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

1. Nov 13, 2011

### ArcanaNoir

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution

I need a point in the right direction.

2. Nov 13, 2011

### 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?

3. Nov 13, 2011

### ArcanaNoir

Thank goodness you're here!

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

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

4. Nov 13, 2011

### I like Serena

Thanks!
It's nice to feel appreciated!

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?

5. Nov 13, 2011

### 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?

6. Nov 13, 2011

### 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

7. Nov 13, 2011

### I like Serena

Yep!

8. Nov 13, 2011

### ArcanaNoir

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

9. Nov 13, 2011

### I like Serena

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