:introduction_to_abstract_algebra: is about finding a more rigorous way to understand maths. The word abstract means to condense something to a small modular level, so that you have the freedom to pick and chose what you want to do with the abstract module. One such abstract module, is the group. %%visits: 2
In the book, How to study for a maths degree, the book describes mathematical objects, which if I recall correctly are informally defined to think of something as it’s own entity that you can apply to many things, and not just in math.
Groups have a lot of geometric uses, as well as a lot of uses in algebra. The reason for its appeal is that it can reduce complicated mathematical ideas into a lot more manageable form. There are a lot of uses in [[platonic_solid]]s, and t
There is a matrix form of symmetry groups.
“Every group encodes {{c1::symmetry}} in an abstract sense” paul cook # intuition In [[set_theory]], if you only had elements and not anything you could do with the elements, then that would be pretty pointless. For example, the notion of adding and taking away is not really understood well or mathematically rigorous without the idea of groups. It is basically a set that can interact with each other, and has a set of rules to make that interaction plausible. For example, there should be a “trivial” element, that is to say, an element that when applied to other elements does nothing, and “does nothing” in mathematical terms is the binary_operation [[function|maps]] the two elements to the non-trivial one.
$dim(G\setminu H) = dim(G) - dim(H)$
Grou := Set G,
with a binary operation. G × G → G
satisfying 1. Identity ∃e ∈ G : ∀ a ∈ G, ea = ae = a
2. Inverses ∀a ∈ G ∃b ∈ G : ab = e = ba
3. Associativity ∀a, b, c ∈ G, (ab)c = a(bc)
4. Closur := S × S → S It must
be Associative, have inverses, have the identity, and have closure
Grou := A group is a set G and a mapping from the Cartesian
product G × G into
G (a law of composition or a
multiplication law) which we denote by juxtaposition (which means we put
it together to show multiplication
Subgrou := H is a subgroups of G iff H is a group with
operation *G and H
is a subset of G
All groups of even order contain at least one non-identity element which squares to the identity. e.g. ℤ2 = {e, a}
Informally speaking, generators of a group are like the “DNA” of a group, that is to say, they can “replicate” and make all the other elements, they contain the instructions on how to build the whole group.
⟨⟩ stand for span. In a similar idea to a vector. Cyclic groups are abelian. Since n, m ∈ ℤ gngm = gn + m = gm + n = gmgm
Consider this conjecture for Zp. If p is prime then ℤp = ⟨n⟩ for any n ∈ {1, 2, 3…p − 1} Proof: Let n ∈ {1, 2, 3…, p − 1}. Now, suppose for contradiction sake, that n does not generate ℤp, then n must have an order less than p. This is tautological to [nk] = [0], k < p. This implies kn is a multiple of p. But if k has to be less than p, then it must be 1 as those are the only two divisors of a prime. Then n must be be by the same logic. But this is a contradiction as we have explicitly defined n ∈ {1, 2, 3, …, p − 1} So our initial assumption, that n does not generate ℤP is false.
Any subgroup of a cyclic group is also cyclic. lagranges theorem
Roots of unity in the complex plane. - $G = \left\{ e^{i \left( \frac{2\pi}{n}k \right) } | k \in \left\{ 0,1,2, \ldots n-1 \right\} \right\}$, the roots of unity is a cyclic group, that is isomorphic to the ℤp
Let G be a cyclic group generated by g0. then - g0n, n = 0, 1, 2, …, |G| − 1 are distinct -Proof by contradiction. Suppose that they is a repeat. - Suppose ∃n1, n2 with 0 ≤ n2 < n1 ≤ |G|−1 and that g0n1 = g0n2 - ⟹ g0n1 − n2 = e - let q = n1 − n2 and note that q > 0 and q > |G| - consider any n ∈ ℤ then n = kq + r - g0n = g0kq + r = g0kqg0r = g0r = g0r - Hence if g0n = g0r then ⟨g0⟩ does not generate G :todo: see the lecture video to finish - g|G| = e∀g ∈ G - by part a g|G| (it has to be distinct) - g|G|−m = e - Now note that if 0 < m < |G| then the statement contradicts part a of the proof. which leaves only m = 0
⟨e, a, b|ea = a, eb = be = b, e2 = e = a2, b2 = e, aba = bab⟩ is another generator notion of some group isomorphic to D3 ## examples of Cyclic groups - ℤp (addition modulo p). ℤ1 = {0} = ⟨0⟩ are generated by ⟨1⟩, intuitively - ℤ2 = {0, 1} = ⟨1⟩ - 20 is like saying add 2 to itself, 0 times. - ℤ3 = {0, 1, 2} = ⟨1⟩ = ⟨2⟩ - ℤ4 = {0, 1, 2, 3} = ⟨1⟩ = ⟨3⟩ - Any factor of p will need to be considered differently.
To prove that x is a group. The first thing to check is closure. Finding identity helps to find inverses. With associativity, it helps to make a(bc)c−1 = (ab) since that implies a(bc) = (ab)c.
you can assume that [[matrix]] multiplication is Associative 1. [[matrix]] Matrices with determinant equal to 1 under matrix multiplication, for example 2x2 matrices
Closure let A, B be two 2 × 2 matrices of unit determinant, i.e. A, B ∈ G. Then AB ∈ G. Then AB is a 2 × 2 matrix and det (AB) = det (A)det (B) = 1 Therefore AB ∈ G Associativity we must check if (AB)C = A(BC) where A, B, C ∈ G. In components (AB)C as
$(AB)C = \sum _{k=1}^{2} \sum _{j=1}^{2} \left( A_{ik}B_{jk} \right) C_{kl}$
(Now note that the Aik term has an no k position, so it is independent to the summation cycling through k, so we can factor it out as a sort of constant, which is really helpful for proving that this is equal to A(BC))
$= \sum _{j=1}^{2} A_{ij} \left( \sum _{k=1}^{2} B_{jk}C_{kl} \right)$
Identity $\mathbb{1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \in G$ as det (𝟙) = 1
inverses All matrices of form $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in G$ have inverses $\begin{pmatrix} d & -b & -c & a \end{pmatrix}$ and since ad − bc = 1 the determinant is also 1. Inverses are in G.
If p is prime, then ℤp = ⟨n⟩ for any n ∈ {1, 2, 3, …, p − 1} (Using little fermat’s theorem to show how many elements there are in the group, as well as the fact that n cannot share common divisors, means that nn = e and ⟨n⟩ has p elements ]
Proof that e is unique Assume e1, e2 are distinct identity, Then e1e2 = e1 by 1. e1e2 = e2
Proof that the inverse is unique. For contradiction sake suppose. So, ∃a, b, b′ : ab′ = ab = ba = b′a = e, b ≠ a So, bab′ = bab So, b′ = b which is our contradiction
:todo: Are permutation groups, primative groups? that is to say most
groups are usually perm groups, and all properties of groups are perm
groups I can see what S3 looks like, but what
does the superset Perm(X)
look like? Symmetric grou := The set of all permutations of
n elements is called the symmetric group. Perm(X) ⊂ Sym(X)
|SN| = n!
S3 can be thought
of as some sort of action on the three points of a triangle, but this
may not extend to S4
:todo: $\frac{n!}{\Pi j^{n_{j}}n_{j}!}$ is the number of n = ∑jnj $\left| A_{n} \right| = \frac{n!}{2}$
Proving isomorphism: - Map generator to generator - compare elements - compare generators
When dealing with subsets to prove it is a subgroup, we do not need to check associativity as it is inherited
see also: [[permutation_group]] [[kliens_four-group]]
tags :math:introduction_to_abstract_algebra:groups_and_symmetries: