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