Metamath Proof Explorer


Theorem conjsubg

Description: A conjugated subgroup is also a subgroup. (Contributed by Mario Carneiro, 13-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 conjsubg ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → ran ⁡ F ∈ SubGrp ⁡ G

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 1 subgss ⊢ S ∈ SubGrp ⁡ G → S ⊆ X
6 5 adantr ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → S ⊆ X
7 df-ima ⊢ x ∈ X ⟼ A + ˙ x - ˙ A S = ran ⁡ x ∈ X ⟼ A + ˙ x - ˙ A ↾ S
8 resmpt ⊢ S ⊆ X → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S = x ∈ S ⟼ A + ˙ x - ˙ A
9 8 4 eqtr4di ⊢ S ⊆ X → x ∈ X ⟼ A + ˙ x - ˙ A ↾ S = F
10 9 rneqd ⊢ S ⊆ X → ran ⁡ x ∈ X ⟼ A + ˙ x - ˙ A ↾ S = ran ⁡ F
11 7 10 eqtrid ⊢ S ⊆ X → x ∈ X ⟼ A + ˙ x - ˙ A S = ran ⁡ F
12 6 11 syl ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A S = ran ⁡ F
13 subgrcl ⊢ S ∈ SubGrp ⁡ G → G ∈ Grp
14 eqid ⊢ x ∈ X ⟼ A + ˙ x - ˙ A = x ∈ X ⟼ A + ˙ x - ˙ A
15 1 2 3 14 conjghm ⊢ G ∈ Grp ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ∈ G GrpHom G ∧ x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 onto X
16 13 15 sylan ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ∈ G GrpHom G ∧ x ∈ X ⟼ A + ˙ x - ˙ A : X ⟶ 1-1 onto X
17 16 simpld ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A ∈ G GrpHom G
18 simpl ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → S ∈ SubGrp ⁡ G
19 ghmima ⊢ x ∈ X ⟼ A + ˙ x - ˙ A ∈ G GrpHom G ∧ S ∈ SubGrp ⁡ G → x ∈ X ⟼ A + ˙ x - ˙ A S ∈ SubGrp ⁡ G
20 17 18 19 syl2anc ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → x ∈ X ⟼ A + ˙ x - ˙ A S ∈ SubGrp ⁡ G
21 12 20 eqeltrrd ⊢ S ∈ SubGrp ⁡ G ∧ A ∈ X → ran ⁡ F ∈ SubGrp ⁡ G