Metamath Proof Explorer


Theorem nsgconj

Description: The conjugation of an element of a normal subgroup is in the subgroup. (Contributed by Mario Carneiro, 4-Feb-2015)

Ref Expression
Hypotheses isnsg3.1 ⊢ X = Base G
isnsg3.2 ⊢ + ˙ = + G
isnsg3.3 ⊢ - ˙ = - G
Assertion nsgconj ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → A + ˙ B - ˙ A ∈ S

Proof

Step Hyp Ref Expression
1 isnsg3.1 ⊢ X = Base G
2 isnsg3.2 ⊢ + ˙ = + G
3 isnsg3.3 ⊢ - ˙ = - G
4 nsgsubg ⊢ S ∈ NrmSGrp ⁡ G → S ∈ SubGrp ⁡ G
5 4 3ad2ant1 ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → S ∈ SubGrp ⁡ G
6 subgrcl ⊢ S ∈ SubGrp ⁡ G → G ∈ Grp
7 5 6 syl ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → G ∈ Grp
8 simp2 ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → A ∈ X
9 1 subgss ⊢ S ∈ SubGrp ⁡ G → S ⊆ X
10 5 9 syl ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → S ⊆ X
11 simp3 ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → B ∈ S
12 10 11 sseldd ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → B ∈ X
13 1 2 3 grpaddsubass ⊢ G ∈ Grp ∧ A ∈ X ∧ B ∈ X ∧ A ∈ X → A + ˙ B - ˙ A = A + ˙ B - ˙ A
14 7 8 12 8 13 syl13anc ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → A + ˙ B - ˙ A = A + ˙ B - ˙ A
15 1 2 3 grpnpcan ⊢ G ∈ Grp ∧ B ∈ X ∧ A ∈ X → B - ˙ A + ˙ A = B
16 7 12 8 15 syl3anc ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → B - ˙ A + ˙ A = B
17 16 11 eqeltrd ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → B - ˙ A + ˙ A ∈ S
18 simp1 ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → S ∈ NrmSGrp ⁡ G
19 1 3 grpsubcl ⊢ G ∈ Grp ∧ B ∈ X ∧ A ∈ X → B - ˙ A ∈ X
20 7 12 8 19 syl3anc ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → B - ˙ A ∈ X
21 1 2 nsgbi ⊢ S ∈ NrmSGrp ⁡ G ∧ B - ˙ A ∈ X ∧ A ∈ X → B - ˙ A + ˙ A ∈ S ↔ A + ˙ B - ˙ A ∈ S
22 18 20 8 21 syl3anc ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → B - ˙ A + ˙ A ∈ S ↔ A + ˙ B - ˙ A ∈ S
23 17 22 mpbid ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → A + ˙ B - ˙ A ∈ S
24 14 23 eqeltrd ⊢ S ∈ NrmSGrp ⁡ G ∧ A ∈ X ∧ B ∈ S → A + ˙ B - ˙ A ∈ S