Let (G, *) and (G,0) be groups, with respective identity elements e and ē. Suppose there is a bijection f:G+ G such that for all a, b e G, f(ab) = f(a) f(b). Strictly speaking, that should be f(a+b) = f(a) o f(b) since f(a), f(b) EG, but we can suppress the operations for conciseness.1. Show that f(e) = ē. Hint: Let a = e in the equation. 2. Show that if G is abelian, then G is abelian. Let (G, *) and (G, 0) be groups, with respective identity elements e and ē. Suppose there is a bijectionf:G → G such that for all a, b e G,f(ab) = f(a)f(b).Strictly speaking, that should be f(a + b) = f(a) o f(b) since f(a), f(b) e G, but we can suppress theoperations for conciseness.1. Show that f(e) = ē. Hint: Let a = e in the equation.2. Show that if G is abelian, then G is abelian.

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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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Let (G, *) and (G,0) be groups, with respective identity elements e and ē. Suppose there is a bijection f:G+ G such that for all a, b e G, f(ab) = f(a) f(b). Strictly speaking, that should be f(a+b) = f(a) o f(b) since f(a), f(b) EG, but we can suppress the operations for conciseness.

1. Show that f(e) = ē. Hint: Let a = e in the equation. 2. Show that if G is abelian, then G is abelian.

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