Internal Direct Product

“Sometimes, I don’t understand what you’re talking about,” I said, shrugging my shoulders. “Most of the time, I don’t understand what I’m talking about. There is nothing more unknown as a sure thing. The future is no more uncertain than the present. The present is a chaotic masquerade; the world, a meaningless merry-go-round, life is a messy game, and I am just a broken, pathetic pawn. I believe that I know nothing, and yet I am not quite sure about that,” Apocalypse, Anawim, #justtothepoint.

Recall. Let G₁, G₂, . . . , G_n be a finite collection of groups. The external direct product of G₁, G₂, . . . , G_n, written as G₁ ⊕ G₂ ⊕ ··· ⊕ G_n, is the set of all n-tuples for which the ith component is an element of G_i and the operation is component-wise, e.g., ℤ₃ ⊕ U(5) ⊕ D₄, (2, 3, sr)(2, 4, sr²) = (2+2, 3·4, (sr)(sr²)) =[3·4 = 12 (mod 5) = 2; (sr)(sr²) = s(rs)r² = s(sr^-1)r² = s(sr³)r² = s²r⁵ = r] (1, 2, r); ℤ₂ ⊕ ℤ₂ ⊕ ℤ₂, (1, 0, 1) + (0, 1, 0) = (1, 1, 1).

Let G be a group, H and H subgroups of G (H, K ≤ G), G is the internal direct product of H and K, G = H x K or G = HK (the notation is unfortunately not standard) if

1. H and K are normal subgroups of G, H, K ◁ G.
2. Their intersection is trivial, H ∩ K = {e}.
3. Every element of G can be expressed uniquely as the product of an element of an element of H and K, i.e, G = HK = {hk | h ∈ H, k ∈ K}.
   Most authors, they only require that G is the subset product of H and K, that is, G = HK.

Examples

• Let G = ℤ₈* (≋ V₄ ≋ ℤ₂ x ℤ₂), H = ⟨3⟩ = {1, 3}, K = ⟨5⟩ = {1, 5}. Then, (ii) H ∩ K = {1}, |HK| = $\frac{|H||K|}{|H∩K|} = \frac{2·2}{1}$ = 4 ⇒ [|G| = 4] G = HK (iii). Besides, since G is Abelian, H, K ◁ G. Therefore, G = H x K is an internal product.
• U(12) = {1, 5, 7, 11}, ⟨5⟩ = {1, 5}, ⟨5⟩ ◁ U(12), ⟨7⟩ = {1, 7}, ⟨7⟩ ◁ U(12), 5·7 =_{mod 12} 11. ⟨5⟩ ∩ ⟨7⟩ = {1}. U(12) = ⟨5⟩ x ⟨7⟩ ≋ ℤ₂ x ℤ₂.
• Let G = D₆ be the dihedral group of order 12, D₆ = ⟨r, s| r⁶ = s² = e, rs = sr⁵⟩ where r³s = sr³. Consider H = ⟨r³⟩ = {e, r³} ≋ ℤ₂. K = ⟨r², s⟩ = {e, r², r⁴, s, r²s, r⁴s} ≋ S₃. H and K are both normal, they only have the identity in common, and G = HK (|HK| = $\frac{|H||K|}{|H∩K|} = \frac{2·6}{1}$ = 12 = |G|). D₆ is an internal product, D₆ = H x K ≋ ℤ₂ x D₃.

Let’s check that H ◁ G ↭ ∀g ∈ D₆, g·H = H·g. Of course, g·e = e·g = g. Besides, g = r^k, r³ g = r³r^k = r^3+k = r^k+3 = r^kr³ = gr³. If g = r^ks, r³g = r³r^ks = r^k(r³s) =[Dihedral group, r^ks = sr^n-k, 1≤k≤n-1. In particular, k = 3, r³s = sr^6-3 = sr³] r^k(sr³) = (r^ks)r³ = gr³

Recall that Z(G) = {z ∈ G| zg = gz ∀g ∈G}. Z(D₆) = ⟨r³⟩ = {e, r³}.

• Counterexample. Let G = S₃, H = A₃ = ⟨(123)⟩ = {1, (123), (132)} ◁ G, K = ⟨(12)⟩ = {1, (12)}, H ∩ K = {e}, G = HK (|HK| = $\frac{|H||K|}{|H∩K|}=\frac{3·2}{1} = 6$ = 3! = |S₃|), but K is not a normal subgroup, e.g., (123)(12) = (13) ≠ (23) = (12)(123). Therefore, S₃ is not an internal direct product of any of its subgroups.

More generally, let H₁, H₂, ···, H_n be a finite collection of subgroups of G. G is the internal direct product of H₁, H₂, ··· H_n, G = H₁ x H₂ x ··· x H_n, if

1. H_i ◁ G, ∀i, 1 ≤ i ≤ n.
2. Every element of G can be expressed as h₁h₂···h_n. G = H₁ H₂ ··· H_n = {h₁h₂···h_n | h_i ∈ H_i}
3. Each subgroup is disjoint (with the exception of the identity) from the product of all previous subgroups, (H₁H₂···H_i) ∩ H_i+1 = {e} ∀i, 1 ≤ i ≤ n-1.

Theorem. If G is the internal direct product of a finite number of subgroups H₁, H₂, ··· H_n, then G is isomorphic to the external direct product of H₁, H₂, ··· H_n. More concisely, G = H₁ x H₂ x ··· x H_n ≋ H₁ ⊕ H₂ ⊕ ··· ⊕ H_n

Proof.

Let G = H₁ x H₂ x ··· x H_n

• Claim: ∀h_i ∈ H_i, h_j ∈ H_j, i ≠ j, h_ih_j = h_ih_j.

[By assumption, H_j ◁ G ↭ gH_jg^-1 ⊆ H_j, let g =_{G=H1xH2x···xHn} (e·e···h_i···e·e) = h_i] h_ih_jh_i^-1 ∈ H_j and (h_ih_jh_i^-1)h_j^-1 ∈ H_jh_j^-1 ⇒ [Ha = H ↭ a ∈ H, h_j^-1 ∈ H_j] h_ih_jh_i^-1h_j^-1 ∈ H_jh_j^-1 = H_j and similarly h_i(h_jh_i^-1h_j^-1) ∈ h_iH_i = H_i.

Therefore, h_ih_jh_i^-1h_j^-1 ∈ H_i ∩ H_j.

i ≠ j, say i < j, let h ∈ H_i ∩ H_j, h = e₁···e_i-1he_i+1···e_j-1, therefore h ∈ H₁ ··· H_i ··· H_j-1 ∩ H_j =[Condition 3. Each subgroup is disjoint from the product of all previous subgroup.] {e}.

h_ih_jh_i^-1h_j^-1 ∈ H_i ∩ H_j = {e} ⇒ h_ih_jh_i^-1h_j^-1 = e ⇒ h_ih_j = h_jh_i

• Next, we claim that each member of G can be expressed uniquely in the form h₁h₂···h_n, ∀h_i ∈ H_i

Suppose, h₁h₂···h_n = h’₁h’₂···h’_n ⇒ h₁h₂···h_nh_n^-1 = h’₁h’₂···h’_nh_n^-1 ⇒ h₁h₂···h_n-1 = h’₁h’₂···h’_nh_n^-1 ⇒ h₁h₂···h_n-1h_n-1^-1 = h’₁h’₂···h’_nh_n^-1h_n-1^-1 ⇒ [∀h_i ∈ H_i, h_j ∈ H_j, i ≠ j, h_ih_j = h_jh_i] h₁h₂···h_n-2 = h’₁h’₂···h_n-1h’_n-1^-1h’_nh_n^-1 ⇒ e = h’₁h₁^-1h’₂h₂^-1 ··· h’_nh_n^-1 ⇒ h_n = h’₁h₁^-1h’₂h₂^-1 ··· h’_n ⇒ [Multiplying both sides by h’_n^-1 and ∀h_i ∈ H_i, h_j ∈ H_j, i ≠ j, h_ih_j = h_jh_i] h’_n^-1h_n = h’₁h₁^-1 ··· h’_n-1h_n-1^-1 ∈ H₁H₂ ··· H_n-1 ∩ H_n = {e} ⇒ h’_n^-1h_n = e ⇒ h’_n = h_n. Therefore, h₁h₂···h_n-1 = h’₁h’₂···h’_n-1 and repeating the argument, we can obtain h_n-1 = h’_n-1, and so on, so forth h_i = h’_i ∀i, 1 ≤ i ≤ n.

• Φ: H₁ x H₂ x ··· x H_n → H₁ ⊕ H₂ ⊕ ··· ⊕ H_n, Φ(h₁h₂···h_n) = (h₁, h₂···, h_n) is an isomorphism.

1. Since every element of G has an unique representation as h₁h₂···h_n, Φ is well defined, onto, and one-to-one.
2. Let a = h₁h₂···h_n, b = h’₁h’₂···h’_n, Φ(ab) = Φ(h₁h₂···h_nh’₁h’₂···h’_n) = [∀h_i ∈ H_i, h_j ∈ H_j, i ≠ j, h_ih_j = h_jh_i] Φ(h₁h’₁h₂h’₂···h_nh’_n) = (h₁h’_1, h₂h’₂,···, h_nh’_n) = (h₁, h₂,···, h_n)(h’_1, h’₂,···, h’_n) = Φ(a)·Φ(b)∎

Examples

• We already know that G = ℤ₆ = {0, 1, 2, 3, 4, 5, 6} ≋ ℤ₂ x ℤ₃ = ⟨(1, 1)⟩ = {(0, 0), (1, 0), (0, 1), (1, 1), (1, 2), (0, 2)}. Let H = ⟨3⟩ = {0, 3} ≋ ℤ₂, K = ⟨2⟩ = {0, 2, 4} ≋ ℤ₃. Then, |H| = 2, |K| = 3, |H ∩ K| = 1 (H ∩ K = {0}), G = HK (|HK| = 3·2/1 = 6 = |G|), and therefore ℤ₆ is the internal direct product of H and K, ℤ₆ = ⟨3⟩ x ⟨2⟩ ≋ ⟨3⟩ ⊕ ⟨2⟩ ≋ ℤ₂ ⊕ ℤ₃.

• U₈ = {1, 3, 5, 7}, H₁ = ⟨3⟩ = {1, 3} ≋ ℤ₂, H₂ = ⟨5⟩ = {1, 5} ≋ ℤ₂. H₁ ∩ H₂ = {1}, by the same reasoning as before, H₁H₂ = U₈, H₁ ◁ U₈, H₂ ◁ U₈ ⇒[By the previous theorem] U₈ ≋ H₁ ⊕ H₂ = ⟨3⟩ ⊕ ⟨5⟩ = ℤ₂ ⊕ ℤ₂.

• n-odd, D_2n = {e, r, r², ···, r^2n-1, s, sr, ···, sr^2n-1}, let H₁ = ⟨s, r²⟩ = {e, r², r⁴, ···, r^2n-2, s, sr², ···, sr^2n-2} so since |D_2n/⟨s, r²⟩| = 2 ⇒[Subgroup of index 2 is normal] H₁ = ⟨s, r²⟩ ◁ D_2n. Futhermore, H₁ ≋ D_n.

Recall that ℤ(D_n) = $ \begin{cases} e, n~is~ odd\\\\ e~ and~ α^{\frac{n}{2}}, n~is~ even \end{cases}$

Let H₂ = ⟨rⁿ⟩ = {e, rⁿ} = ℤ(D_2n) ⇒ [The center of a group is a normal subgroup] H₂ ◁ D_2n. Futhermore, D_2n = H₁H₂, and H₁ ∩ H₂ = {e} ⇒ [By the previous theorem] D_2n ≋ H₁ ⊕ H₂ ≋ ℤ₂ ⊕ D_n, e.g., D₆ ≋ ℤ₂ ⊕ D₃, D₁₄ ≋ ℤ₂ ⊕ D₇.

Bibliography

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.