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.

it could be 1,3 and more others to create a cylinder.

step-by-step explanation:

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42