Metamath Proof Explorer


Theorem subgabl

Description: A subgroup of an abelian group is also abelian. (Contributed by Mario Carneiro, 3-Dec-2014)

Ref Expression
Hypothesis subgabl.h ⊢ H = G ↾ 𝑠 S
Assertion subgabl ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G → H ∈ Abel

Proof

Step Hyp Ref Expression
1 subgabl.h ⊢ H = G ↾ 𝑠 S
2 1 subgbas ⊢ S ∈ SubGrp ⁡ G → S = Base H
3 2 adantl ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G → S = Base H
4 eqid ⊢ + G = + G
5 1 4 ressplusg ⊢ S ∈ SubGrp ⁡ G → + G = + H
6 5 adantl ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G → + G = + H
7 1 subggrp ⊢ S ∈ SubGrp ⁡ G → H ∈ Grp
8 7 adantl ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G → H ∈ Grp
9 simp1l ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → G ∈ Abel
10 simp1r ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → S ∈ SubGrp ⁡ G
11 eqid ⊢ Base G = Base G
12 11 subgss ⊢ S ∈ SubGrp ⁡ G → S ⊆ Base G
13 10 12 syl ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → S ⊆ Base G
14 simp2 ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → x ∈ S
15 13 14 sseldd ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → x ∈ Base G
16 simp3 ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → y ∈ S
17 13 16 sseldd ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → y ∈ Base G
18 11 4 ablcom ⊢ G ∈ Abel ∧ x ∈ Base G ∧ y ∈ Base G → x + G y = y + G x
19 9 15 17 18 syl3anc ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G ∧ x ∈ S ∧ y ∈ S → x + G y = y + G x
20 3 6 8 19 isabld ⊢ G ∈ Abel ∧ S ∈ SubGrp ⁡ G → H ∈ Abel