# Direct Products

“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.

# Definitions

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:

1. Associativity (g, h)[(g’, h’)(g’’, h’’)] = (g, h)(g’◦ g’’, h’∗ h’’) = (g ◦ (g’◦ g’’), h ∗ (h’∗ h’’)) =[Associativity G and H] ((g ◦ g’) ◦ g’’, (h ∗ h’) ∗ h’’) = (g ◦ g’, h ∗ h’)(g’’, h’’) = [(g, h)(g’, h’)](g’’, h’’).
2. Identity, (e, e’)(g, h) = (e ◦ g, e * h) = (g, h) = (g ◦ e, h * e) = (g, h)(e, e’) where e and e’ are the identity elements of G and H respectively, hence (e, e’) is the identity element of G⊕H.
3. Inverses, (g-1, h-1)(g, h) = (g-1 ◦ g, h-1 * h) = (e, e’) = (g ◦ g-1, h * h-1) = (g, h)(g-1, h-1), therefore (g, h)-1 = (g-1, h-1)

Let G1, G2,…, Gn be groups. For (a1, a2, …, an) and (b1, b2, …, bn) in $\prod_{i=1}^{n} G_i$, let’s define (a1, a2, …, an)(b1, b2, …, bn) to be the element (a1b1, a2b2, …, anbn). Then, $\prod_{i=1}^{n} G_i$ is a group, the direct product of the groups Gi, written as G1⊕G2⊕…⊕Gn. In other words, the external direct product of the groups Gi is the group of all n-tuples for which the i-th component is an element of Gi and the group operation is component-wise.

G1⊕G2⊕…⊕Gn = {(a1, a2, …, an)| ai ∈ Gi}

More generally, the direct product G1⊕G2⊕… is the group of sequences (g1, g1, ···) where gi ∈ Gi and the group operation is component-wise, (g1, g2, ···)·(h1, h2, ···) = (g1·h1, g2·h2, ···)

The external direct product is a group where:

• The operation is component-wise, i.e., ai and bi ∈ Gi and Gi is a group, so there is closure ⇒ aibi ∈ Gi. Therefore, $\prod_{i=1}^{n} G_i$ is well defined and closed under this operation.

• Associative is inherited from each Gi ∀i, 1 ≤i ≤ n as it is thrown back onto the associative law in each component of the external direct product.

• If ei is the identity element in Gi, (e1, e2, …, en) is the identity in G1⊕G2⊕…⊕Gn.

• Finally, the inverse of (a1, a2, …, an) is (a1-1, a2-1, …, an-1).

# Examples

• 5⊕ℤ = {(m, n)| m ∈ ℤ5, n ∈ ℤ}. (2, 1) + (3, 3) = (5, 4) = (0, 4), (2, 7) + (4, 21) = (6, 28) = (1, 28).

• x⊕ℤ3 = {(m, n)| m ∈ ℝx, n ∈ ℤ3}. (2, 1) + (3, 1) = (5, 2), (2, 1) + (4, 2) = (6, 0), (4, 2)-1 =[the inverse of (a1, a2, …, an) is (a1-1, a2-1, …, an-1) and 2 + 1=3 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. (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2). In general, ℝ ⊕ ℝ ⊕ ··· ⊕ ℝ is the familiar Euclidean n-space ℝn with usual vector addition, (a1, a2, ···, an) + (b1, b2, ···, bn) = (a1 + b1, a2 + b2, ···, an + bn).

• 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. (x1, y1)·(x2, y2) = (x1·x2, y1·y2)

• 2 ⊕ ℤ3 has 6 elements, namely (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), and (1, 2). It is cyclic, ℤ2 ⊕ ℤ3 = ⟨(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 ℤ2 ⊕ ℤ3 is isomorphic to ℤ6,2 ⊕ ℤ3 ≋ ℤ6.

(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 |ℤ2 ⊕ ℤ3| = 6 ⇒ ℤ2 ⊕ ℤ3 = ⟨(1, 1)⟩.

• Groups forming the direct products may be completely different, e.g., G1 = ℤ, G2 = S3, and G3 = GL2(ℝ) where the group operations are addition, composition, and matrix multiplication respectively. Then, the group operation in G1 ⊕ G2 ⊕ G3 = ℤ ⊕ S3 ⊕ GL2(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, $(\begin{smallmatrix}ap+br & aq+bs\\ cp+dr & cq+ds\end{smallmatrix})$).

• 2 ⊕ ℤ2 = {(0, 0), (1, 0), (0, 1), (1, 1)}. |ℤ2 ⊕ ℤ2| = 4 = 2p, p=2, 2 prime ⇒ [Classification of Groups of Order 2p] G is either isomorphic to ℤ4 (cyclic and Abelian) or the dihedral group D2. Since ℤ2 ⊕ ℤ2 is Abelian, ℤ2 ⊕ ℤ2 ≋ D2 ≋ K4, the Klein-four-group. Their Hasse diagrams is shown in Figure 1.e.

# Classification of Groups of Order 4.

Classification of Groups of Order 4. A group of order 4 is isomorphic to ℤ4 or ℤ2 ⊕ ℤ2

Proof. Let G = {e, a, b, ab}. If G is cyclic, then G ≋ ℤ4 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, a4 = e ⇒ G = ⟨a⟩) divide |G| = 4.

Let be the mapping Φ: G → ℤ2 ⊕ ℤ2, 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 ℤ2 ⊕ ℤ2, e.g. homomorphism, Φ(a·a) = Φ(a2) =[|a| = 1] Φ(e) = (0, 0), Φ(a)·Φ(a) = (1, 0)(1, 0) = (0, 0). Besides, |G| = ℤ2 ⊕ ℤ2, hence Φ is isomorphism.

• ℤ3 ⊕ ℤ4 = {(0, 0),(1, 0),(2, 0),(0, 1),(1, 1),(2, 1), (0, 2),(1, 2),(2, 2),(0, 3),(1, 3),(2, 3)}. |Z3 ⊕ Z4| = 12.

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.

• U(4) ⊕ U(10) = {(1, 1),(1, 3),(1, 7),(1, 9),(3, 1),(3, 3),(3, 7),(3, 9)}, e.g., (3, 7)(3, 9) = (3·3 mod 4, 7·9 mod 10) = (1, 3). |U(4) ⊕ U(10)| = 8 even, but is not cyclic, even though U(4) and U(10) are both cyclic because (1, 9) and (3, 9) are two elements of order 2 in U(4) ⊕ U(10) -Read the following theorem-.

Theorem. Any cyclic group of even order has exactly one element of order 2.

Proof

Let G be cyclic, G = ⟨a⟩ = {e (= a0), a, a2,…, a2n-1}, where aj ≠ ak, ∀j, k, j ≠ k, j, k < 2n. a2n = e. m = 2n (G is a cyclic group of even order).

First, let’s show that g = an has order 2. g2 = (an)2 = a2n = am = e.

Second, let’s show that it is unique. Let g ≠ an, g = aj, j ≠ n, g2 = e ⇒ (aj)2 = e ⇒ a2j = e ⇒[By assumption, |a| = m] 2j ≥ m = 2n ⇒ j ≥ n.

Suppose that j > n ⇒ ∃k: 1 ≤ k < n, j = n + k ⇒ e = a2j = a2(n+k) = a2na2k = ea2k = a2k = e. ⇒ ∃k: 1 ≤ k < n (2k < 2n=m), a2k = e ⊥ (|a| = m)

• U8 ⊕ U10 = {(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 ⊕ ℤ2 is generated by (1, 0) and (0, 1), i.e., ℤ ⊕ ℤ2 = ⟨(1, 0), (0, 1)⟩

# Properties of External Direct Products

Lemma. Let G1, G2,…, Gn be finite groups. |G1⊕G2⊕...⊕Gn| = |G1|·|G2|···|Gn|. 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. (g1, g2,…, gn)(g’1, g’2,…, g’n) = (g1g’1, g2g’2,…, gng’n) = [since each Gi is Abelian] = (g’1g1, g’2g2,…, g’ngn) = (g’1, g’2,…, g’n)(g1, g2,…, gn)∎

Remark. If one group is not Abelian, then the direct product is not also Abelian. Suppose Gi is not Abelian ⇒ ∃gi, gi’ ∈ Gi: gigi’ ≠ gi‘gi ⇒ (e1, e2,…,gi,… en)(e1’, e2’,…,gi’,… en’) ≠ (e1’, e2’,…,gi’,… en’)(e1, e2,…,gi,… en)

Example. ℤ2 ⊕ D3 where D3 is the group of symmetries of an equilateral triangle. D3 = {id (e), rotations (ρ1 -120° around the center anticlockwise-, ρ2 -240°-), reflections or mirror images on all its vertices (μ1, μ2, and μ3)}

Observe that the operation in ℤ2 is addition mod 2, while the operation in D3 is written using multiplicative notation. Thus, when you multiply tow pairs, you add in ℤ2 in the first component and multiply in D3 in the second component, e.g., (1, ρ2)(1, μ2) = (1+1, ρ2·μ2) = (0, μ3). The identity is (0, e). Futhermore, since D3 is not Abelian, ℤ2 ⊕ D3 is not Abelian either, e.g., (1, μ2)(0, ρ2) = (1, μ1), but (0, ρ2)(1, μ2) = (1, μ3).

Theorem. Order of an Element in a Direct Product. Let G1, G2,…, Gn 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, |(g1, g2,..., gn)| = lcm(|g1|, |g2|,...,|gn)|.

Proof.

Let s = lcm(|g1|, |g2|,…,|gn|) ⇒ s is a multiple of each |gi|, ∃a1, a2,…, an: s = a1|g1| = a2|g2| = … = an|gn|.

(g1, g2,…, gn)s = (g1s, g2s,…, gns) ⇒ [s = a1|g1| = a2|g2| = … = an|gn|.] (g1s, g2s,…, gns) = ( (g1|g1|)a1, (g2|g2|)a2,…, (gn|gn|)an) = (e1, e2, …, en) ⇒ s ≤ |(g1, g2,…, gn)| = t.

On the other hand, (g1, g2,…, gn)t = (e1, e2,…, en) ⇒ (g1t, g2t,…, gtt) = (e1, e2,…, en) ⇒ t is a multiple of each |gi| ⇒ t ≥ s = lcm(|g1|, |g2|,…,|gn)|. Therefore, |(g1, g2,…, gn)| = lcm(|g1|, |g2|,…,|gn)|∎

# Examples

• Find the order of (8, 4, 10) in the group ℤ12 ⊕ ℤ60 ⊕ ℤ24.

Cyclic groups. Let a be an arbitrary element of order n in a group G, and let k be a positive integer. Then, ⟨ak⟩=⟨agcd(n,k)⟩ and |⟨ak⟩| = $\frac{n}{gcd(n,k)}.$

12 = ⟨1⟩. d = gcd(12, 8) = 4, b = ak = [Addition notation] k·a = [In ℤ12, 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 ℤ12 ⊕ ℤ60 ⊕ ℤ24.

• How many elements of order 6 are there in ℤ6 ⊕ ℤ9? lcm(|a|, |b|) = 6 ⇒ a) |a| = 6 (1, 5), |b| = 1 (0) or 3 (b=3 or 6 -|b|=3- but by Lagrange’s theorem, |b| cannot be 6). b) |a| = 2 (3), |b| = 3 (3, 6).

The desired elements are {(1, 0),(1, 3),(1, 6),(5, 0),(5, 3),(5, 6),(3, 3),(3, 6)}.

• How many elements of order 25 are there in ℤ25 ⊕ ℤ5? 25 = lcm(|a|, |b|) ⇒ [lcm(25, 5) = lcm(25, 1) = 25, but lcm(5, 5) = 5] There are two possibilities: |a| = 25, |b| = 1 or 5.

Elements of ℤ25 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 ℤ5 of order 5 = {1, 2, 3, 4}, and of order 1, obviously only the identity, 0.

#Elements of order 25 in ℤ25 ⊕ ℤ5 = 20·1 + 20·4 = 100.

• How many elements of order 5 in ℤ25 ⊕ ℤ5? 5 = lcm(|a|, |b|) ⇒ There are three possibilities: a) 5, 1; b) 5, 5; c) 1, 5.

Elements of ℤ25 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|= |aj| 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 ℤ5, e.g., 52 = 5·2 = 10] |5| = |10| = |15| = |20| = 5.

Elements of ℤ5 of order 5 = {1, 2, 3, 4}, and of order 1, the identity (0).

The number of elements of order 5 in ℤ25 ⊕ ℤ5 = 4·1 (a) + 4·4 (b) + 1·4 (c) = 24.

• Let G = S3 ⊕ ℤ5. What are all possible orders of elements in G? Prove that G is not cyclic.
1. Elements of ℤ5 can have orders of 1 and 5. Elements of S3 (|S3| = 6) can have orders of 1, 2, and 3 (S3 is not cyclic).
2. Elements of S3 ⊕ ℤ5 can have orders of 1, 2, 3, 5, 10, and 15.
3. |G| = |S3||ℤ5| = 6·5 = 30.
4. The order of G is 30, but there is no element of order 30 in G (S3 is not cyclic either), so G is not cyclic.
• Calculate the number of cyclic subgroups of order 10 in ℤ100 ⊕ ℤ25. First, we are going to calculate how many elements of order 10.

Elements of ℤ100 can have orders of 1, 2, 4, 5, 10, 20, 25, 50 and 100. Elements of ℤ25 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 ℤ25 of order 2. b) 10, 1; c) 10, 5.

Elements of ℤ100 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|= |aj| 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 ℤ100, e.g., 103 = 10·3 = 30] |10| = |30| = |70| = |90| = 10.

Elements of ℤ25 of order 1 = {0}, and of order 5 = {5, 10, 15, 20} -it was calculated previously-.

The number of elements of order 10 in ℤ100 ⊕ ℤ25 = 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, ⟨ak⟩=⟨agcd(n,k)⟩ and |⟨ak⟩| = $\frac{n}{gcd(n,k)}.$ 10 = |a|= |ak| ↭ 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.

• |ℤ6 ⊕ U(14)| = 6 x 6 = 36. U(14) = {1, 3, 5, 9, 11, 13}. The order of (3, 3) is lcm(|3| in ℤ6, |3| in U(14)) = lcm (2, 6) = 6.

31=3, 32 = 9, 33 = 27 % 14 = 13, 34 = 39 % 14 = 11, 35 = 5, 36 = 1.

• |S4 ⊕ 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}, 41 = 4, 42 = 16 % 9 = 7, 43 = 28 % 9 = 1.

• In SL(2, ℝ) ⊕ ℤ4, the order of $([\bigl(\begin{smallmatrix}cos(2π/3) & -sin(2π/3)\\ sin(2π/3) & cos(2π/3)\end{smallmatrix}\bigr)], 1)$ = lcm(|$([\bigl(\begin{smallmatrix}cos(2π/3) & -sin(2π/3)\\ sin(2π/3) & cos(2π/3)\end{smallmatrix}\bigr)], |1|)$|) = lcm (3, 4) = 12, because the matrix has order 3 (ρ = rotation 2π/3 -120°- has order three, ρ3 = e).

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): $(\begin{smallmatrix}cos(θ) & -sin(θ)\\ sin(θ) & cos(θ)\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}xcosθ -y sinθ\\ xsinθ + ycosθ\end{smallmatrix})$

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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