Metamath Proof Explorer


Theorem conjsubgen

Description: A conjugated subgroup is equinumerous to the original subgroup. (Contributed by Mario Carneiro, 18-Jan-2015)

Ref Expression
Hypotheses conjghm.x ⊢ X = Base G
conjghm.p ⊢ + ˙ = + G
conjghm.m ⊢ - ˙ = - G
conjsubg.f ⊢ F = x ∈ S ⟼ A + ˙ x - ˙ A
Assertion conjsubgen ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → S ≈ ran ⁡ F

Proof

Step Hyp Ref Expression
1 conjghm.x ⊢ X = Base G
2 conjghm.p ⊢ + ˙ = + G
3 conjghm.m ⊢ - ˙ = - G
4 conjsubg.f ⊢ F = x ∈ S ⟼ A + ˙ x - ˙ A
5 subgrcl ⊢ S ∈ SubGrp ⁡ G → G ∈ Grp
6 eqid ⊢ x ∈ X ⟼ A + ˙ x - ˙ A = x ∈ X ⟼ A + ˙ x - ˙ A
7 1 2 3 6 conjghm ⊢ G ∈ Grp ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ∈ G GrpHom G ∧ x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 onto X
8 5 7 sylan ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ∈ G GrpHom G ∧ x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 onto X
9 f1of1 ⊢ x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 onto X → x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 X
10 8 9 simpl2im ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 X
11 1 subgss ⊢ S ∈ SubGrp ⁡ G → S ⊆ X
12 11 adantr ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → S ⊆ X
13 f1ssres ⊢ x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 X ∧ S ⊆ X → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S : S ⟶ 1-1 X
14 10 12 13 syl2anc ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S : S ⟶ 1-1 X
15 12 resmptd ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S = x ∈ S ⟼ A + ˙ x - ˙ A
16 15 4 eqtr4di ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S = F
17 f1eq1 ⊢ x ∈ X ⟼ A + ˙ x - ˙ A ↾ S = F → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S : S ⟶ 1-1 X ↔ F : S ⟶ 1-1 X
18 16 17 syl ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S : S ⟶ 1-1 X ↔ F : S ⟶ 1-1 X
19 14 18 mpbid ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → F : S ⟶ 1-1 X
20 f1f1orn ⊢ F : S ⟶ 1-1 X → F : S ⟶ 1-1 onto ran ⁡ F
21 19 20 syl ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → F : S ⟶ 1-1 onto ran ⁡ F
22 f1oeng ⊢ S ∈ SubGrp ⁡ G ∧ F : S ⟶ 1-1 onto ran ⁡ F → S ≈ ran ⁡ F
23 21 22 syldan ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → S ≈ ran ⁡ F