Do or do not. There is no try, Yoda.
Let G be an arbitrary group, let g_{1}, g_{2} ∈ G. Then, g_{1} is said to be a conjugate of g_{2}, if there exists an a ∈ G such that g_{1} = ag_{2}a^{-1}.
Proposition. Let G be an Abelian group and g_{1}, g_{2} ∈G, then g_{1} is a conjugate of g_{2} if and only if g_{1} = g_{2}.
Proof. (Trivial)
⇒ g_{1} = ag_{2}a^{-1} =[Associative] (ag_{2})a^{-1} = [G Abelian group] (g_{2}a)a^{-1} = [G associate and identity] g_{2}.
⇐ g_{1} = g_{2} then g_{1} = eg_{2}e^{-1} (Recall e^{-1} = e)
Proposition. Let G be a group and let g_{1}, g_{2} ∈ G. If g_{1} is a conjugate of g_{2}, then the orders of g_{1} and g_{2} are the same.
Proof. Conjugate Elements in a group, Mathonline
Definition. Let G be a group and let H_{1} and H_{2} be subgroups of G. Then, H_{1} is said to be a conjugate of H_{2} if there exists an a ∈ G such that H_{1} = aH_{2}a^{-1}.
Proposition. Let G be an Abelian group and H_{1} ≤ G, H_{2}≤ G, then H_{1} is a conjugate of H_{2} if and only if H_{1} = H_{2}.
Proof.
⇒ Suppose H_{1} is a conjugate of H_{2} ⇒ ∃a ∈ G: H_{1} = aH_{2}a^{-1} ↭ H_{1}a = aH_{2} ↭ [G is Abelian] aH_{1} = aH_{2} ↭ [Multiplying both sides by a^{-1}] H_{1} = H_{2}
⇐ Suppose H_{1} = H_{2}. Let e ∈ G be the identity element on G. Then, H_{1} = 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^{-1}Hg = {g^{-1}hg | h ∈ H} ≤ G.
🔑 g^{-1}Hg is not always equal to H. Let G = S_{3}, H= {(1), (12)}, g = (13), g^{-1}Hg = {(1), (23)} ≠ H_{1} 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^{-1}Hg = {g^{-1}hg | h ∈ H} ≤ G ⇒ x = g^{-1}hg, y = g^{-1}h’g.
xy^{-1} = g^{-1}hg(g^{-1}h’g)^{-1} = [Socks and shoes principle Λ (g^{-1})^{-1} = g] g^{-1}hgg^{-1}h’^{-1}g = [Associative] g^{-1}h(gg^{-1})h’^{-1}g = g^{-1}hh’^{-1}g ∈ g^{-1}Hg ∎
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 = {h_{1}h_{2}: h_{1} ∈ H, h_{2} ∈ 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^{-1}h = 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 = S_{3}, 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)} ≤ S_{3}, HK = {1, (123), (132)}{1, (12)} =[(123)(12) = (13), (132)(12) = (23)] {1, (12), (123), (13), (132), (23)} = S_{3}. KH = {1, (12)} {1, (123), (132)} = {1, (123), (132), (12), (23), (13)} = S_{3}.
Let H = ⟨s⟩, K = ⟨rs⟩ ≤ D_{4}, HK = {1, s}{1, rs} =[s·rs = (sr)s = (r^{-1}s)s = r^{3}s^{2} = r^{3}] {1, rs, s, r^{3}}. 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 (r^{3} ∉ KH), r^{3} ∈ HK does not have an inverse (r ∉ HK).
The product of subsets may commute, e.g., H = {i, j, k}, K = {-i, -j, -k} ⊆ Q_{8}. HK = {i, j, k}{-i, -j, -k} = {1, -k, j, k, 1, -i, -j, i, 1} = {1, i, -i, j, -j, k, -k} = Q_{8} \ {-1}. KH = {-i, -j, -k}{i, j, k} = {1, -k, j, k, 1, -i, -j, -i, 1} = Q_{8} \ {-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^{-1}H^{-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^{-1}h^{-1}: h ∈ H, k ∈ K} = K^{-1}H^{-1}
⇐) Suppose that HK = KH. Is HK ≤ G? Let x, y ∈ HK, ∃h_{1}, h_{2} ∈ H, k_{1}, k_{2} ∈ K such that x = h_{1}k_{1} and y = h_{2}k_{2}.
xy = h_{1}k_{1}h_{2}k_{2} =[Associative] h_{1}(k_{1}h_{2})k_{2} = [HK = KH, ∃k’∈ K and h’ ∈ H] h_{1}(h’k’)k_{2} ∈ HK because h_{1}h’ ∈ H and k’k_{2} ∈ 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^{-1}h^{-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.
H ≤ G and K ≤ G ⇒ H∩K ≤ G ⇒ |H∩K| ≥ 1 because e ∈ H∩K, so we are never dividing by zero.
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).
∀t ∈ H∩K, ∀h ∈ H, k ∈ K, hk = ht(t^{-1}k) ∈ HK because ht ∈ H, t^{-1}k ∈ 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’^{-1}hk = k’ ⇒ h’^{-1}h = k’k^{-1} = t ∈ H∩K 🚀
k’k^{-1} = t ⇒ k’ = tk ⇒ k = t^{-1}k’ 🚀. h’^{-1}h = t ⇒ h = h’t 🚀
Therefore, h’k’ = hk =[🚀] (h’t)(t^{-1}k’) 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|.