Metamath Proof Explorer


Theorem lsmidm

Description: Subgroup sum is idempotent. (Contributed by NM, 6-Feb-2014) (Revised by Mario Carneiro, 21-Jun-2014) (Proof shortened by AV, 27-Dec-2023)

Ref Expression
Hypothesis lsmub1.p
|- .(+) = ( LSSum ` G )
Assertion lsmidm
|- ( U e. ( SubGrp ` G ) -> ( U .(+) U ) = U )

Proof

Step Hyp Ref Expression
1 lsmub1.p
 |-  .(+) = ( LSSum ` G )
2 subgsubm
 |-  ( U e. ( SubGrp ` G ) -> U e. ( SubMnd ` G ) )
3 1 smndlsmidm
 |-  ( U e. ( SubMnd ` G ) -> ( U .(+) U ) = U )
4 2 3 syl
 |-  ( U e. ( SubGrp ` G ) -> ( U .(+) U ) = U )