Metamath Proof Explorer


Theorem subg0

Description: A subgroup of a group must have the same identity as the group. (Contributed by Stefan O'Rear, 27-Nov-2014) (Revised by Mario Carneiro, 30-Apr-2015)

Ref Expression
Hypotheses subg0.h ⊢ H = G ↾ 𝑠 S
subg0.i ⊢ 0 ˙ = 0 G
Assertion subg0 ⊢ S ∈ SubGrp ⁡ G → 0 ˙ = 0 H

Proof

Step Hyp Ref Expression
1 subg0.h ⊢ H = G ↾ 𝑠 S
2 subg0.i ⊢ 0 ˙ = 0 G
3 eqid ⊢ + G = + G
4 1 3 ressplusg ⊢ S ∈ SubGrp ⁡ G → + G = + H
5 4 oveqd ⊢ S ∈ SubGrp ⁡ G → 0 H + G 0 H = 0 H + H 0 H
6 1 subggrp ⊢ S ∈ SubGrp ⁡ G → H ∈ Grp
7 eqid ⊢ Base H = Base H
8 eqid ⊢ 0 H = 0 H
9 7 8 grpidcl ⊢ H ∈ Grp → 0 H ∈ Base H
10 6 9 syl ⊢ S ∈ SubGrp ⁡ G → 0 H ∈ Base H
11 eqid ⊢ + H = + H
12 7 11 8 grplid ⊢ H ∈ Grp ∧ 0 H ∈ Base H → 0 H + H 0 H = 0 H
13 6 10 12 syl2anc ⊢ S ∈ SubGrp ⁡ G → 0 H + H 0 H = 0 H
14 5 13 eqtrd ⊢ S ∈ SubGrp ⁡ G → 0 H + G 0 H = 0 H
15 subgrcl ⊢ S ∈ SubGrp ⁡ G → G ∈ Grp
16 eqid ⊢ Base G = Base G
17 16 subgss ⊢ S ∈ SubGrp ⁡ G → S ⊆ Base G
18 1 subgbas ⊢ S ∈ SubGrp ⁡ G → S = Base H
19 10 18 eleqtrrd ⊢ S ∈ SubGrp ⁡ G → 0 H ∈ S
20 17 19 sseldd ⊢ S ∈ SubGrp ⁡ G → 0 H ∈ Base G
21 16 3 2 grpid ⊢ G ∈ Grp ∧ 0 H ∈ Base G → 0 H + G 0 H = 0 H ↔ 0 ˙ = 0 H
22 15 20 21 syl2anc ⊢ S ∈ SubGrp ⁡ G → 0 H + G 0 H = 0 H ↔ 0 ˙ = 0 H
23 14 22 mpbid ⊢ S ∈ SubGrp ⁡ G → 0 ˙ = 0 H