By V. B. Alekseev

Translated by way of Sujit Nair

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Z1 + z2 ) · z3 = z1 · z3 + z2 · z3 for all complex numbers z1 , z2 , z3 . 3) form a field. This is the field of complex numbers. 3) give (a, 0) + (b, 0) = (a + b, 0) (a, 0) · (b, 0) = (a · b, 0) Thus, if we assign to each complex number of the form (a, 0) the real number a, then the operations on the numbers of the form (a, 0) will correspond to the usual operations on real numbers. Therefore we will simply identify the complex number (a, 0) with the real number a and 2 we will say that the field of the complex numbers includes field of real numbers.

For example, any commutative group is solvable: if G is a commutative group, then we already obtain K(G) = {e} at the first step. A group G is solvable if its commutator is commutative, since K2 (G) = {e}. Problem-159 Are the following groups solvable: a) the cyclic group Zn b) the symmetry group of a triangle, c) the symmetry group of a square, g) the group of quaternions, d) the rotation group of a tetrahedron, e) the symmetry group of a tetrahedron, f) the rotation group of a cube. 15 Solvable groups.

We showed that to each normal subgroup, there corresponds a certain homomorphism. Let us now show that conversely, every surjective homomorphism of a group G onto group F can be seen as a natural homomorphism from G to the quotient group G/N by a suitable normal subgroup. Definition 18 Let ϕ : G → F be a group homomorphism. Then the set of elements g of G such that ϕ(g) = eF is called the kernel of the homomorphism ϕ and is denoted Ker ϕ. Problem-140 Prove that Ker ϕ is a subgroup of the group G.