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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 )