Isomorphism of C(x)-axa^-1 Function in Group G

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Homework Help Overview

The discussion revolves around the function defined on a group G, specifically examining the mapping c_a(x) = axa^{-1} for a fixed element a in G. Participants are tasked with demonstrating that this function is an isomorphism by verifying properties such as one-to-one, onto, and homomorphic behavior.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the steps needed to show that the function is one-to-one by assuming c(x) = c(y) and manipulating the equation to demonstrate x = y. There is also exploration of how to show that the function is onto and homomorphic, with some participants expressing confusion about specific steps in the reasoning.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the proof for one-to-one and onto properties. There is a collaborative effort to clarify the homomorphic aspect of the function, although some confusion remains regarding the details of the calculations.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the extent of direct solutions provided. The nature of the problem requires careful handling of group properties and function definitions.

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Homework Statement


Let G be any group and let a be a fixed element of G. Define a function [tex]c_{a}[/tex]:G-->G by [tex]c_{a}[/tex](x)=ax[tex]a^{-1}[/tex] for all x in G. Show that c is an isomorphism


The Attempt at a Solution


Need to show 1-1, onto and c(ab)=c(a)c(b)
I guess my biggest problem is starting because I get to c(a)=c(b) for 1-1 and don't know what c(a) is.
 
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I'll get you started on 1-1. So take [tex]c(x)=axa^{-1}[/tex].
Assume that c(x)=c(y). Then by definition

[tex]axa^{-1}=aya^{-1}[/tex]

Do you see how to continue?
 
Ok we want to show x=y
So axa^-1=aya^-1
axa^-1a=aya^-1a
ax=ay
a^-1ax=a^-1ay
x=y, so 1-1

For onto we need to c(x)=axa^-1=y
We need to be able to solve for x, I think
y=axa^-1
ya=axa^-1a
ya=ax
a^-1ya=a^-1ax
a^-1ya=x
 
Yes!

And now you just need to show that c is a homomorphism...
 
c(ab)=abx(ab)^-1
=abxa^-1b^-1
A bit confused...
 
No. Let's get you started:

[tex]c(x)c(y)=(axa^{-1})(aya^{-1})[/tex]
 
ax(a^-1a)(ya^-1)
axya^-1=c(xy)
 
Yes!
 

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