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PROPOSITION 9.2. Let f :G G' be an isomorphism of groups. For any element a of G, we have the following. (1) (f(a)) = {f (x) : x E (a)}. %3D

PROPOSITION 9.2. Let f :G G' be an isomorphism of groups. For any element a of G, we have the following. (1) (f(a)) = {f (x) : x E (a)}. %3D

Advanced Engineering Mathematics
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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Please prove the proposition (1) in the picture given below:

**Proposition 9.2**: Let \( f : G \rightarrow G' \) be an isomorphism of groups. For any element \( a \) of \( G \), we have the following: 1. \( \langle f(a) \rangle = \{ f(x) : x \in \langle a \rangle \} \). This proposition states that if there is an isomorphism \( f \) between groups \( G \) and \( G' \), then the image of the subgroup generated by an element \( a \) under \( f \) is the subgroup generated by \( f(a) \) in \( G' \).
Transcribed Image Text:**Proposition 9.2**: Let \( f : G \rightarrow G' \) be an isomorphism of groups. For any element \( a \) of \( G \), we have the following: 1. \( \langle f(a) \rangle = \{ f(x) : x \in \langle a \rangle \} \). This proposition states that if there is an isomorphism \( f \) between groups \( G \) and \( G' \), then the image of the subgroup generated by an element \( a \) under \( f \) is the subgroup generated by \( f(a) \) in \( G' \).
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