Show isomorphism for element g in group G

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Homework Help Overview

The discussion revolves around demonstrating that a specific map, defined by ig(x) = gxg' for a fixed element g in a group G, is an isomorphism of G with itself. Participants are exploring the properties of this map in the context of group theory.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants express confusion about the problem's requirements, questioning whether they need to find a function or show that the map preserves group structure. Others clarify the definitions of g and g' and discuss the necessary properties for the map to be an isomorphism, including bijection and structure preservation.

Discussion Status

Participants are actively engaging with the problem, with some providing attempts at proofs for the map being a bijection and preserving the group operation. There is a mix of interpretations and approaches being explored, but no consensus has been reached on the correctness of the reasoning presented.

Contextual Notes

There are questions regarding the definitions of g and g', as well as the assumptions about the group operation, which may affect the clarity of the discussion. Some participants express uncertainty about the implications of non-commutativity in their reasoning.

bennyska
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Homework Statement


let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.

Homework Equations


The Attempt at a Solution


not even really understanding the question. can someone break it down for me, and explain what the question is asking? am i trying to find a function, or rather show that i(x) preserves the structure?
 
Last edited:
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Does it specify what is F? Or is g an element of G?

Also, does it specify what exactly is g' in ig(x) = gxg'?
 


i'm sorry, it's supposed to be:
let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.
g' is the inverse of g, i think.
 


I am assuming that g is an element of G and that g' is the inverse of g.

The problem is giving you a function and asking you to show that it is an isomorphism. To show that it is an isomorphism, you must show it is a bijection and that for all x, y in G, ig(xy) = ig(x)ig(y)
 


so for the latter part:
ig(xy)=gxyg'
=gxg'gyg'
=ig(x)*ig(y)
is that right?
 
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and for the bijection:

injection
let ig(x) = ig(y) for x,y in G. then
gxg'=gyg'
g'gxg'=g'gy'g inverses
xg'=yg'
x=y by right cancellation

surjection
let x,y in G. then
y=gxg'
yg=gxg'g
yg=gx
g'yg=g'gx
g'yg=x

is this right? i kind of feel like I'm using normal multiplication to do this, ignoring commutativity.
 

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