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