Metamath Proof Explorer


Theorem nsgsubg

Description: A normal subgroup is a subgroup. (Contributed by Mario Carneiro, 18-Jan-2015)

Ref Expression
Assertion nsgsubg ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) → 𝑆 ∈ ( SubGrp ‘ 𝐺 ) )

Proof

Step Hyp Ref Expression
1 eqid ( Base ‘ 𝐺 ) = ( Base ‘ 𝐺 )
2 eqid ( +g𝐺 ) = ( +g𝐺 )
3 1 2 isnsg ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) ↔ ( 𝑆 ∈ ( SubGrp ‘ 𝐺 ) ∧ ∀ 𝑥 ∈ ( Base ‘ 𝐺 ) ∀ 𝑦 ∈ ( Base ‘ 𝐺 ) ( ( 𝑥 ( +g𝐺 ) 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 ( +g𝐺 ) 𝑥 ) ∈ 𝑆 ) ) )
4 3 simplbi ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) → 𝑆 ∈ ( SubGrp ‘ 𝐺 ) )