If x=x^-1 for all x in G, then it's the same in H, for isomorphisms.

  • Thread starter ArcanaNoir
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  • #1
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Homework Statement



G and H are isomorphic.
Prove that if x-1=x for all x in G, then x-1=x for all x in H.


Homework Equations



G is isomorphic to H means there is an operation preserving bijection from G to H.

The Attempt at a Solution





I need a point in the right direction.
 

Answers and Replies

  • #2
I like Serena
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Pick an operation involving x and x^-1 and map it?

Or perhaps you should look up the exact definition of an inverse?
 
  • #3
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Thank goodness you're here!

Okay, so um, [tex] \theta (x) = y [/tex] ?

inverse means [tex] x*x^{-1} = e [/tex]
 
  • #4
I like Serena
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Thank goodness you're here!
Thanks!
It's nice to feel appreciated! :blushing:


Okay, so um, [tex] \theta (x) = y [/tex] ?

inverse means [tex] x*x^{-1} = e [/tex]
What about [tex] \theta (x*x^{-1}) ? [/tex]


Inverse in a group G means according to the axioms:
For each a in G, there exists an element b in G such that a • b = b • a = e​



So let's pick an element a in H.
Can you find an (the!) inverse b that satisfies the axiom, with the help of the isomorphism you have?
 
  • #5
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[tex] \theta (x*x^{-1})=\theta (x)*\theta (x^{-1}) [/tex]
[tex] \theta (e_G)=e_H=\theta (x)*\theta (x^{-1}) [/tex]
let [itex] \theta (x)=a [/itex]
then [itex] e_H=a*\theta (x^{-1}) [/itex]
thus, the inverse of [itex] \theta (x)=a [/itex] in H is [itex] \theta (x^{-1}) [/itex]

Yes?
 
  • #6
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Oh wait, but since [itex] x=x^{-1} , \theta (x^{-1})=\theta (x)= a [/itex]
Thus, each a in H, a=a-1

Did I prove it? Did I prove it? :D
 
  • #7
I like Serena
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Yep! :approve:
 
  • #8
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Hooray! Thank you very much. That was the most painless problem ever :)
 
  • #9
I like Serena
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Ah, but then, when things *click*, most of them are painless.
Perhaps you should revisit older problems in time...
 

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