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.
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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Question
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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