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.

Proof. Conjugate Elements in a group, Mathonline

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|.

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