Metamath Proof Explorer


Theorem primrootsunit

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

Ref Expression
Hypotheses primrootsunit.1
|- ( ph -> R e. CMnd )
primrootsunit.2
|- ( ph -> K e. NN )
primrootsunit.3
|- U = { a e. ( Base ` R ) | E. i e. ( Base ` R ) ( i ( +g ` R ) a ) = ( 0g ` R ) }
Assertion primrootsunit
|- ( ph -> ( ( R PrimRoots K ) = ( ( R |`s U ) PrimRoots K ) /\ ( R |`s U ) e. Abel ) )

Proof

Step Hyp Ref Expression
1 primrootsunit.1
 |-  ( ph -> R e. CMnd )
2 primrootsunit.2
 |-  ( ph -> K e. NN )
3 primrootsunit.3
 |-  U = { a e. ( Base ` R ) | E. i e. ( Base ` R ) ( i ( +g ` R ) a ) = ( 0g ` R ) }
4 nfv
 |-  F/ j ( i ( +g ` R ) a ) = ( 0g ` R )
5 nfv
 |-  F/ i ( j ( +g ` R ) a ) = ( 0g ` R )
6 oveq1
 |-  ( i = j -> ( i ( +g ` R ) a ) = ( j ( +g ` R ) a ) )
7 6 eqeq1d
 |-  ( i = j -> ( ( i ( +g ` R ) a ) = ( 0g ` R ) <-> ( j ( +g ` R ) a ) = ( 0g ` R ) ) )
8 4 5 7 cbvrexw
 |-  ( E. i e. ( Base ` R ) ( i ( +g ` R ) a ) = ( 0g ` R ) <-> E. j e. ( Base ` R ) ( j ( +g ` R ) a ) = ( 0g ` R ) )
9 8 rabbii
 |-  { a e. ( Base ` R ) | E. i e. ( Base ` R ) ( i ( +g ` R ) a ) = ( 0g ` R ) } = { a e. ( Base ` R ) | E. j e. ( Base ` R ) ( j ( +g ` R ) a ) = ( 0g ` R ) }
10 3 9 eqtri
 |-  U = { a e. ( Base ` R ) | E. j e. ( Base ` R ) ( j ( +g ` R ) a ) = ( 0g ` R ) }
11 1 2 10 primrootsunit1
 |-  ( ph -> ( ( R PrimRoots K ) = ( ( R |`s U ) PrimRoots K ) /\ ( R |`s U ) e. Abel ) )