5. Let G be a group and define z" = z * z * *.. * z for n factors c, where z E G and n E Z+.(a) Suppose that G is abelian. Give a mathematical induction proof that (x * y)" = (x") * (y") forall x, y E G.(b) Now, for G not necessarily abelian, prove the following: if (x*y)² = (x²) * (y²) then the elementsx and y commute, i.e. x * y = y * x.

Answered: 5. Let G be a group and define z" z *…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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5. Let G be a group and define z" = z * z * *.. * z for n factors c, where z E G and n E Z+.
(a) Suppose that G is abelian. Give a mathematical induction proof that (x * y)" = (x") * (y") for
all x, y E G.
(b) Now, for G not necessarily abelian, prove the following: if (x*y)² = (x²) * (y²) then the elements
x and y commute, i.e. x * y = y * x.

Expert Solution

Step 1

Given that G is a group.

Also $z^{n}$ is defined as follows.

$z^{n} = z * z * z * . . . * z$

(a) Use mathematical induction to prove ${(x * y)}^{n} = (x^{n}) * (y^{n}) for all x, y \in G$ .

$P (n) : {(x * y)}^{n} = (x^{n}) * (y^{n}) for all x, y \in G . Base Case : n = 1 . {(x * y)}^{1} = x * y (x^{1}) * (y^{1}) = x * y Thus, {(x * y)}^{1} = (x^{1}) * (y^{1}) . Induction Step : Assume that P (k) is true . That is assume that {(x * y)}^{k} = (x^{k}) * (y^{k}) . Assume that n = k is true and show that P (n) is true for n = k + 1 . Consider the LHS of the statement P (k + 1) : {(x * y)}^{k + 1} = (x^{k + 1}) * (y^{k + 1}) . {(x * y)}^{k + 1} = {(x * y)}^{k} * (x * y) = (x^{k} * y^{k}) * (x * y) [By assumption that P (k) is true .] = x^{k} * x * y^{k} * y [G is abelian] = x^{k + 1} * y^{k + 1} Thus the statement is true for n = k + 1 .$

Thus by the principle of mathematical induction the given statement is true.

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