Metamath Proof Explorer


Theorem f1ghm0to0

Description: If a group homomorphism F is injective, it maps the zero of one group (and only the zero) to the zero of the other group. (Contributed by AV, 24-Oct-2019) (Revised by Thierry Arnoux, 13-May-2023)

Ref Expression
Hypotheses f1ghm0to0.a A = Base R
f1ghm0to0.b B = Base S
f1ghm0to0.n N = 0 S
f1ghm0to0.1 0 ˙ = 0 R
Assertion f1ghm0to0 F R GrpHom S F : A 1-1 B X A F X = N X = 0 ˙

Proof

Step Hyp Ref Expression
1 f1ghm0to0.a A = Base R
2 f1ghm0to0.b B = Base S
3 f1ghm0to0.n N = 0 S
4 f1ghm0to0.1 0 ˙ = 0 R
5 4 3 ghmid F R GrpHom S F 0 ˙ = N
6 5 3ad2ant1 F R GrpHom S F : A 1-1 B X A F 0 ˙ = N
7 6 eqeq2d F R GrpHom S F : A 1-1 B X A F X = F 0 ˙ F X = N
8 simp2 F R GrpHom S F : A 1-1 B X A F : A 1-1 B
9 simp3 F R GrpHom S F : A 1-1 B X A X A
10 ghmgrp1 F R GrpHom S R Grp
11 1 4 grpidcl R Grp 0 ˙ A
12 10 11 syl F R GrpHom S 0 ˙ A
13 12 3ad2ant1 F R GrpHom S F : A 1-1 B X A 0 ˙ A
14 f1veqaeq F : A 1-1 B X A 0 ˙ A F X = F 0 ˙ X = 0 ˙
15 8 9 13 14 syl12anc F R GrpHom S F : A 1-1 B X A F X = F 0 ˙ X = 0 ˙
16 7 15 sylbird F R GrpHom S F : A 1-1 B X A F X = N X = 0 ˙
17 fveq2 X = 0 ˙ F X = F 0 ˙
18 17 6 sylan9eqr F R GrpHom S F : A 1-1 B X A X = 0 ˙ F X = N
19 18 ex F R GrpHom S F : A 1-1 B X A X = 0 ˙ F X = N
20 16 19 impbid F R GrpHom S F : A 1-1 B X A F X = N X = 0 ˙