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Metamath Proof Explorer

Theorem subrgsubg

Description: A subring is a subgroup. (Contributed by Mario Carneiro, 3-Dec-2014)

Ref Expression
Assertion subrgsubg ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝐴 ∈ ( SubGrp ‘ 𝑅 ) )

Proof

Step Hyp Ref Expression
1 subrgrcl ⊢ ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝑅 ∈ Ring )
2 ringgrp ⊢ ( 𝑅 ∈ Ring → 𝑅 ∈ Grp )
3 1 2 syl ⊢ ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝑅 ∈ Grp )
4 eqid ⊢ ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
5 4 subrgss ⊢ ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → 𝐴 ⊆ ( Base ‘ 𝑅 ) )
6 eqid ⊢ ( 𝑅 ↾s 𝐴 ) = ( 𝑅 ↾s 𝐴 )
7 6 subrgring ⊢ ( 𝐴 ∈ ( SubRing ‘ 𝑅 ) → ( 𝑅 ↾s 𝐴 ) ∈ Ring )
8 ringgrp ⊢ ( ( 𝑅 ↾s 𝐴 ) ∈ Ring → ( 𝑅 ↾s 𝐴 ) ∈ Grp )
9 7 8 syl ⊢ ( 𝐴 ∈ ( SubRing ‘ 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 ∈ ( SubRing ‘ 𝑅 ) → 𝐴 ∈ ( SubGrp ‘ 𝑅 ) )