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Metamath Proof Explorer

Theorem subg0cl

Description: The group identity is an element of any subgroup. (Contributed by Mario Carneiro, 2-Dec-2014)

Ref Expression
Hypothesis subg0cl.i 
|- .0. = ( 0g ` G )
Assertion subg0cl 
|- ( S e. ( SubGrp ` G ) -> .0. e. S )

Proof

Step Hyp Ref Expression
1 subg0cl.i 
 |-  .0. = ( 0g ` G )
2 eqid 
 |-  ( G |`s S ) = ( G |`s S )
3 2 subggrp 
 |-  ( S e. ( SubGrp ` G ) -> ( G |`s S ) e. Grp )
4 eqid 
 |-  ( Base ` ( G |`s S ) ) = ( Base ` ( G |`s S ) )
5 eqid 
 |-  ( 0g ` ( G |`s S ) ) = ( 0g ` ( G |`s S ) )
6 4 5 grpidcl 
 |-  ( ( G |`s S ) e. Grp -> ( 0g ` ( G |`s S ) ) e. ( Base ` ( G |`s S ) ) )
7 3 6 syl 
 |-  ( S e. ( SubGrp ` G ) -> ( 0g ` ( G |`s S ) ) e. ( Base ` ( G |`s S ) ) )
8 2 1 subg0 
 |-  ( S e. ( SubGrp ` G ) -> .0. = ( 0g ` ( G |`s S ) ) )
9 2 subgbas 
 |-  ( S e. ( SubGrp ` G ) -> S = ( Base ` ( G |`s S ) ) )
10 7 8 9 3eltr4d 
 |-  ( S e. ( SubGrp ` G ) -> .0. e. S )