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=BaseR
f1ghm0to0.b B=BaseS
f1ghm0to0.n N=0R
f1ghm0to0.0 0˙=0S
Assertion f1ghm0to0 FRGrpHomSF:A1-1BXAFX=0˙X=N

Proof

Step Hyp Ref Expression
1 f1ghm0to0.a A=BaseR
2 f1ghm0to0.b B=BaseS
3 f1ghm0to0.n N=0R
4 f1ghm0to0.0 0˙=0S
5 3 4 ghmid FRGrpHomSFN=0˙
6 5 3ad2ant1 FRGrpHomSF:A1-1BXAFN=0˙
7 6 eqeq2d FRGrpHomSF:A1-1BXAFX=FNFX=0˙
8 simp2 FRGrpHomSF:A1-1BXAF:A1-1B
9 simp3 FRGrpHomSF:A1-1BXAXA
10 ghmgrp1 FRGrpHomSRGrp
11 1 3 grpidcl RGrpNA
12 10 11 syl FRGrpHomSNA
13 12 3ad2ant1 FRGrpHomSF:A1-1BXANA
14 f1veqaeq F:A1-1BXANAFX=FNX=N
15 8 9 13 14 syl12anc FRGrpHomSF:A1-1BXAFX=FNX=N
16 7 15 sylbird FRGrpHomSF:A1-1BXAFX=0˙X=N
17 fveq2 X=NFX=FN
18 17 6 sylan9eqr FRGrpHomSF:A1-1BXAX=NFX=0˙
19 18 ex FRGrpHomSF:A1-1BXAX=NFX=0˙
20 16 19 impbid FRGrpHomSF:A1-1BXAFX=0˙X=N