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