Metamath Proof Explorer


Theorem cycsubgcld

Description: The cyclic subgroup generated by A is a subgroup. Deduction related to cycsubgcl . (Contributed by Rohan Ridenour, 3-Aug-2023)

Ref Expression
Hypotheses cycsubgcld.1
|- B = ( Base ` G )
cycsubgcld.2
|- .x. = ( .g ` G )
cycsubgcld.3
|- F = ( n e. ZZ |-> ( n .x. A ) )
cycsubgcld.4
|- ( ph -> G e. Grp )
cycsubgcld.5
|- ( ph -> A e. B )
Assertion cycsubgcld
|- ( ph -> ran F e. ( SubGrp ` G ) )

Proof

Step Hyp Ref Expression
1 cycsubgcld.1
 |-  B = ( Base ` G )
2 cycsubgcld.2
 |-  .x. = ( .g ` G )
3 cycsubgcld.3
 |-  F = ( n e. ZZ |-> ( n .x. A ) )
4 cycsubgcld.4
 |-  ( ph -> G e. Grp )
5 cycsubgcld.5
 |-  ( ph -> A e. B )
6 1 2 3 cycsubgcl
 |-  ( ( G e. Grp /\ A e. B ) -> ( ran F e. ( SubGrp ` G ) /\ A e. ran F ) )
7 4 5 6 syl2anc
 |-  ( ph -> ( ran F e. ( SubGrp ` G ) /\ A e. ran F ) )
8 7 simpld
 |-  ( ph -> ran F e. ( SubGrp ` G ) )