Metamath Proof Explorer


Theorem subginvcl

Description: The inverse of an element is closed in a subgroup. (Contributed by Mario Carneiro, 2-Dec-2014)

Ref Expression
Hypothesis subginvcl.i
|- I = ( invg ` G )
Assertion subginvcl
|- ( ( S e. ( SubGrp ` G ) /\ X e. S ) -> ( I ` X ) e. S )

Proof

Step Hyp Ref Expression
1 subginvcl.i
 |-  I = ( invg ` G )
2 eqid
 |-  ( G |`s S ) = ( G |`s S )
3 2 subggrp
 |-  ( S e. ( SubGrp ` G ) -> ( G |`s S ) e. Grp )
4 simpr
 |-  ( ( S e. ( SubGrp ` G ) /\ X e. S ) -> X e. S )
5 2 subgbas
 |-  ( S e. ( SubGrp ` G ) -> S = ( Base ` ( G |`s S ) ) )
6 5 adantr
 |-  ( ( S e. ( SubGrp ` G ) /\ X e. S ) -> S = ( Base ` ( G |`s S ) ) )
7 4 6 eleqtrd
 |-  ( ( S e. ( SubGrp ` G ) /\ X e. S ) -> X e. ( Base ` ( G |`s S ) ) )
8 eqid
 |-  ( Base ` ( G |`s S ) ) = ( Base ` ( G |`s S ) )
9 eqid
 |-  ( invg ` ( G |`s S ) ) = ( invg ` ( G |`s S ) )
10 8 9 grpinvcl
 |-  ( ( ( G |`s S ) e. Grp /\ X e. ( Base ` ( G |`s S ) ) ) -> ( ( invg ` ( G |`s S ) ) ` X ) e. ( Base ` ( G |`s S ) ) )
11 3 7 10 syl2an2r
 |-  ( ( S e. ( SubGrp ` G ) /\ X e. S ) -> ( ( invg ` ( G |`s S ) ) ` X ) e. ( Base ` ( G |`s S ) ) )
12 2 1 9 subginv
 |-  ( ( S e. ( SubGrp ` G ) /\ X e. S ) -> ( I ` X ) = ( ( invg ` ( G |`s S ) ) ` X ) )
13 11 12 6 3eltr4d
 |-  ( ( S e. ( SubGrp ` G ) /\ X e. S ) -> ( I ` X ) e. S )