%%visits: 3 Creating groups of higher order, from smaller groups. ## intuition direct_products are [[group]]s with a coordinate system instead of just one element. For example G = G1 × G2 where (e1, e2) is the identity
Theorem ℤp × ℤq ≡ ℤpq⇔ p and q are relatively prime.
proof. Let ℤp = < a> and ℤq find the generator and show it’s order
Remember the operation of g1*G1g1′
If G1 and G2 are abelian groups then G1 × G2 is abelian
Proof: $\left( g_1,g_2 \right) \left( g_1',g_2' \right) = \left( g_1g_1',g_2g_2' \right) = \left( g_1'g_1,g_2'g_2 \right)= \left( g_1',g_2' \right\left( g_1,g_2 \right) )$ hence proved.
|G1 × G2| = |G1||G2| as Let gi ∈ G1 and hi ∈ G2. This is implied from the cartesian product and elements in a group being distinct.
G1 ≡ G1 × {e2} ⊂ G1 × G2 Think of it as 1-d slice in a 2d plane! ## rigour Definition 7.1 in the lecture notes.
direct_produc := Let G1 and G2 be two groups. Then
G = G1 × G2 = {(g1, g2) : g1 ∈ G1, g2 ∈ G2}
with the multiplication law (g1, g2)(g1′, g2′) = (g1g1′, g2g2′).
G1 × G2
is called the direct_product of G1 and G2. The action is sort of
[[dot_product]]. ## exam clinic ## examples 1. Let G1 = {e1, a1}
with a12 = e1
and G2 = {e2, a2}
with a22 = e2
since the cardinality of the group is 4, it is isomorphic to the
kliens_four group or ℤ4.
Recall that ℤ4 = ⟨b⟩ with b4 = e then
ℤ4 = {e, b, b2, b3},
|e| = 1, |b| = 4, |b2| = 2, |b3| = 4
(since 3 and 4 are coprime), Returning to G1 × G2
all of its elements are of the form: (a1n, a2m).
(a1n, a2m)2 = (a12n, a22m) = (e1, e2).
All non-identity elements in G1 × G2 have order 2. G1 × G2 is not isomorphic to Z4 as the elements order should be preserved
So it must be isomorphic to G1 × G2 ≡ V4
ℤ2 × ℤ3 ≡ ℤ6 with elements (a, b) with a ∈ {0, 1}, b{0, 1, 2}. If we find an element that generates ℤ2 × ℤ3, this would be the first question you should ask, since it makes life a lot easier. It is clear that (1, 1) generates, since 2, and 3 are coprime.
ℤ2 × ℤ4 Forms a discretised torus.
ℝ2 :todo:D3 is not a direct_product, since ## resources tags :math: