“The race to build more friendly and better idiot-proof programs against those lazy and stupid bastards could never be won. It’s a lost de-evolution’s race to the bottom,” Dad said, raising his voice while checking the time. “Mind your tone, dad!” Mum protested energetically. “Whatever, AI, big data, and robots are going to eradicate all their fork jobs, maybe that’s why they are so lazy, dumb, and apathetic, they have nothing to lose anyway. Enjoy your bugs, have nothing and be happy!” Dad replied. Apocalypse, Anawim, #justtothepoint.
Informally, the direct product is a way to combine two groups into a new, larger group.
Let (G, ◦) and (H, ∗) be groups. Then, G × H can be made into a group, called the direct product of G and H, written or expressed as G⊕H using component–wise multiplication, that is, (g, h)(g’, h’)=(g◦g, h*h')
Proof that G⊕H is a group:
Let G_{1}, G_{2},…, G_{n} be groups. For (a_{1}, a_{2}, …, a_{n}) and (b_{1}, b_{2}, …, b_{n}) in $\prod_{i=1}^{n} G_i$, let’s define (a_{1}, a_{2}, …, a_{n})(b_{1}, b_{2}, …, b_{n}) to be the element (a_{1}b_{1}, a_{2}b_{2}, …, a_{n}b_{n}). Then, $\prod_{i=1}^{n} G_i$ is a group, the direct product of the groups G_{i}, written as G_{1}⊕G_{2}⊕…⊕G_{n}. In other words, the external direct product of the groups G_{i} is the group of all n-tuples for which the i-th component is an element of G_{i} and the group operation is component-wise.
G_{1}⊕G_{2}⊕…⊕G_{n} = {(a_{1}, a_{2}, …, a_{n})| a_{i} ∈ G_{i}}
More generally, the direct product G_{1}⊕G_{2}⊕… is the group of sequences (g_{1}, g_{1}, ···) where g_{i} ∈ G_{i} and the group operation is component-wise, (g_{1}, g_{2}, ···)·(h_{1}, h_{2}, ···) = (g_{1}·h_{1}, g_{2}·h_{2}, ···)
The external direct product is a group where:
The operation is component-wise, i.e., a_{i} and b_{i} ∈ G_{i} and G_{i} is a group, so there is closure ⇒ a_{i}b_{i} ∈ G_{i}. Therefore, $\prod_{i=1}^{n} G_i$ is well defined and closed under this operation.
Associative is inherited from each G_{i} ∀i, 1 ≤i ≤ n as it is thrown back onto the associative law in each component of the external direct product.
If e_{i} is the identity element in G_{i}, (e_{1}, e_{2}, …, e_{n}) is the identity in G_{1}⊕G_{2}⊕…⊕G_{n}.
Finally, the inverse of (a_{1}, a_{2}, …, a_{n}) is (a_{1}^{-1}, a_{2}^{-1}, …, a_{n}^{-1}).
ℤ_{5}⊕ℤ = {(m, n)| m ∈ ℤ_{5}, n ∈ ℤ}. (2, 1) + (3, 3) = (5, 4) = (0, 4), (2, 7) + (4, 21) = (6, 28) = (1, 28).
ℝ^{x}⊕ℤ_{3} = {(m, n)| m ∈ ℝ^{x}, n ∈ ℤ_{3}}. (2, 1) + (3, 1) = (5, 2), (2, 1) + (4, 2) = (6, 0), (4, 2)^{-1} =[the inverse of (a_{1}, a_{2}, …, a_{n}) is (a_{1}^{-1}, a_{2}^{-1}, …, a_{n}^{-1}) and 2 + 1=_{ℤ3} 0] (1/4, 1).
Let ℝ be the group of real numbers under addition. Then, the direct product ℝ ⊕ ℝ is the group of all two-component vectors (x, y) under vector addition. (x_{1}, y_{1}) + (x_{2}, y_{2}) = (x_{1} + x_{2}, y_{1} + y_{2}). In general, ℝ ⊕ ℝ ⊕ ··· ⊕ ℝ is the familiar Euclidean n-space ℝ^{n} with usual vector addition, (a_{1}, a_{2}, ···, a_{n}) + (b_{1}, b_{2}, ···, b_{n}) = (a_{1} + b_{1}, a_{2} + b_{2}, ···, a_{n} + b_{n}).
Let ℝ^{+} be the group of positive real numbers under multiplication. Then, the direct product ℝ^{+} ⊕ ℝ^{+} is the group of all two-component vectors (x, y) in the first quadrant under component-wise multiplication. (x_{1}, y_{1})·(x_{2}, y_{2}) = (x_{1}·x_{2}, y_{1}·y_{2})
ℤ_{2} ⊕ ℤ_{3} has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). It is cyclic, ℤ_{2} ⊕ ℤ_{3} = ⟨(1, 1)⟩ = {(1, 1), (0, 2), (1, 0), (0, 1), (1, 2), (0, 0)}. Since we know that there is, up to isomorphism, only one cyclic structure of any given orden, then ℤ_{2} ⊕ ℤ_{3} is isomorphic to ℤ_{6}, ℤ_{2} ⊕ ℤ_{3} ≋ ℤ_{6}.
(1, 1) is an element of order 6: 2(1, 1) = (1, 1) + (1, 1) = (0, 2), 3(1, 1) = (0, 2) + (1, 1) = (1, 3) = (1, 0), 4(1, 1) = (2, 1) = (0, 1), 5(1, 1) = (1, 2), 6(1, 1) = (0, 0) ⇒ |⟨(1, 1⟩| = 6 and |ℤ_{2} ⊕ ℤ_{3}| = 6 ⇒ ℤ_{2} ⊕ ℤ_{3} = ⟨(1, 1)⟩.
Groups forming the direct products may be completely different, e.g., G_{1} = ℤ, G_{2} = S_{3}, and G_{3} = GL_{2}(ℝ) where the group operations are addition, composition, and matrix multiplication respectively. Then, the group operation in G_{1} ⊕ G_{2} ⊕ G_{3} = ℤ ⊕ S_{3} ⊕ GL_{2}(ℝ) is defined by, (n, σ, $(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})$)(m, τ, $(\begin{smallmatrix}p & q\\ r & s\end{smallmatrix})$) = (n+m, σ∘τ, $(\begin{smallmatrix}ap+br & aq+bs\\ cp+dr & cq+ds\end{smallmatrix})$).
ℤ_{2} ⊕ ℤ_{2} = {(0, 0), (1, 0), (0, 1), (1, 1)}. |ℤ_{2} ⊕ ℤ_{2}| = 4 = 2p, p=2, 2 prime ⇒ [Classification of Groups of Order 2p] G is either isomorphic to ℤ_{4} (cyclic and Abelian) or the dihedral group D_{2}. Since ℤ_{2} ⊕ ℤ_{2} is Abelian, ℤ_{2} ⊕ ℤ_{2} ≋ D_{2} ≋ K_{4}, the Klein-four-group. Their Hasse diagrams is shown in Figure 1.e.
Classification of Groups of Order 4. A group of order 4 is isomorphic to ℤ_{4} or ℤ_{2} ⊕ ℤ_{2}
Proof. Let G = {e, a, b, ab}. If G is cyclic, then G ≋ ℤ_{4} and we know that there is, up to isomorphism, only one cyclic structure of any given orden.
Otherwise, it follows from Lagrange’s Theorem (The order of an element of a finite group divides the order of the group) that |a| = |b| = |ab| = 2 because only 1 (e is the only element of order 1), 2, and 4 (G is not cyclic ⊥; a ≠ e, a^{4} = e ⇒ G = ⟨a⟩) divide |G| = 4.
Let be the mapping Φ: G → ℤ_{2} ⊕ ℤ_{2}, defined as Φ(e) = (0, 0), Φ(a) = (1, 0), Φ(b) = (0, 1), Φ(ab) = (1, 1). Since we know that |a| = |b| = |ab| = 2, Φ is an isomorphism from G onto ℤ_{2} ⊕ ℤ_{2}, e.g. homomorphism, Φ(a·a) = Φ(a^{2}) =[|a| = 1] Φ(e) = (0, 0), Φ(a)·Φ(a) = (1, 0)(1, 0) = (0, 0). Besides, |G| = ℤ_{2} ⊕ ℤ_{2}, hence Φ is isomorphism.
The order of (1, 0) is 3, since (1, 0) + (1, 0) + (1, 0) = (0, 0), 3 ≡ 0 mod 3. The order of (0, 2) is 2, since (0, 2) + (0, 2) = (0, 0), and 4 ≡ 0 mod 4. The order of (1, 3) is 12 because 2(1, 3) = (2, 2), 3(1, 3) = (0, 1), and 4(1, 3) = (1, 0), 5(1, 3) = (2, 3) and 6(1, 3) = (0, 2) ≠ (0, 0). However, the order of (1, 3) must divide 12, so the only choice left is 12. Thus, ℤ3 ⊕ ℤ4 = ⟨(1, 3)⟩ and ℤ3 ⊕ ℤ4 ≋ ℤ12.
Theorem. Any cyclic group of even order has exactly one element of order 2.
Proof
Let G be cyclic, G = ⟨a⟩ = {e (= a^{0}), a, a^{2},…, a^{2n-1}}, where a^{j} ≠ a^{k}, ∀j, k, j ≠ k, j, k < 2n. a^{2n} = e. m = 2n (G is a cyclic group of even order).
First, let’s show that g = a^{n} has order 2. g^{2} = (a^{n})^{2} = a^{2n} = a^{m} = e.
Second, let’s show that it is unique. Let g ≠ a^{n}, g = a^{j}, j ≠ n, g^{2} = e ⇒ (a^{j})^{2} = e ⇒ a^{2j} = e ⇒[By assumption, |a| = m] 2j ≥ m = 2n ⇒ j ≥ n.
Suppose that j > n ⇒ ∃k: 1 ≤ k < n, j = n + k ⇒ e = a^{2j} = a^{2(n+k)} = a^{2n}a^{2k} = ea^{2k} = a^{2k} = e. ⇒ ∃k: 1 ≤ k < n (2k < 2n=m), a^{2k} = e ⊥ (|a| = m)
U_{8} ⊕ U_{10} = {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3), (3, 7), (3, 9), (5, 1), (5, 3), (5, 7), (5, 9), (7, 1),(7, 3), (7, 7), (7, 9)}, e.g., (5, 7)(7, 9) = (5·7 (mod 8), 7·9 (mod 10)) = (3, 3)
ℤ ⊕ ℤ_{2} is generated by (1, 0) and (0, 1), i.e., ℤ ⊕ ℤ_{2} = ⟨(1, 0), (0, 1)⟩
Lemma. Let G_{1}, G_{2},…, G_{n} be finite groups. |G_{1}⊕G_{2}⊕...⊕G_{n}| = |G_{1}|·|G_{2}|···|G_{n}|. This follows from the formula for the cardinality of the Cartesian product of sets.
Lemma. The direct product of Abelian groups is also Abelian.
Proof. (g_{1}, g_{2},…, g_{n})(g’_{1}, g’_{2},…, g’_{n}) = (g_{1}g’_{1}, g_{2}g’_{2},…, g_{n}g’_{n}) = [since each G_{i} is Abelian] = (g’_{1}g_{1}, g’_{2}g_{2},…, g’_{n}g_{n}) = (g’_{1}, g’_{2},…, g’_{n})(g_{1}, g_{2},…, g_{n})∎
Remark. If one group is not Abelian, then the direct product is not also Abelian. Suppose G_{i} is not Abelian ⇒ ∃g_{i}, g_{i}’ ∈ G_{i}: g_{i}g_{i}’ ≠ g_{i}‘g_{i} ⇒ (e_{1}, e_{2},…,g_{i},… e_{n})(e_{1}’, e_{2}’,…,g_{i}’,… e_{n}’) ≠ (e_{1}’, e_{2}’,…,g_{i}’,… e_{n}’)(e_{1}, e_{2},…,g_{i},… e_{n})
Example. ℤ_{2} ⊕ D_{3} where D_{3} is the group of symmetries of an equilateral triangle. D_{3} = {id (e), rotations (ρ_{1} -120° around the center anticlockwise-, ρ_{2} -240°-), reflections or mirror images on all its vertices (μ_{1}, μ_{2}, and μ_{3})}
Observe that the operation in ℤ_{2} is addition mod 2, while the operation in D_{3} is written using multiplicative notation. Thus, when you multiply tow pairs, you add in ℤ_{2} in the first component and multiply in D_{3} in the second component, e.g., (1, ρ_{2})(1, μ_{2}) = (1+1, ρ_{2}·μ_{2}) = (0, μ_{3}). The identity is (0, e). Futhermore, since D_{3} is not Abelian, ℤ_{2} ⊕ D_{3} is not Abelian either, e.g., (1, μ_{2})(0, ρ_{2}) = (1, μ_{1}), but (0, ρ_{2})(1, μ_{2}) = (1, μ_{3}).
Theorem. Order of an Element in a Direct Product. Let G_{1}, G_{2},…, G_{n} be groups. The order of an element in a direct product of a finite number of finite groups is the least common multiple of the orders of the components of the element. Formally, |(g_{1}, g_{2},..., g_{n})| = lcm(|g_{1}|, |g_{2}|,...,|g_{n})|.
Proof.
Let s = lcm(|g_{1}|, |g_{2}|,…,|g_{n}|) ⇒ s is a multiple of each |g_{i}|, ∃a_{1}, a_{2},…, a_{n}: s = a_{1}|g_{1}| = a_{2}|g_{2}| = … = a_{n}|g_{n}|.
(g_{1}, g_{2},…, g_{n})^{s} = (g_{1}^{s}, g_{2}^{s},…, g_{n}^{s}) ⇒ [s = a_{1}|g_{1}| = a_{2}|g_{2}| = … = a_{n}|g_{n}|.] (g_{1}^{s}, g_{2}^{s},…, g_{n}^{s}) = ( (g_{1}^{|g1|})^{a1}, (g_{2}^{|g2|})^{a2},…, (g_{n}^{|gn|})^{an}) = (e_{1}, e_{2}, …, e_{n}) ⇒ s ≤ |(g_{1}, g_{2},…, g_{n})| = t.
On the other hand, (g_{1}, g_{2},…, g_{n})^{t} = (e_{1}, e_{2},…, e_{n}) ⇒ (g_{1}^{t}, g_{2}^{t},…, g_{t}^{t}) = (e_{1}, e_{2},…, e_{n}) ⇒ t is a multiple of each |g_{i}| ⇒ t ≥ s = lcm(|g_{1}|, |g_{2}|,…,|g_{n})|. Therefore, |(g_{1}, g_{2},…, g_{n})| = lcm(|g_{1}|, |g_{2}|,…,|g_{n})|∎
Cyclic groups. Let a be an arbitrary element of order n in a group G, and let k be a positive integer. Then, ⟨a^{k}⟩=⟨a^{gcd(n,k)}⟩ and |⟨a^{k}⟩| = $\frac{n}{gcd(n,k)}.$
ℤ_{12} = ⟨1⟩. d = gcd(12, 8) = 4, b = a^{k} = [Addition notation] k·a = [In ℤ_{12}, a = 1] k·1 = k, |⟨8⟩| = 12/4 = 3. Similarly, |⟨4⟩| = 60/gcd(60, 4) = 60/4 = 15 and |⟨10⟩| = 24/gcd(24, 10)= 24/2 = 12. lcm (3, 15, 12) = 60, therefore (8, 4, 10) is of order 60 in ℤ_{12} ⊕ ℤ_{60} ⊕ ℤ_{24}.
The desired elements are {(1, 0),(1, 3),(1, 6),(5, 0),(5, 3),(5, 6),(3, 3),(3, 6)}.
Elements of ℤ_{25} of order 25 = {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24}
Elements of ℤ_{5} of order 5 = {1, 2, 3, 4}, and of order 1, obviously only the identity, 0.
#Elements of order 25 in ℤ_{25} ⊕ ℤ_{5} = 20·1 + 20·4 = 100.
Elements of ℤ_{25} of order 5 = {5, 10, 15, 20}, and of order 1, the identity (0).
5 is clearly an element of order 5 (5·5 = 25 (mod 25)= 0). |a|= |a^{j}| iff gcd(n (=5), j) = 1. Therefore gcd(5, 1) = gcd(5, 2) = gcd(5, 3) = gcd(5, 4) = 1 ⇒[Be careful, we are using additive notation in ℤ_{5}, e.g., 5^{2} = 5·2 = 10] |5| = |10| = |15| = |20| = 5.
Elements of ℤ_{5} of order 5 = {1, 2, 3, 4}, and of order 1, the identity (0).
The number of elements of order 5 in ℤ_{25} ⊕ ℤ_{5} = 4·1 (a) + 4·4 (b) + 1·4 (c) = 24.
Elements of ℤ_{100} can have orders of 1, 2, 4, 5, 10, 20, 25, 50 and 100. Elements of ℤ_{25} can have orders of 1, 5, 25.
10 = lcm(|a|, |b|) ⇒ a) 2, 5; b) 5… By Lagrange’s theorem, it is not possible, there are no elements of ℤ_{25} of order 2. b) 10, 1; c) 10, 5.
Elements of ℤ_{100} of order 2 = {50}, and of order 10 = {10, 30, 70, 90}
10 is clearly an element of order 10 (10·10 = 100 (mod 100)= 0). |a|= |a^{j}| iff gcd(n (=10), j) = 1. Therefore gcd(10, 1) = gcd(10, 3) = gcd(10, 7) = gcd(10, 9) = 1 ⇒[Be careful, we are using additive notation in ℤ_{100}, e.g., 10^{3} = 10·3 = 30] |10| = |30| = |70| = |90| = 10.
Elements of ℤ_{25} of order 1 = {0}, and of order 5 = {5, 10, 15, 20} -it was calculated previously-.
The number of elements of order 10 in ℤ_{100} ⊕ ℤ_{25} = 1·4 (a) + 4·1 (b) + 4·4 (c) = 24.
Every cyclic subgroup of order 10 have four generators (Let a be an element of order n, ⟨a^{k}⟩=⟨a^{gcd(n,k)}⟩ and |⟨a^{k}⟩| = $\frac{n}{gcd(n,k)}.$ 10 = |a|= |a^{k}| ↭ 10 = 10/gcd(10, k) ↭ gcd(10, k) = 1, that is, k = 1, 2, 3, 4). Besides, no two of the cyclic subgroups can have an element of order 10 in common, because this element would have to generate both subgroups. Therefore, each of this subgroup have four of its generators. So we have 24 elements of order 10 / 4 generators for each one of these cyclic subgroups = 6 cyclic subgroups of order 10.
|ℤ_{6} ⊕ U(14)| = 6 x 6 = 36. U(14) = {1, 3, 5, 9, 11, 13}. The order of (3, 3) is lcm(|3| in ℤ_{6}, |3| in U(14)) = lcm (2, 6) = 6.
3^{1}=3, 3^{2} = 9, 3^{3} = 27 % 14 = 13, 3^{4} = 39 % 14 = 11, 3^{5} = 5, 3^{6} = 1.
|S_{4} ⊕ U(9)|= 24 x 6 = 144. |(24)(31), 4| = lcm (|(24)(31)|, |4|) = lcm (2, 3) = 6.
|(24)(31)| = 2 because (24)(31)·(24)(31) =[Disjoint cycles commute] (24)((31)·(31))(24) = (24)(24) = (). U(9) = {1, 2, 4, 5, 7, 8}, 4^{1} = 4, 4^{2} = 16 % 9 = 7, 4^{3} = 28 % 9 = 1.
The matrix $(\begin{smallmatrix}cos(θ) & -sin(θ)\\ sin(θ) & cos(θ)\end{smallmatrix})$ rotates points in the xy plane counterclockwise through an angle θ about the origin of a two-dimensional Cartesian coordinate system. To perform the rotation on a plane with standard coordinates (x, y): $(\begin{smallmatrix}cos(θ) & -sin(θ)\\ sin(θ) & cos(θ)\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}xcosθ -y sinθ\\ xsinθ + ycosθ\end{smallmatrix})$