    # Conjugate of subgroups. Product of group subsets.

Do or do not. There is no try, Yoda. Let G be an arbitrary group, let g1, g2 ∈ G. Then, g1 is said to be a conjugate of g2, if there exists an a ∈ G such that g1 = ag2a-1.

Proposition. Let G be an Abelian group and g1, g2 ∈G, then g1 is a conjugate of g2 if and only if g1 = g2.

Proof. (Trivial)

⇒ g1 = ag2a-1 =[Associative] (ag2)a-1 = [G Abelian group] (g2a)a-1 = [G associate and identity] g2.

⇐ g1 = g2 then g1 = eg2e-1 (Recall e-1 = e)

Proposition. Let G be a group and let g1, g2 ∈ G. If g1 is a conjugate of g2, then the orders of g1 and g2 are the same.

Definition. Let G be a group and let H1 and H2 be subgroups of G. Then, H1 is said to be a conjugate of H2 if there exists an a ∈ G such that H1 = aH2a-1.

Proposition. Let G be an Abelian group and H1 ≤ G, H2≤ G, then H1 is a conjugate of H2 if and only if H1 = H2.

Proof.

⇒ Suppose H1 is a conjugate of H2 ⇒ ∃a ∈ G: H1 = aH2a-1 ↭ H1a = aH2 ↭ [G is Abelian] aH1 = aH2 ↭ [Multiplying both sides by a-1] H1 = H2

⇐ Suppose H1 = H2. Let e ∈ G be the identity element on G. Then, H1 = eHe-1

Let G be a group, H ≤ G be a subgroup of G, and g an arbitrary element of G. Then, the conjugate of H by g is a subgroup of G, that is, g-1Hg = {g-1hg | h ∈ H} ≤ G.

🔑 g-1Hg is not always equal to H. Let G = S3, H= {(1), (12)}, g = (13), g-1Hg = {(1), (23)} ≠ H1 because (13)-1(12)(13) = (13)(12)(13) = (23), (13)-1(1)(13) = (13)(1)(13) = (13)(13) = (1).

Proof. (Two-Step Subgroup Test)

Suppose x, y ∈ g-1Hg = {g-1hg | h ∈ H} ≤ G ⇒ x = g-1hg, y = g-1h’g.

xy-1 = g-1hg(g-1h’g)-1 = [Socks and shoes principle Λ (g-1)-1 = g] g-1hgg-1h’-1g = [Associative] g-1h(gg-1)h’-1g = g-1hh’-1g ∈ g-1Hg ∎

# Product of group subsets

Definition. One can define a product of groups subsets in a natural way. If S and T are subsets of a group G, S, T ⊆ G, then their product is the subset of G defined by ST = {st: s∈ S and t ∈ T}.

Definition. Let G be a group, and let H and K be two subgroups of G. The Product of H and K is HK = {hk: h ∈ H and k ∈ K}

H-1 = {h-1: h ∈ H}
HH = {h1h2: h1 ∈ H, h2 ∈ H}

Proposition. If H is a subgroup of G, H ≤ G, then H-1 = H.

Proof.

H ⊆ H-1. Since H ≤ G, ∀h ∈ H, ∃h-1∈ H such that h·h-1 = h-1h = e and h = (h-1)-1 ∈ H-1.

H-1 ⊆ H. ∀h-1 ∈ H-1 where h ∈ H. Since H ≤ G and h ∈ H, then h-1 ∈ H∎

The converse is not true, e.g., H = {i, -i} is a subset of a group G = {1, -1, i, -i}. H-1 = {i-1, -i-1} = {-i, i} = H, but H is not a subgroup of G.

There are cases where G is a group, H and K are two subgroups of G, and yet HK is not a subgroup of G.

Example.

• Let G = S3, H = ⟨(1, 2)⟩, and K = ⟨(2, 3)⟩. Notice that H = {(1, 2), id} and K = {(2, 3), id}. HK = {id, (1, 2, 3), (1, 2), (2, 3)}. |HK| = 4 is not a divisor of 6 (=|G|), so HK is not a subgroup of G by Lagrange’s Theorem (If H ≤ G ⇒ |H| | |G|).

• Let H = ⟨(123)⟩ = {1, (123), (132)}, K = ⟨(12)⟩ = {1, (12)} ≤ S3, HK = {1, (123), (132)}{1, (12)} =[(123)(12) = (13), (132)(12) = (23)] {1, (12), (123), (13), (132), (23)} = S3. KH = {1, (12)} {1, (123), (132)} = {1, (123), (132), (12), (23), (13)} = S3.

• Let H = ⟨s⟩, K = ⟨rs⟩ ≤ D4, HK = {1, s}{1, rs} =[s·rs = (sr)s = (r-1s)s = r3s2 = r3] {1, rs, s, r3}. KH = {1, rs}{1, s} = {1, s, rs, r}. Notice that HK ≠ KH. Futhermore, they are not subgroups, e.g., r ∈ KH does not have an inverse (r3 ∉ KH), r3 ∈ HK does not have an inverse (r ∉ HK).

• The product of subsets may commute, e.g., H = {i, j, k}, K = {-i, -j, -k} ⊆ Q8. HK = {i, j, k}{-i, -j, -k} = {1, -k, j, k, 1, -i, -j, i, 1} = {1, i, -i, j, -j, k, -k} = Q8 \ {-1}. KH = {-i, -j, -k}{i, j, k} = {1, -k, j, k, 1, -i, -j, -i, 1} = Q8 \ {-1}.

Proposition. Let G be a group and let H and K be two subgroups of G. Then, HK is a subgroup of G (HK ≤ G) if and only if HK = KH.

Proof.

⇒) Suppose that HK is a subgroup.

(HK)-1 = [HK ≤ G, H ≤ G ⇒ H-1 = H.] HK ⇒🔑 K-1H-1 = HK ⇒ [H, K ≤ G, H ≤ G ⇒ H-1 = H.] KH = HK.

🔑 HK = {hk: h ∈ H, k ∈ K}, (HK)-1 = {(hk)-1: h ∈ H, k ∈ K} = [The socks and shoes principle] { k-1h-1: h ∈ H, k ∈ K} = K-1H-1

⇐) Suppose that HK = KH. Is HK ≤ G? Let x, y ∈ HK, ∃h1, h2 ∈ H, k1, k2 ∈ K such that x = h1k1 and y = h2k2.

xy = h1k1h2k2 =[Associative] h1(k1h2)k2 = [HK = KH, ∃k’∈ K and h’ ∈ H] h1(h’k’)k2 ∈ HK because h1h’ ∈ H and k’k2 ∈ K (H, K ≤ G, closure).

Associativity is inherited from G. e = e·e ∈ HK (identity). ∀x ∈ HK, ∃h ∈ H, k ∈ K, x = hk. x-1 = (hk)-1 = [The Socks and Shoes Principle] k-1h-1 = [k-1 ∈ K, h-1 ∈ H, KH = HK, ∃k’∈ K and h’ ∈ H] h’k’ ∈ HK (inverses)∎

The Counting Formula of Products of Subgroups. Let G be a group and let H and K be two finite subgroups of G, i.e., H, K ≤ G, |H|, |K| < ∞. Then, $|HK|=\frac{|H||K|}{|H∩K|}$

Proof:

It is important to take two facts into consideration.

1. H ≤ G and K ≤ G ⇒ H∩K ≤ G ⇒ |H∩K| ≥ 1 because e ∈ H∩K, so we are never dividing by zero.

2. The set HK has clearly |H||K| symbols. However, not all symbols need to represent distinct (group) elements, that is, we may have hk = h’k’ although h ≠ h’ and k ≠ k’ (hk and h’k’ share the same “group element” in HK).

3. ∀t ∈ H∩K, ∀h ∈ H, k ∈ K, hk = ht(t-1k) ∈ HK because ht ∈ H, t-1k ∈ K, so each group element in HK can be written or represented by at least |H∩K| different ways as an element from H times an element from K or, alternatively, hk is duplicated in the product HK at least |H ∩ K| times (*)

hk = h’k’ ⇒ [Multiplying by h’-1 in both sides of the equation] h’-1hk = k’ ⇒ h’-1h = k’k-1 = t ∈ H∩K 🚀

k’k-1 = t ⇒ k’ = tk ⇒ k = t-1k’ 🚀. h’-1h = t ⇒ h = h’t 🚀

Therefore, h’k’ = hk =[🚀] (h’t)(t-1k’) for some t ∈ H∩K, so every element in HK can be represented or written in exactly |H∩K| different ways or products (or, alternatively, all duplications were already accounted for in (*)), and so, the equality $|HK|=\frac{|H||K|}{|H∩K|}$ holds.

In particular, |H∩K| divides |H||K|.

# 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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