Metamath Proof Explorer


Theorem opprnsg

Description: Normal subgroups of the opposite ring are the same as the original normal subgroups. (Contributed by Thierry Arnoux, 13-Mar-2025)

Ref Expression
Hypothesis oppreqg.o 𝑂 = ( oppr𝑅 )
Assertion opprnsg ( NrmSGrp ‘ 𝑅 ) = ( NrmSGrp ‘ 𝑂 )

Proof

Step Hyp Ref Expression
1 oppreqg.o 𝑂 = ( oppr𝑅 )
2 1 opprsubg ( SubGrp ‘ 𝑅 ) = ( SubGrp ‘ 𝑂 )
3 2 eleq2i ( 𝑔 ∈ ( SubGrp ‘ 𝑅 ) ↔ 𝑔 ∈ ( SubGrp ‘ 𝑂 ) )
4 3 anbi1i ( ( 𝑔 ∈ ( SubGrp ‘ 𝑅 ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝑅 ) ∀ 𝑦 ∈ ( Base ‘ 𝑅 ) ( ( 𝑥 ( +g𝑅 ) 𝑦 ) ∈ 𝑔 → ( 𝑦 ( +g𝑅 ) 𝑥 ) ∈ 𝑔 ) ) ↔ ( 𝑔 ∈ ( SubGrp ‘ 𝑂 ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝑅 ) ∀ 𝑦 ∈ ( Base ‘ 𝑅 ) ( ( 𝑥 ( +g𝑅 ) 𝑦 ) ∈ 𝑔 → ( 𝑦 ( +g𝑅 ) 𝑥 ) ∈ 𝑔 ) ) )
5 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
6 eqid ( +g𝑅 ) = ( +g𝑅 )
7 5 6 isnsg2 ( 𝑔 ∈ ( NrmSGrp ‘ 𝑅 ) ↔ ( 𝑔 ∈ ( SubGrp ‘ 𝑅 ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝑅 ) ∀ 𝑦 ∈ ( Base ‘ 𝑅 ) ( ( 𝑥 ( +g𝑅 ) 𝑦 ) ∈ 𝑔 → ( 𝑦 ( +g𝑅 ) 𝑥 ) ∈ 𝑔 ) ) )
8 1 5 opprbas ( Base ‘ 𝑅 ) = ( Base ‘ 𝑂 )
9 1 6 oppradd ( +g𝑅 ) = ( +g𝑂 )
10 8 9 isnsg2 ( 𝑔 ∈ ( NrmSGrp ‘ 𝑂 ) ↔ ( 𝑔 ∈ ( SubGrp ‘ 𝑂 ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝑅 ) ∀ 𝑦 ∈ ( Base ‘ 𝑅 ) ( ( 𝑥 ( +g𝑅 ) 𝑦 ) ∈ 𝑔 → ( 𝑦 ( +g𝑅 ) 𝑥 ) ∈ 𝑔 ) ) )
11 4 7 10 3bitr4i ( 𝑔 ∈ ( NrmSGrp ‘ 𝑅 ) ↔ 𝑔 ∈ ( NrmSGrp ‘ 𝑂 ) )
12 11 eqriv ( NrmSGrp ‘ 𝑅 ) = ( NrmSGrp ‘ 𝑂 )