Let G be a group (written additively), S a nonempty set and M (S, G) the set of all functions from S in G. We define addiction in M (S, G) as follows: For ?, ? ∈ ? (S, G) and for each s ∈ S, (f + g) (s) = f (s) + g (s). Prove that M (S, G) is a group with this operation, and that it is abelian if and only if G
Let G be a group (written additively), S a nonempty set and M (S, G) the set of all functions from S in G. We define addiction in M (S, G) as follows: For ?, ? ∈ ? (S, G) and for each s ∈ S, (f + g) (s) = f (s) + g (s). Prove that M (S, G) is a group with this operation, and that it is abelian if and only if G
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Let G be a group (written additively), S a nonempty set and M (S, G) the set
of all functions from S in G. We define addiction in M (S, G) as follows: For
?, ? ∈ ? (S, G) and for each s ∈ S, (f + g) (s) = f (s) + g (s). Prove that M (S, G)
is a group with this operation, and that it is abelian if and only if G is
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