Metamath Proof Explorer


Theorem primrootsunit

Description: Primitive roots have left inverses. (Contributed by metakunt, 25-Apr-2025)

Ref Expression
Hypotheses primrootsunit.1 ( 𝜑𝑅 ∈ CMnd )
primrootsunit.2 ( 𝜑𝐾 ∈ ℕ )
primrootsunit.3 𝑈 = { 𝑎 ∈ ( Base ‘ 𝑅 ) ∣ ∃ 𝑖 ∈ ( Base ‘ 𝑅 ) ( 𝑖 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) }
Assertion primrootsunit ( 𝜑 → ( ( 𝑅 PrimRoots 𝐾 ) = ( ( 𝑅s 𝑈 ) PrimRoots 𝐾 ) ∧ ( 𝑅s 𝑈 ) ∈ Abel ) )

Proof

Step Hyp Ref Expression
1 primrootsunit.1 ( 𝜑𝑅 ∈ CMnd )
2 primrootsunit.2 ( 𝜑𝐾 ∈ ℕ )
3 primrootsunit.3 𝑈 = { 𝑎 ∈ ( Base ‘ 𝑅 ) ∣ ∃ 𝑖 ∈ ( Base ‘ 𝑅 ) ( 𝑖 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) }
4 nfv 𝑗 ( 𝑖 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 )
5 nfv 𝑖 ( 𝑗 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 )
6 oveq1 ( 𝑖 = 𝑗 → ( 𝑖 ( +g𝑅 ) 𝑎 ) = ( 𝑗 ( +g𝑅 ) 𝑎 ) )
7 6 eqeq1d ( 𝑖 = 𝑗 → ( ( 𝑖 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) ↔ ( 𝑗 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) ) )
8 4 5 7 cbvrexw ( ∃ 𝑖 ∈ ( Base ‘ 𝑅 ) ( 𝑖 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) ↔ ∃ 𝑗 ∈ ( Base ‘ 𝑅 ) ( 𝑗 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) )
9 8 rabbii { 𝑎 ∈ ( Base ‘ 𝑅 ) ∣ ∃ 𝑖 ∈ ( Base ‘ 𝑅 ) ( 𝑖 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) } = { 𝑎 ∈ ( Base ‘ 𝑅 ) ∣ ∃ 𝑗 ∈ ( Base ‘ 𝑅 ) ( 𝑗 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) }
10 3 9 eqtri 𝑈 = { 𝑎 ∈ ( Base ‘ 𝑅 ) ∣ ∃ 𝑗 ∈ ( Base ‘ 𝑅 ) ( 𝑗 ( +g𝑅 ) 𝑎 ) = ( 0g𝑅 ) }
11 1 2 10 primrootsunit1 ( 𝜑 → ( ( 𝑅 PrimRoots 𝐾 ) = ( ( 𝑅s 𝑈 ) PrimRoots 𝐾 ) ∧ ( 𝑅s 𝑈 ) ∈ Abel ) )