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 ∈ ( SubGrp ‘ 𝐺 ) → ( 𝑈 𝑈 ) = 𝑈 )

Proof

Step Hyp Ref Expression
1 lsmub1.p = ( LSSum ‘ 𝐺 )
2 subgsubm ( 𝑈 ∈ ( SubGrp ‘ 𝐺 ) → 𝑈 ∈ ( SubMnd ‘ 𝐺 ) )
3 1 smndlsmidm ( 𝑈 ∈ ( SubMnd ‘ 𝐺 ) → ( 𝑈 𝑈 ) = 𝑈 )
4 2 3 syl ( 𝑈 ∈ ( SubGrp ‘ 𝐺 ) → ( 𝑈 𝑈 ) = 𝑈 )