G Let G be a not necessarily abelian group with normal subgroups H and K such that H contains K (i. e., K ✂ G, H ✂ G, K ≤ H) and such that G/K is an abelian group. Prove that G/H is an abelian group also.

Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, is equal to , thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.

Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, . Since a and b were generic elements of H, then H/G is abelian.

answer: procent 54 wynosi 33,33%

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6