Proof of a linear operator acting on an inverse of a group element

  • Thread starter Dixanadu
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  • #1
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Main Question or Discussion Point

Hey guys!

Basically, I was wondering how to prove the following statement. Ive seen it in the Hamermesh textbook without proof, so I wanted to know how you go about doing it.

Let's say you have a group element [itex]g_{1}[/itex], which has a corresponding inverse [itex]g_{1}^{-1}[/itex]. Let's also define a linear transformation D for this group element.

So what im trying to prove is that

[itex]D(g_{1}^{-1}) = [D(g_{1})]^{-1} [/itex]

Can u guys point me in the right direction?

Thanks!
 

Answers and Replies

  • #2
Office_Shredder
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By linear transformation do you mean group homomorphism? If so just consider
[tex] D(g_1 g_1^{-1}) = D(g_1) D(g_1^{-1}) [/tex]
 
  • #3
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yes it is a homomorphism. But what u wrote in the previous message, where do I go from there? cos [itex]D(g_{1}g_{1}^{-1})[/itex] is just [itex]D(E)[/itex] where E is the identity, right? o.o
 
  • #4
Office_Shredder
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Yes, and what is D(E) equal to?
 
  • #5
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OHH I get it. Basically, because [itex]D(E) = E[/itex], we can say that

[itex]E = D(g_{1})D(g_{1}^{-1})[/itex]
Then by multiplying both sides on the left by [itex][D(g_{1})]^{-1}[/itex] we get

[itex][D(g_{1})]^{-1} = D(g_{1}^{-1})[/itex]....I think...
 

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