    # 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, . . . , Gn be a finite collection of groups. The external direct product of G1, G2, . . . , Gn, written as G1 ⊕ G2 ⊕ ··· ⊕ Gn, is the set of all n-tuples for which the ith component is an element of Gi 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 = rk, r3 g = r3rk = r3+k = rk+3 = rkr3 = gr3. If g = rks, r3g = r3rks = rk(r3s) =[Dihedral group, rks = srn-k, 1≤k≤n-1. In particular, k = 3, r3s = sr6-3 = sr3] rk(sr3) = (rks)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, ···, Hn be a finite collection of subgroups of G. G is the internal direct product of H1, H2, ··· Hn, G = H1 x H2 x ··· x Hn, if

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

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

Proof.

Let G = H1 x H2 x ··· x Hn

• Claim: ∀hi ∈ Hi, hj ∈ Hj, i ≠ j, hihj = hihj.

[By assumption, Hj ◁ G ↭ gHjg-1 ⊆ Hj, let g =G=H1xH2x···xHn (e·e···hi···e·e) = hi] hihjhi-1 ∈ Hj and (hihjhi-1)hj-1 ∈ Hjhj-1 ⇒ [Ha = H ↭ a ∈ H, hj-1 ∈ Hj] hihjhi-1hj-1 ∈ Hjhj-1 = Hj and similarly hi(hjhi-1hj-1) ∈ hiHi = Hi.

Therefore, hihjhi-1hj-1 ∈ Hi ∩ Hj.

i ≠ j, say i < j, let h ∈ Hi ∩ Hj, h = e1···ei-1hei+1···ej-1, therefore h ∈ H1 ··· Hi ··· Hj-1 ∩ Hj =[Condition 3. Each subgroup is disjoint from the product of all previous subgroup.] {e}.

hihjhi-1hj-1 ∈ Hi ∩ Hj = {e} ⇒ hihjhi-1hj-1 = e ⇒ hihj = hjhi

• Next, we claim that each member of G can be expressed uniquely in the form h1h2···hn, ∀hi ∈ Hi

Suppose, h1h2···hn = h’1h’2···h’n ⇒ h1h2···hnhn-1 = h’1h’2···h’nhn-1 ⇒ h1h2···hn-1 = h’1h’2···h’nhn-1 ⇒ h1h2···hn-1hn-1-1 = h’1h’2···h’nhn-1hn-1-1 ⇒ [∀hi ∈ Hi, hj ∈ Hj, i ≠ j, hihj = hjhi] h1h2···hn-2 = h’1h’2···hn-1h’n-1-1h’nhn-1 ⇒ e = h’1h1-1h’2h2-1 ··· h’nhn-1 ⇒ hn = h’1h1-1h’2h2-1 ··· h’n ⇒ [Multiplying both sides by h’n-1 and ∀hi ∈ Hi, hj ∈ Hj, i ≠ j, hihj = hjhi] h’n-1hn = h’1h1-1 ··· h’n-1hn-1-1 ∈ H1H2 ··· Hn-1 ∩ Hn = {e} ⇒ h’n-1hn = e ⇒ h’n = hn. Therefore, h1h2···hn-1 = h’1h’2···h’n-1 and repeating the argument, we can obtain hn-1 = h’n-1, and so on, so forth hi = h’i ∀i, 1 ≤ i ≤ n.

• Φ: H1 x H2 x ··· x Hn → H1 ⊕ H2 ⊕ ··· ⊕ Hn, Φ(h1h2···hn) = (h1, h2···, hn) is an isomorphism.
1. Since every element of G has an unique representation as h1h2···hn, Φ is well defined, onto, and one-to-one.
2. Let a = h1h2···hn, b = h’1h’2···h’n, Φ(ab) = Φ(h1h2···hnh’1h’2···h’n) = [∀hi ∈ Hi, hj ∈ Hj, i ≠ j, hihj = hjhi] Φ(h1h’1h2h’2···hnh’n) = (h1h’1, h2h’2,···, hnh’n) = (h1, h2,···, hn)(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, D2n = {e, r, r2, ···, r2n-1, s, sr, ···, sr2n-1}, let H1 = ⟨s, r2⟩ = {e, r2, r4, ···, r2n-2, s, sr2, ···, sr2n-2} so since |D2n/⟨s, r2⟩| = 2 ⇒[Subgroup of index 2 is normal] H1 = ⟨s, r2⟩ ◁ D2n. Futhermore, H1 ≋ Dn.

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

Let H2 = ⟨rn⟩ = {e, rn} = ℤ(D2n) ⇒ [The center of a group is a normal subgroup] H2 ◁ D2n. Futhermore, D2n = H1H2, and H1 ∩ H2 = {e} ⇒ [By the previous theorem] D2n ≋ H1 ⊕ H2 ≋ ℤ2 ⊕ Dn, e.g., D6 ≋ ℤ2 ⊕ D3, D14 ≋ ℤ2 ⊕ D7.

# 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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