Direct Products II

I have yet to see any problem, however complicated, which, when looked at in the right way did not become still more complicated, Paul Anderson.

Theorem. If r is a divisor of m and s is a divisor of n, ℤ_m ⊕ ℤ_n has a subgroup isomorphic to ℤ_r ⊕ ℤ_s.

Lemma 1. Suppose G₁ and G₂ are groups, H₁ ≤ G₁ and H₂ ≤ G₂. Then, (H₁ ⊕ H₂) ≤ (G₁ ⊕ G₂)

Proof.

∀ a, b ∈ H₁ ⊕ H₂, a = (a₁, a₂), b = (b₁, b₂), a₁, b₁ ∈ H₁, a₂, b₂ ∈ H₂ ⇒ [H₁ ≤ G₁ and H₂ ≤ G₂] ∃b₁^-1 ∈ H₁, b₂^-1 ∈ H₂ such that a·b^-1 = (a₁, a₂)·(b₁^-1, b₂^-1) = (a₁·b₁^-1, a₂·b₂^-1) ∈ H₁ ⊕ H₂∎

Lemma 2. If G₁, G₂, H₁, H₂ are groups such that G₁ ≋ H₁ and G₂ ≋ H₂, then G₁ ⊕ G₂ ≋ H₁ ⊕ H₂

Proof.

Since G₁ ≋ H₁ and G₂ ≋ H₂ ⇒ ∃Φ₁: G₁ → H₁, Φ₂: G₂ → H₂ isomorphism.

Define a function Φ: G₁ ⊕ G₂ → H₁ ⊕ H₂, by Φ(g₁, g₂) = (Φ₁(g₁), Φ₂(g₂)). It is pretty easy to show that Φ is indeed an isomorphism.

Proof. (Main Theorem)

Recall the following Theorem. 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⟩| = $\frac{n}{gcd(n,k)}.$

If r is a divisor of m and s is a divisor of n, the subgroup ⟨m/r⟩ ≤ ℤ_m is cyclic of order r (so it is isomorphic to ℤ_r) and the subgroup ⟨n/s⟩ ≤ ℤ_n is cyclic of order s (so it is isomorphic to ℤ_s) ⇒ [Lemma 1] ⟨m/r⟩ ⊕ ⟨n/s⟩ ≤ ℤ_m ⊕ ℤ_m and [By Lemma 2] isomorphic to ℤ_r ⊕ ℤ_s ∎

• Let’s find a subgroup of ℤ₃₀ ⊕ ℤ₁₂ isomorphic to ℤ₆ ⊕ ℤ₄.

1. We notice that ℤ₃₀ ⊕ ℤ₁₂ has a subgroup isomorphic to ℤ₆ ⊕ ℤ₄.
2. ⟨5⟩ is a subgroup of ℤ₃₀, |⟨5⟩| = 6. ⟨3⟩ is a subgroup of ℤ₁₂, |⟨3⟩| = 4.
3. By the previous result, ⟨5⟩ ⊕ ⟨3⟩ is such a subgroup.

Criterion for the direct product to be cyclic. Let G and H be finite cyclic groups. Then, the external direct product of both groups G ⊕ H is cyclic if and only if their orders are relatively prime,i.e., |G| and |H| are relatively prime, (|G|, |H|) = 1.

Proof.

⇒) |G| = m, |H| = n, |G ⊕ H| = nm. Let’s assume G ⊕ H is cyclic, ∃ g∈ G, h ∈H such that G ⊕ H = ⟨(g, h)⟩, let d = gcd(n,m).

$(g, h)^{\frac{mn}{d}}=(g^{\frac{mn}{d}}, h^{\frac{mn}{d}})=((g^{m})^{\frac{n}{d}},(h^{n})^{\frac{m}{d}})=(e^{\frac{n}{d}},e’^{\frac{m}{d}})$ = (e, e’) where e and e’ are the identity in G and H respectively, and mn/d, n/d and m/d are integers by the gcd definition ⇒ [|G ⊕ H| = nm and by assumption, G ⊕ H = ⟨(g, h)⟩] |(g, h)| = nm ≤ mn/d ⇒ d = 1.

⇐) Suppose |G| = m, |H| = n, d = gcd(n, m) = 1, G = ⟨g⟩, H = ⟨h⟩ ⇒ [Order of an Element in a Direct Group] ⟨(g, h)⟩ = lcm(n, m) ⇒ [gcd(n, m) = 1] = nm, |G ⊕ H| = nm, and therefore (g, h) is a generator of G ⊕ H.

Using an induction argument and this criterion, we can generalize this result.

Theorem. Let G₁, G₂, and G_n be finite cyclic groups. Then, their external direct product G₁ ⊕ G₂ ⊕ ··· ⊕ G_n is cyclic if and only if their orders are coprime(relatively prime),i.e., |G_i| and |G_j| are relatively prime ∀i ≠ j.

Corollary. Let m = n₁n₂···n_k. Then, ℤ_m, that is, (ℤ_n1n2···nk) is isomorphic to ℤ_n1 ⊕ ℤ_n2 ⊕ ··· ⊕ ℤ_nk if and only if n_i and n_j are relatively prime, ∀i ≠ j, i.e., ℤ_n1n2···nk ≋ ℤ_n1 ⊕ ℤ_n2 ⊕ ··· ⊕ ℤ_nk ↭ (n_i, n_j) = 1, ∀i ≠ j.

Examples:

• ℤ₂ ⊕ ℤ₃ is cyclic, ℤ₂ ⊕ ℤ₃ ≋ ℤ₆ and (1, 1) is its generator.

The only two proper subgroups are {(0, 0), (1, 0)} and {(0, 0), (0, 1), (0, 2)}

• Counterexamples (Not cyclic): ℤ₂ ⊕ ℤ₂ (≇ ℤ₄), ℤ₂ ⊕ ℤ₃₀ (≇ ℤ₆₀), ℤ₃ ⊕ ℤ₉ (≇ ℤ₂₇)
• ℤ₂ ⊕ ℤ₂ ⊕ ℤ₃ ⊕ ℤ₅ ≈ [Criterion for the direct product to be cyclic, gcd(2, 3) = 1] ℤ₂ ⊕ ℤ₆ ⊕ ℤ₅ ≈ [Criterion for the direct product to be cyclic, gcd(6, 5) = 1] ℤ₂⊕ ℤ₃₀ ≇ ℤ₆₀ because gcd(2, 30) = 2 ≠ 1.
• ℤ₃₀ ≈ [Criterion for the direct product to be cyclic, gcd(2, 15) = 1] ℤ₂ ⊕ ℤ₁₅ ≈ [Criterion for the direct product to be cyclic, gcd(3, 5) = 1]ℤ₂ ⊕ ℤ₃ ⊕ ℤ₅ ⇒ ℤ₂ ⊕ ℤ₃₀ ≈ ℤ₂ ⊕ ℤ₂ ⊕ ℤ₃ ⊕ ℤ₅.
• ℤ₃₀ ≈ ℤ₃ ⊕ ℤ₁₀ ⇒ ℤ₂ ⊕ ℤ₃₀ ≈ ℤ₂ ⊕ ℤ₃ ⊕ ℤ₁₀ ≈ ℤ₆ ⊕ ℤ₁₀.

The Group of Units Modulo n as an External Direct Product

Theorem. Suppose s and t are relatively prime. Then, U(st)≈U(s)⊕U(t), i.e., U(st) is isomorphic to the external direct product of U(s) and U(t). Futhermore, U_s(st) ≈ U(t) and U_t(st) ≈ U(s)

Recall. U_n = {[k]| gcd(k, n) = 1}, that is, U_n consists of all [k], where k is relatively prime to n, e.g., U₈ = {[1], [3], [5], [7]}, [3]² = [9] = [1], [5]² = [25] = [1], and [7]² = [49] = [1], so all its elements (except the identity) have order two.

Recall. U_k(n) = {x ∈ U(n): x mod k = 1}, U_k(n) ≤ U(n), e.g., U₁₀₅(5) = {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101}.

Proof. Let’s define an isomorphism, Φ: U(st) → U(s)⊕U(t), Φ(x) = (x mod s, x mod t). Analogously, Φ’ and Φ’’ are both isomorphisms, Φ’: U_s(st) → U(t), Φ’(x) = (x mod t), Φ’’: U_s(st) → U(s), Φ’’(x) = (x mod s).

Φ: U(st) → U(s)⊕U(t), Φ(x) = (x mod s, x mod t).

1. Obviously, Φ is well-defined and Φ is operation preserving.
2. Φ is one-to-one because Φ(x) = Φ(y) ⇒ x mod s = y mod s, x mod t = y mod t ⇒ x -y ≡ 0 mod s, and x -y ≡ 0 mod t ⇒ s | x-y and t | x -y ⇒ [By assumption, (s, t) = 1] st | x-y ⇒ x ≡ y mod st.
3. Φ is onto. |U(st)| = θ(st) =[(s, t) = 1] θ(s)θ(t) = |U(s)||U(t)|. Therefore, Φ is one-to-one between two finite sets of the same order ⇒ Φ is onto.

   θ is Euler’s totient function, it counts the positive integers up to a given integer n that are relatively prime to n. It is a multiplicative function, meaning that if two numbers m and n are relatively prime, then θ(mn) = θ(m)θ(n).

Corollary. Let m be a integer that can be factored as n₁n₂···n_k, where each of the factors is relatively prime to all of the others (∀i, j, i ≠ j, gcd(n_i, n_j) = 1). Then, U(m) ≋ U(n₁)⊕U(n₂)⊕···⊕U(n_k).

Examples:

• U(105) ≋ U(7) ⊕ U(15) ≋ U(7)⊕U(3)⊕U(5).
• U(7) ≋ [t = 7, s = 15, st = 105] U₁₅(105), and U(15) ≋ U₇(105)
• U(21) = {1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20} ≋ [t = 21, s = 5, st = 105] U₅(105), and U(5) ≋ U₂₁(105)

|U(21)| = 12. U(3) = {1, 2}, |U(3)| = 2; and U(7) = {1, 2, 3, 4, 5, 6}, |U(7)| = 6;

Φ: U(st) → U(s)⊕U(t), Φ(x) = (x mod s, x mod t). In particular, Φ: U(21) → U(3)⊕U(7), Φ(x) = (x mod 3, x mod 7).

U₇(21) = {1, 8}, U₃(21) = {1, 4, 10, 13, 16, 19}. Φ’: U₇(21) → U(3), Φ’(x) = x mod 3. Φ’’: U₃(21) → U(7), Φ’’(x) = x mod 7.

Theorem (Gauss, 1801). U(2) = {1} ≋ {0}, U(4) = {1, 3} ≋ ℤ₂. U(2ⁿ) ≋ U(2²) ⊕ U(2^n-2) ≋ ℤ₂⊕ℤ_2^n-2 ∀n ≥ 3, U(pⁿ) ≋ ℤ_pⁿ-p^n-1 for p a positive integer and an odd prime.

Theorem. Every group U(n) is isomorphic to the external direct product of cyclic groups.

Examples

• U(55) ≋[Criterion for the direct product to be cyclic] U(5) ⊕ U(11) ≋ ℤ₄⊕ℤ₁₀ ≋ ℤ₄⊕ℤ₂⊕ℤ₅

• U(8) = U(2³) = {1, 3, 5, 7} has 4 elements and is non-cyclic since each element (except the identity 1) has order 2. U(8) ≋ ℤ₂⊕ℤ₂.

• If p = 5, n = 2, U(5²) ≋[U(pⁿ) ≋ ℤ_pⁿ-p^n-1 for p a positive integer and an odd prime] ℤ_5²-5¹, i.e., U(25) ≋ ℤ₂₀.

• U(16) ≋[U(2ⁿ) ≋ ℤ₂⊕ℤ_2^n-2 ∀n ≥ 3] ℤ₂ ⊕ ℤ₄, U(9) ≈[9 = 3², 2 ≱ 3] ℤ₆, U(7) ≋ ℤ₆, U(5) ≋ ℤ₄, U(3) ≈ ℤ₂.

• U(105) ≋ [105 = 3·5·7] U(3)⊕U(5)⊕U(7) ≋ ℤ₂ ⊕ ℤ₄ ⊕ ℤ₆. |U(105)| = 2 × 4 × 6 = 48.

• U(100) ≋ U(4) ⊕ U(25) ≋ ℤ₂ ⊕ ℤ₂₀

• U(168) ≋ [168 = 3·7·8] U(3)⊕U(7)⊕U(8) ≋ ℤ₂ ⊕ ℤ₆ ⊕ ℤ₂ ⊕ ℤ₂.

• U(720) ≋ [720 = 16·9·5] U(16)⊕U(9)⊕U(5) ≋ ℤ₂ ⊕ ℤ₄ ⊕ ℤ₆ ⊕ ℤ₄.

• What are the possible orders of U(720)? U(720) ≋ ℤ₂ ⊕ ℤ₄ ⊕ ℤ₆ ⊕ ℤ₄, so |(a, b, c, d)| = lcm(|a|, |b|, |c|, |d|) = {1, 2, 3, 4, 6, 12} because |a|= 1 or 2, |b| = |d| = 1, 2, or 4, |c| = 1, 2, 3 or 6.

• Prove that ℝ* ⊕ ℝ* is not isomorphic to ℂ*. For the sake of contradiction, let’s suppose there exists an isomorphism Φ: ℂ* → ℝ* ⊕ ℝ*, |i| = 4 ⇒ [Isomorphisms preserve order of elements] |Φ(i)| = 4 ⊥ because we claim that the only elements of finite order in ℝ* ⊕ ℝ* are {(1, 1), (-1, 1), (1, -1), (-1, -1)} and all these elements have order either 1 or 2.

Claim. Let’s Suppose (a, b) ∈ ℝ* ⊕ ℝ* has finite order ⇒ ∃n: (a, b)ⁿ = (1, 1) ⇒ (aⁿ, bⁿ) = (1, 1) ⇒[The only elements of finite order in ℝ* are 1 and -1] aⁿ = bⁿ = 1 ⇒ a = ±1, b = ±1.

• How many elements of order 12 are there in U(720)? U(720) ≋ ℤ₂ ⊕ ℤ₄ ⊕ ℤ₆ ⊕ ℤ₄. 12 = lcm(|a|, |b|, |c|, |d|)

Case 1. |b| = 4 (there are two choices, i.e., 1 and 3), |c| = 3 (2, 4) or 6 (1, 5). We need to count 2 (|a|)·2(|b|)·4(|c|)·4(|d|) = 64.

Case 2. |d| = 4, |c| = 3 or 6. Besides, we need to add 2 (|a|)·2(|b| because we have already counted |b|=4, so |b|=1 or |b|=2, i.e., b = 0 or 2)·4(|c|)·2(|d|) = 32. Therefore, 64+32= 96 elements of order 12.

• Let H ={$\left(\begin{smallmatrix}1 & a & b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$| a, b ∈ ℤ₃}. Show that H is an Abelian group of order 9. Is H isomorphic to ℤ₉ or ℤ₃ ⊕ ℤ₃

Let $\left(\begin{smallmatrix}1 & a & b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & c & d\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$ ∈ H, then $\left(\begin{smallmatrix}1 & a & b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)\left(\begin{smallmatrix}1 & c & d\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)=\left(\begin{smallmatrix}1 & c+a & d+b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$ ∈ H, so it is closed under matrix multiplication.

Futhermore, $\left(\begin{smallmatrix}1 & c & d\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)\left(\begin{smallmatrix}1 & a & b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)=\left(\begin{smallmatrix}1 & a+c & b+d\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right) = \left(\begin{smallmatrix}1 & c+a & d+b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$, and therefore H is Abelian.

Matrix multiplication is associative, $\left(\begin{smallmatrix}1 & 0 & 0\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$ is the identity element, and $\left(\begin{smallmatrix}1 & -a & -b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$ is the inverse of $\left(\begin{smallmatrix}1 & a & b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$. Therefore, H is an Abelian group.

Futhermore, Φ: ℤ₃ ⊕ ℤ₃ → H defined by Φ(a, b) = $\left(\begin{smallmatrix}1 & a & b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$ is an isomorphism. It is clear that Φ is a bijection, and Φ((a, b) + (c, d)) = Φ(a + c, b + d) = $\left(\begin{smallmatrix}1 & a + c & b + d\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right) = \left(\begin{smallmatrix}1 & a & b\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)\left(\begin{smallmatrix}1 & c & d\\ 0 & 1 & 0\\0 & 0 & 1\end{smallmatrix}\right)$ = Φ(a, b)Φ(c, d) ⇒ Φ is an homomorphism ⇒ Φ is an isomorphism ⇒ H ≋ ℤ₃ ⊕ ℤ₃.

• How many automorphisms of ℤ₁₀₅ are there with order 12?

Isomorphisms preserves order, i.e., ∀a ∈ ℤ₁₀₅, |a| = |Φ(a)|, and we also know that Aut(ℤ₁₀₅) ≈ U(105).

How many elements of order 12 are there in U₁₀₅ = ℤ₂ ⊕ ℤ₄ ⊕ ℤ₆? If (a, b, c) ∈ ℤ₂ ⊕ ℤ₄ ⊕ ℤ₆ has order 12 ⇒ lcm (|a|, |b|, |c|) = 12 ⇒ |b| = 4 (b=1 or 3 -gcd(4, 3) = 1), |c| = 3 (c=2 or 4) or |c| = 6 (c = 1 or 5). Total elements of order 12 = 2 (|a|) x 2 (|b|) x 4 (|c|) = 16. There are 16 automorphisms of order 12.

• How many automorphisms of ℤ₇₂₀ are there of order 12?

Isomorphisms preserves order, i.e., ∀a ∈ ℤ₇₂₀, |a| = |Φ(a)|, and we also know that Aut(ℤ₇₂₀) ≈ U(720). Therefore (previous exercise), there are 96 elements of order 12 in U(720), and 96 automorphism of ℤ₇₂₀ of order 12.

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