(Given) Let G be a finite group and H1, H2, ..., Hk be subgroups of G. (Question) If H;< Hj, show that [G: H;] = [G : H;] [Hj : H;]. (Answer) We know that [G : H;] |G| |Hi| Given Hi< Hj →Hj N Hị = Hị Solution: =IG| |Hj n Hi| |Hi| [G: H;] |G| |G| |Hj] _ |G| |Hj| [G: H;] [Hj : H;] |Hj n Hi| |Hj| |Hj| |Hi|
(Given) Let G be a finite group and H1, H2, ..., Hk be subgroups of G. (Question) If H;< Hj, show that [G: H;] = [G : H;] [Hj : H;]. (Answer) We know that [G : H;] |G| |Hi| Given Hi< Hj →Hj N Hị = Hị Solution: =IG| |Hj n Hi| |Hi| [G: H;] |G| |G| |Hj] _ |G| |Hj| [G: H;] [Hj : H;] |Hj n Hi| |Hj| |Hj| |Hi|
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.7: Direct Sums (optional)
Problem 24E
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