By V. B. Alekseev
Translated by way of Sujit Nair
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Additional resources for Abel’s Theorem in Problems & Solutions
It is possible to write down an arbitrary permutation of 12 . . n degree n in the form where im is the image of element m under the given permutation. i1 i2 . . in Recall that a permutation is a one-to-one mapping; therefore all the elements in the lower line are different. Problem-175 How many different permutations of degree n do we have ? e. composition) of permutations 12 is called the symmetric group of degree n and are denoted by Sn . Problem-176 Prove that for n ≥ 3 the group Sn is non-commutative.
Problem-241 Prove that the function with complex argument f (z) = z 2 is continuous with all values of the argument z. Definition 33 Let f (z) and g(z) be two functions of a complex (or real) argument. The function with complex (or real) argument h(z), which is called the sum of the functions f (z) and g(z), satisfies at each point z0 the equation h(z0 ) = f (z0 ) + g(z0 ) holds. In case the value of f (z0 ) or g(z0 ) is not defined then the value of h(z0 ) is also not defined. In the same way one defines the difference, product and quotient of two functions.
E. z1 · z2 = z2 · z1 and (z1 · z2 ) · z3 = z1 · (z2 · z3 ) for any complex numbers z1 , z2 , z3 . It is easy to verify that (a, b) · (1, 0) = (1, 0) · (a, b) = (a, b) for any complex number (a, b). Thus, the complex number (1, 0) is the identity element in the set of the complex numbers under multiplication. Problem-207 Let z be an arbitrary complex number and z = (0, 0). Prove that there exists complex number z −1 such that z · z −1 = z −1 · z = (1, 0). The results of problems 203 and 204 show that the complex numbers form a commutative group under multiplication.