sqrt(2)] is a field
Prove that φ(m) is even for m≥3
- for m≥3, -1≠1 in ℤ/mℤ - (-1)²=1 so -1 has order 2 - Lagrange's theorem says order of element divides order of group - so order of (ℤ/mℤ)^=φ(m) is even QED
Prove the unit group of the gaussian integers is the fourth roots of unity
- gaussian integers are ℤ[i] - let α∈ℤ[i] be a unit then there is some β∈ℤ[i] st αβ=βα=1 - define the norm map N:ℤ[i]→ℤ by N(a+bi)=a²+b² and you can easily show that N(αβ)=N(α)N(β) - then N(1)=1 so N(α)=N(β)=1 or N(α)=N(β)=-1 - so a²+b²=1 or -1 but -1 is impossible so a²+b²=1 so a=0 and b=±1 or a=±1 and b=0 giving the solutions 1, i, -1, i
Show that the binomial coefficient is an integer
- if we can find a subgroup H of Sₙ of order m!(n-m)! then nCm = |Sₙ|/|H| - by Lagrange's theorem RHS is [Sₙ:H] (number of cosets of H in Sₙ) and is therefore an integer - let H be the subset of Sₙ consisting of permutations σ such that σ permutes A={1, 2, ..., m} and permutes B={m+1, m+2, ..., n} - elements of Sₙ are bijections from {1, 2, ..., n} to itself - the elements of H are those elements σ of Sₙ that satisfy σ(a)∈A ∀a∈A and σ(b)∈B ∀b∈B - let σ∈H then σ(1) can be any one of 1, 2, ..., m so m choices - σ(2) can be any of of 1, 2, ..., m except σ(1) so m-1 choices - until σ(m+1) which can be any element of B so n-m choices - σ(m+2) has n-m-1 choices... σ(n) has 1 choice - so order of H is m(m-1)...(1)(n-m)(n-m-1)...(1)=m!(n-m)!
How to show the group (D₄,∘) satisfies associativity?
- label the vertices of the square 1, 2, 3, 4 going anticlockwise - then you can think each element of D₄ as functions taking [1, 2, 3, 4] to itself - then ∘ just becomes a composition of functions from D₄ to itself - we know composition of functions is associative and therefore the binary operation ∘ on D₄ is
Prove that cyclic groups are abelian
- let G be a cyclic group generated by g - let a,b∈G. WTS ab=ba - a=g^m and b=g^n for some integers m, n - ab=g^(m+n) and ba=g^(n+m) - as addition of reals is associative ab=ba QED
Prove every subgroup of a cyclic group is cyclic
- let G be a cyclic group generated by g∈G so G=<g> and let H be a subgroup of G - every element of G and hence of H has the form gˢ - let m be the smallest positive integer st gᵐ∈H - let a∈H then a∈G so a=gⁿ for some integer n - using division algorithm, n=mq+r for 0<=r<m - so gⁿ=(gᵐ)^qgʳ so gʳ=gⁿ(gᵐ)^-q - gⁿ∈H and gᵐ∈H and H is a subgroup so gʳ∈H - since m was the smallest positive integer st gᵐ∈H we must have r=0 so n=qm and a=(gᵐ)^q showing that all elements in H are powers of gᵐ so H is cyclic QED
Proof of Lagrange's theorem V3
- let G be a finite group and H a subgroup - let g₁H, g₂H, ..., gₘH be the distinct left cosets of H in G - because they are distinct they are disjoint - suppose g∈G then gH must equal one of the gᵢH - but g∈gH because 1∈H - therefore the cosets are not only distinct but each element of G belongs to exactly one of them - therefore #G=#g₁H+#g₂H+...+#gₘH = #H+#H+...+#H = m•#H because size of coset of subgroup = size of subgroup - but m=[G:H] is the index of H in G QED
Prove that two cosets of a subgroup H in a group G are either equal or disjoint
- let G be a group and H a subgroup and g₁,g₂∈G - suppose g₁H and g₂H aren't disjoint, then they have an element in common - all elements in g₁H and g₂H are of form g₁h₁, g₂h₂ respectively where h₁,h₂∈H - there is at least one pair h₁, h₂ st g₁h₁=g₂h₂ ⇒ g₁=g₂h₂h₁⁻¹ - take an element g₁h from g₁H - note that g₁h = (g₂h₂h₁⁻¹)h = g₂(h₂h₁⁻¹h) ∈g₂H because h₂h₁⁻¹h∈H as H is a group - therefore g₁H⊆g₂H and similarly g₂H⊆g₁H so g₁H=g₂H QED
Define order of an element
- let G be a group and g be an element of the group - order of g is the smallest positive integer n st g^n=1 - if there is no such n then g has infinite order
Prove that addition is well-defined in quotient groups
- let G be an additive group and H a subgroup - let a, a', b, b' be elements of G st in G/H we have [a]=[a'] and [b]=[b'] - then a-a'∈H and b-b'∈H - as H is a group the sum must also be in H so (a-a')+(b-b')=(a+b)-(a'+b')∈H so the classes [a+b] and [a'+b'] are equal
What will order of element in an additive group be?
- let G be an additive group and g an element of G - order of g will be the smallest positive integer n st ng=1 - the identity element 1 is usually some form of 0
How to show that ℤ has no subrings other than itself?
- let S be a subring of ℤ - by definition, 0,1∈S - so 1+1∈S, etc so all naturals are in S - also, by definition, S is closed under negation so all integers are in S
Theorem to do with units of ℤ/mℤ and hcf(a, m) where [a]∈ℤ/mℤ + proof
- let [a]∈ℤ/mℤ then [a] is a unit iff hcf(a,m)=1 Proof - suppose [a] is a unit then ∃[b]∈ℤ/mℤ st [a][b]=[b][a]=1 so ab≡1 (mod m) - so m|(ab-1) so ∃k∈ℤ st mk=ab-1 so ab-mk=1 - write h=hcf(a,m) then h|ab and h|mk so h|1 hence h=1 - conversely suppose hcf(a,m)=1 then by Euclidean algorithm we can write 1=ab+cm for some b,c∈ℤ therefore ab≡1 (mod m) so [a] is a unit QED
Find all the elements of order 4 in ℝ/ℤ
- let a-bar∈ℝ/ℤ have order 4 - can suppose 0≤a<1 - then 4a-bar=0-bar ⇒ (4a)-bar = 0-bar - 4a≡0 (mod ℤ) ⇒ 4a-0∈ℤ ⇒ 4a∈ℤ - as 0≤4a<4 4a=0 or 1 or 2 or 3 - so a=0 or 0.25 or 0.5 or 0.75 - the class 0-bar has order 1 and the class 0.5-bar has order 2 - 0.25-bar and 0.75-bar are the elements with order 4
How to prove that every permutation can be written as a product of disjoint cycles?
- let ρ∈Sₙ - consider the sequence 1, ρ1, ρ²1, ... - every term in this finite set is contained in {1, 2, ..., n} thus the sequence must contain repetition - let ρᵘ1 be the first term in the sequence that has already appeared thus ρᵘ1=ρᵛ1 for some 0≤v<u - apply ρ⁻ᵛ to both sides to get ρᵘ⁻ᵛ1=1 - as 0<u-v≤u so if u-v<u then ρᵘ⁻ᵛ1 is the first term in the list that's already appeared which contradicts our assumption that ρᵘ1 was the first one - therefore u-v=u so v=0 hence ρᵘ1=1 and 1, ρ1, ρ²1, ..., ρᵘ⁻¹1 are distinct - let μ₁ be the cycle of length u μ₁=(1, ρ1, ρ²1, ..., ρᵘ⁻¹1) - it is clear that μ₁ has the same effect on the list 1, ρ1, ρ²1, ..., ρᵘ⁻¹1 as ρ - let a be the first element of the set {1, 2, ..., n} not appearing in the list 1, ρ1, ρ²1, ..., ρᵘ⁻¹1 - repeat the above argument with the sequence a, ρa, ρ²a, ... - then there must be a cycle μ₂ st μ₂ and ρ have the same effect on the elements a, ρa, ..., ρᵛ⁻¹a - now WTS that μ₁ and μ₂ are disjoint - suppose otherwise, then ρⁱ1=ρʲa for some 0≤i<u and 0≤j<v - apply ρᵛ⁻ʲ to both sides to get ρⁱ⁺ᵛ⁻ʲ1=ρʲ⁺ᵛ⁻ʲa=a so ρᵏ=a where k=i+v-j - this contradicts our assumption that a doesn't appear in the list 1, ρ1, ρ²1, ..., ρᵘ⁻¹1 hence μ₁ and μ₂ are disjoint - so now the product μ₁μ₂ has the same effect as ρ on the elements 1, ρ1, ρ²1, ..., ρᵘ⁻¹1, a, ρa, ..., ρᵛ⁻¹a - repeat this process starting with the first element of {1, 2, ..., n} not appearing in either cycle μ₁ or μ₂ to construct a μ₃ that is disjoint from both μ₁ and μ₂, etc - as the set {1, 2, ..., n} is finite the process must terminate eventually with some μᵣ - the product μ₁μ₂...μᵣ has the same effect as ρ on {1, 2, ..., n} therefore ρ=μ₁μ₂...μᵣ QED
How to prove that disjoint cycles commute?
- let σ and τ be disjoint cycles in Sₙ and write σ=(a₁, a₂, ..., aₖ) and τ=(b₁, b₂, ..., bₗ) - since σ and τ are disjoint aᵢ≠bⱼ ∀i=1, 2, ..., k and j=1, 2, ..., l Case x not equal to any of the aᵢ or bⱼ - then στx=σx=x=τx=τσx Case x=aᵢ for some i=1, 2, ..., k - so x≠bⱼ ∀j=1, 2, ..., l so τx=x - so στx=σx=σaᵢ=aᵢ₊₁ where aₖ₊₁ is interpreted as a₁ - τσx=τσaᵢ=τaᵢ₊₁=aᵢ₊₁ because aᵢ₊₁ doesn't equal any of the bⱼ - so τσx=στx Case x=bⱼ for some i=1, 2, ..., l - similar to previous case QED
What happens to sign of Pₙ if you apply a permutation from Sₙ on Pₙ + why?
- let σ∈Sₙ - if σ is the product of an even number of transpositions then σ(Pₙ)=Pₙ - if σ is the product of an odd number of transpositions then σ(Pₙ)=-Pₙ - because any permutation can be written as a product of transpositions and every transposition changes the sign of Pₙ
Differential equations and subgroups
- let 𝓁 be the set of infinitely differentiable real functions - the identity element is clearly 0 - can define a set e.g. H={x(t)∈𝓁 : tdx/dt-2x=0} - this set contains the identity element 0 because it satisfies the differential equation - suppose x_1 and x_2 are in H then you can show that their sum is also in H - suppose x_1 is in H then you can show that -x_1 is in H - therefore H is a subgroup of 𝓁
How to write a permutation as a product of disjoint cycles?
- start with 1 and see where the pemutation takes this element and repeat until you get back to 1. This forms one cycle - then start with an element that isn't in the previous cycle and see what cycle that makes - repeat until all elements are part of a cycle
Prove the group (G, ∘) has a unique identity element
- suppose e and e' are the identity elements - then for all a in G we have a∘e=e∘a=a and a∘e'=e'∘a=a - in first equation let a=e' then e'∘e=' - in second equation let a=e then e'∘e=e - so e'=e is a unique identity element QED
Prove that g^m=1 iff g has a finite order d with d|m
- suppose g has finite order d with d|m - then g^d=1 and m=qd for some integer q - so g^m = (g^d)^q = 1 - conversely suppose g^m=1 then g^|m|=1 and |m| is a positive integer - therefore g has a finite order - using division with remainder can write m=qd+r with q a integer and 0<=r<d - so 1=g^m=((g^d)^q)g^r = g^r - as d is the order it is the least positive integer such that g^d=1 so g^r=1 is possible iff r=0 so m=qd so d|m QED
Prove that g has order 1 iff g is the identity element
- suppose g has order 1 then by definition of order g^1=1 which is only possible when g=1 - conversely the order of the identity element is clearly 1 QED
Show that ℝ^ is not cyclic (using ^ instead of asterisk)
- suppose it is cyclic then it has a generator a - -1 is in the group - so there is a natural n s.t a^n=-1 but then a^2n=1 meaning order of a is finite - meaning the group is finite - this is a contradiction
Why is multiplying a vector in set ℝⁿ by a scalar not a binary operation?
- the vector and the result are in ℝⁿ - but the scalar is in ℝ, not in ℝⁿ
Is 𝕊={α∈ℂ : |α|=1} a subgroup?
- this is the unit circle centred at the origin in the complex plane - it is clearly a subset of ℂ* - identity element is 1 which is indeed in 𝕊 - suppose α,β∈𝕊 then |αβ|=|α||β| = 1x1 = 1 so αβ∈𝕊 - suppose α∈𝕊 then |α^-1| = 1/|α| = 1/1=1 so a^-1∈𝕊 QED
Prove that (Sym(A), ∘) is a group
- we know composition of functions is associative - we know this group is closed under composition because the function obtained by composing two bijections is also a bijection (foundations) - all the elements have inverses because they are bijections (foundations) - there is an identity element id:A→A defined by id(x)=x. Given any other element f∈A, f∘id(x)=f(x) and id∘f(x) = f(x) QED
Formula for calculating φ(m)
- write m as product of powers of prime factors - m=p_1^r_1...p_k^r_k - then φ(m)=(p_1^r_1-p_1^(r_1-1))...(p_k^r_k-p_k^(r_k-1))
Can you have abelian subgroup of a non-abelian group + e.g. with (D₃, ∘)?
- yes - e.g. D₃ has the subgroup ({ρ₀, ρ₁, ρ₂}, ∘) which is group of all rotations - this is abelian since it doesn't matter which order you do multiple rotations
Let G be a group with an element g. g has order n. Prove that <g> = {1, g, g², ..., gⁿ⁻¹}
- {1, g, g², ..., gⁿ⁻¹} ⊆ <g> = {..., g⁻², g⁻¹, 1, g, g², ...} obvious - suppose h∈<g>, then h=gᵐ - using division algorithm, m=qn+r where 0≤r≤n-1 - gᵐ = ((gⁿ)^q)gʳ = (1^q)gʳ = gʳ∈{1, g, g², ..., gⁿ⁻¹} because 0≤r≤n-1 - <g> ⊆ {1, g, g², ..., gⁿ⁻¹} and therefore the sets are equal QED
Prove |Aₙ|=|Sₙ|/2
- |Sₙ|=n! so WTS |Aₙ|=n!/2 - by Lagrange's theorem |Sₙ|=[Sₙ:Aₙ]|Aₙ| so suffices to show that [Sₙ:Aₙ]=2 - let τ∈Sₙ be a transposition - we know Aₙ is subset (and subgroup) of Sₙ containing all even permutations therefore the coset τAₙ contains only odd permutations but does it contain them all? - suppose σ∈Sₙ is odd then τσ is even and hence in Aₙ therefore τ(τσ) is in the coset τAₙ - but τ(τσ)=τ²σ=σ so σ∈Aₙ so τAₙ contains all odd permutations - there are no other cosets because they would have to overlap with either Aₙ or τAₙ but cosets are either disjoint or equal so Aₙ and τAₙ are the only two cosets and as all cosets have the same size, |Aₙ|=n!/2 QED
Let τ∈Sₙ be a transposition. Then what is τ(Pₙ)? + Explain
- τ(Pₙ)=-Pₙ - let the transposition be τ=(i,j) where i<j WLOG - there are n-i terms where xᵢ is at the start (i.e. terms of the form (xᵢ-...) (because this term can subtract n-i numbers bigger than it) - there are n-j terms of the form (xⱼ-...) - there are i-1 terms of the form (...-xᵢ) - there are j-1 terms of the form (...-xⱼ) - in total there are n-i+i-1=n-1 terms with xᵢ in them and n-1 terms with xⱼ in them - the only term where both appear is (xᵢ-xⱼ) (since i<j) which becomes negated (xⱼ-xᵢ) so there are n-2 other terms for each - each time a swap for xᵢ to xⱼ is made the sign flips, but for each swap of xᵢ there is also a swap for xⱼ to xᵢ so result is that the sign doesn't change - the sign becomes negated because of the (xᵢ-xⱼ) term
Describe congruence classes and ℤ/mℤ
- ℤ/mℤ is set of all congruence classes modulo m - e.g [a] is the congruence class of a modulo m so [a] = {..., a-2m, a-m, a, a+m, a+2m, ...} - ℤ/mℤ = {[0], [1], ..., [m-1]} - each class is distinct so there are exactly m classes
Elements with finite order in (ℝ*, x)
1 has order 1 -1 has order 2 Rest have infinite order
Value of [D₄:<ρ₁>] + why?
2 - because the only 2 cosets of R=<ρ₁> are 1R and σ₀R
How to tell if a permutation is odd or even?
A cycle of length n can be written as a product of n-1 transpositions So a cycle of length n is even if n is odd and it's odd if n is even
Define field
A field (F, +, •) is a commutative ring such that every non-zero element is a unit. So a commutative ring F is a field iff F^={a∈F : a≠0}
Abelian group definition
A group (G, ∘) is abelian if it also satisfies - commutativity ∀a,b∈S a∘b=b∘a
Define a group
A group is a pair (G, ∘) where G is a set and ∘ is a binary operation on G st the following properties hold - closure i.e. ∀a,b∈G a∘b∈G - associativity ∀a,b,c∈G (a∘b)∘c = a∘(b∘c) - existence of the identity element ∃e∈G st ∀a∈G e∘a=a∘e=a - existence of an inverse ∀a∈G ∃b∈G st a∘b=b∘a=e
Geometrically describe subgroup in ℝ³
A plane is a subgroup in ℝ³ iff it passes through the origin so i.e. has vector equation of the form r.n=0
What does it mean for a ring to be commutative?
A ring (R, +, •) is commutative if for all a, b in R, a•b=b•a
What is a field?
A set on which addition, subtraction, multiplication and division by non-zero numbers are defined and behave
Cosets of a line passing through the origin
All lines parallel to it
Units in ℝ
All the non-zero elements
How can every permutation be written?
As a product of disjoint cycles
Prove each element in group (G, ∘) has a unique inverse
Assume a has two inverses, b and c b=b∘e = b∘(a∘c) = (b∘a)∘c = e∘c = c Therefore b=c is the unique inverse of a QED
Define nth alternating group
Aₙ = {σ∈Sₙ:σ is even}
Why can you drop e.g. (3) in a product of cycles?
Because (3) is just the identity - it takes 3 to itself and fixes all other elements
Why is ℤ not a field?
Because e.g. 2∈ℤ but 2 isn't a unit
Why would the permutation the product of permutations (1, 4, 6)(3,2) give same result in S₆ as it would in S₇?
Because neither permutation contains numbers bigger than 6 so all numbers bigger than 6 would just stay fixed
Why does closure hold for all groups?
Because the operation associated with group is defined to be binary which ensures closure automatically
Why isn't ℝ[x] a field?
Because x∈ℝ[x] but x isn't a unit
Permutation
Element of Sₙ
How to find inverse of a residue class in ℤ/mℤ?
Euclidean algorithm
What's an even and odd permutation?
Even permutation can be written as product of even number of transpositions Odd permutation can be written as product of odd number of transpositions
What is ℝ/ℤ?
Every real number is congruent mod ℤ to a unique number in the interval [0, 1) The set of all congruence classes {a-bar : a∈[0,1)}
General linear group + its importance?
GL₂(ℝ) = {A∈M₂(ℝ) : detA≠0} - (M₂(ℝ), x) is not a group because not all matrices have inverses - so GL₂ is set of all invertible matrices - GL₂(ℝ) is a group under multiplication of matrices
How to find product (compose) two permutations in Sₙ?
If a and b are the two permutations in Sₙ and you want to find ab, then apply b first to the numbers 1, 2, ..., n and then apply a to this new set
Explain whether or not D₄ is a cyclic group
It's not because taking any element a∈ℤ and creating the cyclic subgroup <a> doesn't generate the whole of D₄
How to find inverse of a permutation written in cycle notation?
Just reverse the cycle
Define subgroup
Let (G, ∘) be a group and let H⊆G and suppose (H,∘) is also a group. Then we say (H,∘) is a subgroup of (G,∘)
Define isomorphism
Let (G,∘) and (H,✳) be groups We say the function φ:G→H is an isomorphism if it is a bijection and it satisfies φ(g₁∘g₂) = φ(g₁)✳φ(g₂) ∀g₁,g₂∈G i.e G and H are isomorphic
Define subring
Let (R, +, •) be a ring and S⊆R. Suppose that (S, +, •) is also a ring with the same multiplicative identity. Then S is a subring of R
Define congruence modulo subgroups
Let G be a an additive abelian group and H a subgroup of G. Let a,b∈G. We say a and b are congruent modulo H if a-b∈H and write a≡b (mod H)
Lagrange's Theorem V2
Let G be a finite group and H a subgroup of G. Then the order of H divides the order of G
Lagrange's theorem V3
Let G be a finite group and H a subgroup then #G=[G:H]•#H
Lagrange's Theorem V1
Let G be a finite group and g an element of G. The order of g divides the order of G
Define left and right coset
Let G be a group and H be a subgroup. Let g∈G - gH={gh: h∈H} is a left coset of H in G - Hg={hg: h∈H} is a right coset of H in G
Define index
Let G be a group and H be a subgroup. We define index [G:H] to be the number of left cosets of H in G
Fact about intersection and union of subgroups + proofs
Let G be a group and H₁, H₂ subgroups Then H₁∩H₂ is a subgroup (almost trivial proof) H₁∪H₂ is not always a subgroup e.g (4Z, +) and (3Z, +) are subgroups of (Z,+)
Explain cyclic groups and cyclic subgroups
Let G be a group and g be an element of G and define <g> = {g^n : n∈ℤ} = {..., g^-2, g^-1, 1, g, g^2, ...} This is known as the cyclic subgroup of G generated by g. If G=<g> then we call G a cyclic group and that g is a generator of G
Corollary from Lagrange's Theorem
Let G be a group of order n and g an element of G. Then g^n=1
Define order of a group
Let G be a group. The order of G is the number of elements that G has. Denote by |G| or #G
Define proper and trivial subgroups
Let G be a group. {1} is known as the trivial subgroup of G whereas all other subgroups are non-trivial Subgroups that aren't equal to G are proper subgroups
Criterion for a subgroup
Let G be a subgroup. A subset H of G is a subgroup iff it satisfies the following three properties - 1∈H - if a,b∈H then ab∈H - if a∈H then a^-1∈H (note these are the same properties required to be a group, minus associativity)
Prove relationship between cosets and congruence classes
Let G be an additive abelian group and H a subgroup - a-bar is set of elements in G that are congruent to a (mod H) - this means b-a∈H ⇒ b∈a+H - so a-bar = a+H i.e. the congruence class of a modulo H is a left coset of H QED
Define quotient group
Let G be an additive abelian group and H a subgroup We define the quotient group G/H to be the set of congruence classes G/H = {a-bar: a∈G} = {a+H : a∈G} with addition being defined by a-bar + b-bar = (a+b)-bar
Define congruence class of a modulo H
Let G be an additive abelian group and H a subgroup. Let a∈G. a-bar denotes the congruence class of a modulo H a-bar = {b∈G : b≡a (mod H)}
Define unit
Let R be a ring and u∈R Then u is called a unit if ∃v∈R st uv=vu=1
Criterion for subring
Let R be a ring. A subset S⊆R is a subring iff it satisfies - 0, 1∈S (additive and multiplicative identities) - if a, b∈S then a+b∈S (closed under addition) - if a∈S then -a∈S (additive inverses) - if a, b∈S then ab∈S (closed under multiplication)
Define unit group
Let R be a ring. We define the unit group of R to be R^={a∈R:a is a unit in R}
Define binary operation
Let S be a set. A binary operation on S is a rule which for every two elements from S gives another element of S
Prove that composition of functions is associative
Let S₁, S₂, S₃, S₄ be sets and let f, g, h be functions h: S₁→S₂, g:S₂→S₃, f:S₃→S₄ Let k=g∘h and l=f∘g - f∘(g∘h)(x) = f∘k(x) = f(k(x)) = f(g(h(x))) - (f∘g)∘h(x) = l∘h(x) = l(h(x)) = f(g(h(x))) - therefore composition of functions is associative QED
Euler-totient function definition
Let m≥1 be an integer. We denote the order of the group (ℤ/mℤ)^ by φ(m)
Euler's theorem
Let m≥2 be an integer. Let a be an integer st hcf(a, m)=1 then a^φ(m)≡1 (mod m)
Define nth alternating polynomial
Let n≥2 be an integer and x₁, x₂, ..., xₙ be variables and let Pₙ be the polynomial ∏(xᵢ-xⱼ) over 1≤i<j≤n
Fermat's little theorem
Let p be a prime and a an integer st p∤a. Then a^(p-1)≡1 (mod p)
How to represent subgroups in a diagram?
Like a tree, with the group at the top and subgroups branching out from below
How to represent an nxm matrix of real numbers
M_(nxm)(ℝ)
Show that the modular group is a group
Modular group is SL₂(ℤ) - this is clearly a subset of the special linear group SL₂(ℝ) - it contains the identity matrix - multiplying any two matrices in here gives a matrix with integer elements and determinant 1 so it also belongs to the group - inverses of all matrices have integer entries and determinant 1 - so it's a subgroup of SL₂(ℝ) and therefore is a group itself
Find unit group of M₂(R) where R is a commutative ring
M₂(R)^ = GL₂(R) = {A∈M₂(R) : det(A)∈R^}
Unit group of M₂(ℝ)
M₂(ℝ)^ = GL₂(ℝ) i.e all matrices with non-zero determinants
Can a permutation be written as a product of odd and even number of transpositions?
No
Is (ab)² always equal to a²b² for groups?
No because (ab)² = abab and a²b²=aabb and these aren't necessarily equal if the groups aren't abelian
Is (ℝ³, +, x) a ring?
No because cross product is not associative
Explain whether ℝ* is a subgroup of ℝ or not
No because ℝ is a group under addition but ℝ is a group under multiplication. For it to be a subgroup, require same binary operation
What is the zero ring?
Ring with just the element 0 in it. It is the additive and multiplicative identity
What are symmetries of a shape?
Rotations and reflections that keep the shape occupying the same position but the vertex ordering may change
Special linear group + its importance
SL₂(ℝ) = {A∈M₂(ℝ) : detA=1} this is a subgroup of M₂(ℝ)
Special orthogonal group
SO₂(ℝ) = {R_θ : θ∈ℝ} i.e. all rotation matrices - this is a subgroup of SL₂(ℝ)
What is Sym(A)
Set of bijections from A to itself
What does ℝ[x] represent?
Set of polynomials in x with real coefficients
Explain whether or nht roots of unity are cyclic subgroups
They are because there is always some element a whose cyclic subgroup <a> generates all the other nth roots of unity in the group
Let G be a group and H a finite subgroup. Prove that if g∈G then the #gH = #H
To show they have the same size it is sufficient to show there exists a bijection between the two sets - let φ:H→gH defined by φ(h)=gh Injectiveness: - suppose φ(h₁)=φ(h₂) then gh₁=gh₂ - premultiply by g⁻¹ to get h₁=h₂ so φ is injective Surjectiveness: - suppose k∈gH then WTS k=φ(h) for some h∈H - but by definition of coset gH={gh:h∈H} so k=gh=φ(h) for some h in H so φ is surjective QED
Why does V3 imply V2?
V2 says #H divides #G V3 tells us the ratio is actually [G:H]
Is ({1, -1}, x) a group?
Yes
Is ℤ a subset of ℝ?
Yes
Is ℤ/mℤ a ring?
Yes
Rules on adding and multiplying congruence classes
[a]+[b] = [a+b] [a][b] = [ab]
How would you denote cosets for an additive group?
g+H and H+g for left and right cosets respectively
Explain whether the identity element in Sₙ is odd or even
id(Pₙ)=+Pₙ so id must be even as the result is positive so we must be able to write it as a product of an even number of transpositions
Roots of unity are subgroups of what
nth roots of unity are Uₙ={1, ζ, ζ², ..., ζⁿ⁻¹} - this is clearly a subset of ℂ* - it is also a group and the criteria can be easily checked - therefore it is a subgroup
Unit group of ℤ
{-1, +1}
Examples of fields
ℚ, ℝ, ℂ
What set is cross product a binary operation on + why?
ℝ³ because it is the only set for which it is defined
Euclidean n-space definition
ℝⁿ = {(x₁, x₂, ..., xₙ) : x₁, x₂, ..., xₙ∈ℝ}
Examples of rings
ℤ, ℂ, ℝ, ℚ, ℝ[x]
Theorem about when is ℤ/mℤ a field? Proof
ℤ/pℤ is a field when p is prime i.e. (ℤ/pℤ)^={[1], [2], ..., [p-1]} Proof - we know ℤ/mℤ is a commutative ring for integer m≥2 - to show that ℤ/pℤ is a field, all non-zero elements [a]∈ℤ/pℤ need to have an inverse - a is non-zero so a=1, 2, ..., p-1 none of which are divisible by p - since p is prime, hcf(a, p)=1 for all these a so by previous theorem, [a] is invertible in ℤ/pℤ therefore ℤ/pℤ is a field
Value of [ℤ:{0}] + why?
∞ - cosets of a+{0} = {a} so there are infinely many cosets of {0} in ℤ
