Metamath Proof Explorer


Theorem 1nsgtrivd

Description: A group with exactly one normal subgroup is trivial. (Contributed by Rohan Ridenour, 3-Aug-2023)

Ref Expression
Hypotheses 1nsgtrivd.1 𝐵 = ( Base ‘ 𝐺 )
1nsgtrivd.2 0 = ( 0g𝐺 )
1nsgtrivd.3 ( 𝜑𝐺 ∈ Grp )
1nsgtrivd.4 ( 𝜑 → ( NrmSGrp ‘ 𝐺 ) ≈ 1o )
Assertion 1nsgtrivd ( 𝜑𝐵 = { 0 } )

Proof

Step Hyp Ref Expression
1 1nsgtrivd.1 𝐵 = ( Base ‘ 𝐺 )
2 1nsgtrivd.2 0 = ( 0g𝐺 )
3 1nsgtrivd.3 ( 𝜑𝐺 ∈ Grp )
4 1nsgtrivd.4 ( 𝜑 → ( NrmSGrp ‘ 𝐺 ) ≈ 1o )
5 1 nsgid ( 𝐺 ∈ Grp → 𝐵 ∈ ( NrmSGrp ‘ 𝐺 ) )
6 3 5 syl ( 𝜑𝐵 ∈ ( NrmSGrp ‘ 𝐺 ) )
7 2 0nsg ( 𝐺 ∈ Grp → { 0 } ∈ ( NrmSGrp ‘ 𝐺 ) )
8 3 7 syl ( 𝜑 → { 0 } ∈ ( NrmSGrp ‘ 𝐺 ) )
9 en1eqsn ( ( { 0 } ∈ ( NrmSGrp ‘ 𝐺 ) ∧ ( NrmSGrp ‘ 𝐺 ) ≈ 1o ) → ( NrmSGrp ‘ 𝐺 ) = { { 0 } } )
10 8 4 9 syl2anc ( 𝜑 → ( NrmSGrp ‘ 𝐺 ) = { { 0 } } )
11 6 10 eleqtrd ( 𝜑𝐵 ∈ { { 0 } } )
12 snex { 0 } ∈ V
13 elsn2g ( { 0 } ∈ V → ( 𝐵 ∈ { { 0 } } ↔ 𝐵 = { 0 } ) )
14 12 13 mp1i ( 𝜑 → ( 𝐵 ∈ { { 0 } } ↔ 𝐵 = { 0 } ) )
15 11 14 mpbid ( 𝜑𝐵 = { 0 } )