Metamath Proof Explorer


Theorem cnmsubglem

Description: Lemma for rpmsubg and friends. (Contributed by Mario Carneiro, 21-Jun-2015)

Ref Expression
Hypotheses cnmgpabl.m M = mulGrp fld 𝑠 0
cnmsubglem.1 x A x
cnmsubglem.2 x A x 0
cnmsubglem.3 x A y A x y A
cnmsubglem.4 1 A
cnmsubglem.5 x A 1 x A
Assertion cnmsubglem A SubGrp M

Proof

Step Hyp Ref Expression
1 cnmgpabl.m M = mulGrp fld 𝑠 0
2 cnmsubglem.1 x A x
3 cnmsubglem.2 x A x 0
4 cnmsubglem.3 x A y A x y A
5 cnmsubglem.4 1 A
6 cnmsubglem.5 x A 1 x A
7 eldifsn x 0 x x 0
8 2 3 7 sylanbrc x A x 0
9 8 ssriv A 0
10 5 ne0ii A
11 4 ralrimiva x A y A x y A
12 cnfldinv x x 0 inv r fld x = 1 x
13 2 3 12 syl2anc x A inv r fld x = 1 x
14 13 6 eqeltrd x A inv r fld x A
15 11 14 jca x A y A x y A inv r fld x A
16 15 rgen x A y A x y A inv r fld x A
17 1 cnmgpabl M Abel
18 ablgrp M Abel M Grp
19 difss 0
20 eqid mulGrp fld = mulGrp fld
21 cnfldbas = Base fld
22 20 21 mgpbas = Base mulGrp fld
23 1 22 ressbas2 0 0 = Base M
24 19 23 ax-mp 0 = Base M
25 cnex V
26 difexg V 0 V
27 cnfldmul × = fld
28 20 27 mgpplusg × = + mulGrp fld
29 1 28 ressplusg 0 V × = + M
30 25 26 29 mp2b × = + M
31 cnfld0 0 = 0 fld
32 cndrng fld DivRing
33 21 31 32 drngui 0 = Unit fld
34 eqid inv r fld = inv r fld
35 33 1 34 invrfval inv r fld = inv g M
36 24 30 35 issubg2 M Grp A SubGrp M A 0 A x A y A x y A inv r fld x A
37 17 18 36 mp2b A SubGrp M A 0 A x A y A x y A inv r fld x A
38 9 10 16 37 mpbir3an A SubGrp M