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=BaseG
isnsg3.2 +˙=+G
isnsg3.3 -˙=-G
Assertion nsgconj SNrmSGrpGAXBSA+˙B-˙AS

Proof

Step Hyp Ref Expression
1 isnsg3.1 X=BaseG
2 isnsg3.2 +˙=+G
3 isnsg3.3 -˙=-G
4 nsgsubg SNrmSGrpGSSubGrpG
5 4 3ad2ant1 SNrmSGrpGAXBSSSubGrpG
6 subgrcl SSubGrpGGGrp
7 5 6 syl SNrmSGrpGAXBSGGrp
8 simp2 SNrmSGrpGAXBSAX
9 1 subgss SSubGrpGSX
10 5 9 syl SNrmSGrpGAXBSSX
11 simp3 SNrmSGrpGAXBSBS
12 10 11 sseldd SNrmSGrpGAXBSBX
13 1 2 3 grpaddsubass GGrpAXBXAXA+˙B-˙A=A+˙B-˙A
14 7 8 12 8 13 syl13anc SNrmSGrpGAXBSA+˙B-˙A=A+˙B-˙A
15 1 2 3 grpnpcan GGrpBXAXB-˙A+˙A=B
16 7 12 8 15 syl3anc SNrmSGrpGAXBSB-˙A+˙A=B
17 16 11 eqeltrd SNrmSGrpGAXBSB-˙A+˙AS
18 simp1 SNrmSGrpGAXBSSNrmSGrpG
19 1 3 grpsubcl GGrpBXAXB-˙AX
20 7 12 8 19 syl3anc SNrmSGrpGAXBSB-˙AX
21 1 2 nsgbi SNrmSGrpGB-˙AXAXB-˙A+˙ASA+˙B-˙AS
22 18 20 8 21 syl3anc SNrmSGrpGAXBSB-˙A+˙ASA+˙B-˙AS
23 17 22 mpbid SNrmSGrpGAXBSA+˙B-˙AS
24 14 23 eqeltrd SNrmSGrpGAXBSA+˙B-˙AS