Metamath Proof Explorer


Theorem subrngsubg

Description: A subring is a subgroup. (Contributed by AV, 14-Feb-2025)

Ref Expression
Assertion subrngsubg ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝐴 ∈ ( SubGrp ‘ 𝑅 ) )

Proof

Step Hyp Ref Expression
1 subrngrcl ⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝑅 ∈ Rng )
2 rnggrp ⊢ ( 𝑅 ∈ Rng → 𝑅 ∈ Grp )
3 1 2 syl ⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝑅 ∈ Grp )
4 eqid ⊢ ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
5 4 subrngss ⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝐴 ⊆ ( Base ‘ 𝑅 ) )
6 eqid ⊢ ( 𝑅 ↾s 𝐴 ) = ( 𝑅 ↾s 𝐴 )
7 6 subrngrng ⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → ( 𝑅 ↾s 𝐴 ) ∈ Rng )
8 rnggrp ⊢ ( ( 𝑅 ↾s 𝐴 ) ∈ Rng → ( 𝑅 ↾s 𝐴 ) ∈ Grp )
9 7 8 syl ⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → ( 𝑅 ↾s 𝐴 ) ∈ Grp )
10 4 issubg ⊢ ( 𝐴 ∈ ( SubGrp ‘ 𝑅 ) ↔ ( 𝑅 ∈ Grp ∧ 𝐴 ⊆ ( Base ‘ 𝑅 ) ∧ ( 𝑅 ↾s 𝐴 ) ∈ Grp ) )
11 3 5 9 10 syl3anbrc ⊢ ( 𝐴 ∈ ( SubRng ‘ 𝑅 ) → 𝐴 ∈ ( SubGrp ‘ 𝑅 ) )