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

[ArcanaNoir]

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.

 

[I like Serena]

Pick an operation involving x and x^-1 and map it?

Or perhaps you should look up the exact definition of an inverse?

 

[ArcanaNoir]

Thank goodness you're here!

Okay, so um, \theta (x) = y ?

inverse means x*x^{-1} = e

 

[I like Serena]

Thank goodness you're here!

Thanks!
It's nice to feel appreciated!

Okay, so um, \theta (x) = y ?

inverse means x*x^{-1} = e

What about \theta (x*x^{-1}) ?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?

 

[ArcanaNoir]

\theta (x*x^{-1})=\theta (x)*\theta (x^{-1})
\theta (e_G)=e_H=\theta (x)*\theta (x^{-1})
let \theta (x)=a
then e_H=a*\theta (x^{-1})
thus, the inverse of \theta (x)=a in H is \theta (x^{-1})

Yes?

 

[ArcanaNoir]

Oh wait, but since x=x^{-1} , \theta (x^{-1})=\theta (x)= a
Thus, each a in H, a=a^-1

Did I prove it? Did I prove it? :D

 

[I like Serena]

Yep!

 

[ArcanaNoir]

Hooray! Thank you very much. That was the most painless problem ever :)

 

[I like Serena]

Ah, but then, when things *click*, most of them are painless.
Perhaps you should revisit older problems in time...