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 ˙ = 0 G
1nsgtrivd.3 ⊢ φ → G ∈ Grp
1nsgtrivd.4 ⊢ φ → NrmSGrp ⁡ G ≈ 1 𝑜
Assertion 1nsgtrivd ⊢ φ → B = 0 ˙

Proof

Step Hyp Ref Expression
1 1nsgtrivd.1 ⊢ B = Base G
2 1nsgtrivd.2 ⊢ 0 ˙ = 0 G
3 1nsgtrivd.3 ⊢ φ → G ∈ Grp
4 1nsgtrivd.4 ⊢ φ → NrmSGrp ⁡ G ≈ 1 𝑜
5 1 nsgid ⊢ G ∈ Grp → B ∈ NrmSGrp ⁡ G
6 3 5 syl ⊢ φ → B ∈ 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 ≈ 1 𝑜 → NrmSGrp ⁡ G = 0 ˙
10 8 4 9 syl2anc ⊢ φ → NrmSGrp ⁡ G = 0 ˙
11 6 10 eleqtrd ⊢ φ → B ∈ 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 ⊢ φ → B = 0 ˙