If a and b are arbitrary distinct elements of a  group G and H is any subgroup of G, then           Ha=Hb⇔ab\power{-1}∈H           aH=bH⇔b\power{-1}a∈H.

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.4: Cosets Of A Subgroup
Problem 25E: If H and K are arbitrary subgroups of G, prove that HK=KH if and only if HK is a subgroup of G.
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If a and b are arbitrary distinct elements of a
 group G and H is any subgroup of G, then
          Ha=Hb⇔ab\power{-1}∈H
          aH=bH⇔b\power{-1}a∈H.

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