“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 G_{1}, G_{2}, . . . , G_{n} be a finite collection of groups. The external direct product of G_{1}, G_{2}, . . . , G_{n}, written as G_{1} ⊕ G_{2} ⊕ ··· ⊕ 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., ℤ_{3} ⊕ U(5) ⊕ D_{4}, (2, 3, sr)(2, 4, sr^{2}) = (2+2, 3·4, (sr)(sr^{2})) =[3·4 = 12 (mod 5) = 2; (sr)(sr^{2}) = s(rs)r^{2} = s(sr^{-1})r^{2} = s(sr^{3})r^{2} = s^{2}r^{5} = r] (1, 2, r); ℤ_{2} ⊕ ℤ_{2} ⊕ ℤ_{2}, (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
Let’s check that H ◁ G ↭ ∀g ∈ D_{6}, g·H = H·g. Of course, g·e = e·g = g. Besides, g = r^{k}, r^{3 }g = r^{3}r^{k} = r^{3+k} = r^{k+3} = r^{k}r^{3} = gr^{3}. If g = r^{k}s, r^{3}g = r^{3}r^{k}s = r^{k}(r^{3}s) =[Dihedral group, r^{k}s = sr^{n-k}, 1≤k≤n-1. In particular, k = 3, r^{3}s = sr^{6-3} = sr^{3}] r^{k}(sr^{3}) = (r^{k}s)r^{3} = gr^{3}
Recall that Z(G) = {z ∈ G| zg = gz ∀g ∈G}. Z(D_{6}) = ⟨r^{3}⟩ = {e, r^{3}}.
More generally, let H_{1}, H_{2}, ···, H_{n} be a finite collection of subgroups of G. G is the internal direct product of H_{1}, H_{2}, ··· H_{n}, G = H_{1} x H_{2} x ··· x H_{n}, if
Theorem. If G is the internal direct product of a finite number of subgroups H_{1}, H_{2}, ··· H_{n}, then G is isomorphic to the external direct product of H_{1}, H_{2}, ··· H_{n}. More concisely, G = H_{1} x H_{2} x ··· x H_{n} ≋ H_{1} ⊕ H_{2} ⊕ ··· ⊕ H_{n}
Proof.
Let G = H_{1} x H_{2} x ··· x H_{n}
[By assumption, H_{j} ◁ G ↭ gH_{j}g^{-1} ⊆ H_{j}, let g =_{G=H1xH2x···xHn} (e·e···h_{i}···e·e) = h_{i}] h_{i}h_{j}h_{i}^{-1} ∈ H_{j} and (h_{i}h_{j}h_{i}^{-1})h_{j}^{-1} ∈ H_{j}h_{j}^{-1} ⇒ [Ha = H ↭ a ∈ H, h_{j}^{-1} ∈ H_{j}] h_{i}h_{j}h_{i}^{-1}h_{j}^{-1} ∈ H_{j}h_{j}^{-1} = H_{j} and similarly h_{i}(h_{j}h_{i}^{-1}h_{j}^{-1}) ∈ h_{i}H_{i} = H_{i}.
Therefore, h_{i}h_{j}h_{i}^{-1}h_{j}^{-1} ∈ H_{i} ∩ H_{j}.
i ≠ j, say i < j, let h ∈ H_{i} ∩ H_{j}, h = e_{1}···e_{i-1}he_{i+1}···e_{j-1}, therefore h ∈ H_{1} ··· H_{i} ··· H_{j-1} ∩ H_{j} =[Condition 3. Each subgroup is disjoint from the product of all previous subgroup.] {e}.
h_{i}h_{j}h_{i}^{-1}h_{j}^{-1} ∈ H_{i} ∩ H_{j} = {e} ⇒ h_{i}h_{j}h_{i}^{-1}h_{j}^{-1} = e ⇒ h_{i}h_{j} = h_{j}h_{i}
Suppose, h_{1}h_{2}···h_{n} = h’_{1}h’_{2}···h’_{n} ⇒ h_{1}h_{2}···h_{n}h_{n}^{-1} = h’_{1}h’_{2}···h’_{n}h_{n}^{-1} ⇒ h_{1}h_{2}···h_{n-1} = h’_{1}h’_{2}···h’_{n}h_{n}^{-1} ⇒ h_{1}h_{2}···h_{n-1}h_{n-1}^{-1} = h’_{1}h’_{2}···h’_{n}h_{n}^{-1}h_{n-1}^{-1} ⇒ [∀h_{i} ∈ H_{i}, h_{j} ∈ H_{j}, i ≠ j, h_{i}h_{j} = h_{j}h_{i}] h_{1}h_{2}···h_{n-2} = h’_{1}h’_{2}···h_{n-1}h’_{n-1}^{-1}h’_{n}h_{n}^{-1} ⇒ e = h’_{1}h_{1}^{-1}h’_{2}h_{2}^{-1} ··· h’_{n}h_{n}^{-1} ⇒ h_{n} = h’_{1}h_{1}^{-1}h’_{2}h_{2}^{-1} ··· h’_{n} ⇒ [Multiplying both sides by h’_{n}^{-1} and ∀h_{i} ∈ H_{i}, h_{j} ∈ H_{j}, i ≠ j, h_{i}h_{j} = h_{j}h_{i}] h’_{n}^{-1}h_{n} = h’_{1}h_{1}^{-1} ··· h’_{n-1}h_{n-1}^{-1} ∈ H_{1}H_{2} ··· H_{n-1} ∩ H_{n} = {e} ⇒ h’_{n}^{-1}h_{n} = e ⇒ h’_{n} = h_{n}. Therefore, h_{1}h_{2}···h_{n-1} = h’_{1}h’_{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.
We already know that G = ℤ_{6} = {0, 1, 2, 3, 4, 5, 6} ≋ ℤ_{2} x ℤ_{3} = ⟨(1, 1)⟩ = {(0, 0), (1, 0), (0, 1), (1, 1), (1, 2), (0, 2)}. Let H = ⟨3⟩ = {0, 3} ≋ ℤ_{2}, K = ⟨2⟩ = {0, 2, 4} ≋ ℤ_{3}. Then, |H| = 2, |K| = 3, |H ∩ K| = 1 (H ∩ K = {0}), G = HK (|HK| = 3·2/1 = 6 = |G|), and therefore ℤ_{6} is the internal direct product of H and K, ℤ_{6} = ⟨3⟩ x ⟨2⟩ ≋ ⟨3⟩ ⊕ ⟨2⟩ ≋ ℤ_{2} ⊕ ℤ_{3}.
U_{8} = {1, 3, 5, 7}, H_{1} = ⟨3⟩ = {1, 3} ≋ ℤ_{2}, H_{2} = ⟨5⟩ = {1, 5} ≋ ℤ_{2}. H_{1} ∩ H_{2} = {1}, by the same reasoning as before, H_{1}H_{2} = U_{8}, H_{1} ◁ U_{8}, H_{2} ◁ U_{8} ⇒[By the previous theorem] U_{8} ≋ H_{1} ⊕ H_{2} = ⟨3⟩ ⊕ ⟨5⟩ = ℤ_{2} ⊕ ℤ_{2}.
n-odd, D_{2n} = {e, r, r^{2}, ···, r^{2n-1}, s, sr, ···, sr^{2n-1}}, let H_{1} = ⟨s, r^{2}⟩ = {e, r^{2}, r^{4}, ···, r^{2n-2}, s, sr^{2}, ···, sr^{2n-2}} so since |D_{2n}/⟨s, r^{2}⟩| = 2 ⇒[Subgroup of index 2 is normal] H_{1} = ⟨s, r^{2}⟩ ◁ D_{2n}. Futhermore, H_{1} ≋ 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 H_{2} = ⟨r^{n}⟩ = {e, r^{n}} = ℤ(D_{2n}) ⇒ [The center of a group is a normal subgroup] H_{2} ◁ D_{2n}. Futhermore, D_{2n} = H_{1}H_{2}, and H_{1} ∩ H_{2} = {e} ⇒ [By the previous theorem] D_{2n} ≋ H_{1} ⊕ H_{2} ≋ ℤ_{2} ⊕ D_{n}, e.g., D_{6} ≋ ℤ_{2} ⊕ D_{3}, D_{14} ≋ ℤ_{2} ⊕ D_{7}.