Metamath Proof Explorer


Theorem gicrcl

Description: Isomorphism implies the right side is a group. (Contributed by Mario Carneiro, 6-May-2015)

Ref Expression
Assertion gicrcl ( 𝑅 ≃𝑔 𝑆 → 𝑆 ∈ Grp )

Proof

Step Hyp Ref Expression
1 brgic ⊢ ( 𝑅 ≃𝑔 𝑆 ↔ ( 𝑅 GrpIso 𝑆 ) ≠ ∅ )
2 n0 ⊢ ( ( 𝑅 GrpIso 𝑆 ) ≠ ∅ ↔ ∃ 𝑓 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) )
3 1 2 bitri ⊢ ( 𝑅 ≃𝑔 𝑆 ↔ ∃ 𝑓 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) )
4 gimghm ⊢ ( 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) → 𝑓 ∈ ( 𝑅 GrpHom 𝑆 ) )
5 ghmgrp2 ⊢ ( 𝑓 ∈ ( 𝑅 GrpHom 𝑆 ) → 𝑆 ∈ Grp )
6 4 5 syl ⊢ ( 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) → 𝑆 ∈ Grp )
7 6 exlimiv ⊢ ( ∃ 𝑓 𝑓 ∈ ( 𝑅 GrpIso 𝑆 ) → 𝑆 ∈ Grp )
8 3 7 sylbi ⊢ ( 𝑅 ≃𝑔 𝑆 → 𝑆 ∈ Grp )