Determine whether the given map $$\phi$$ is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? $$\ with\ \ where\ \phi(x)=x^{2}\ for\ x \in \mathbb{Q}$$ I have the one-to-one requirement $$If\ \phi(x)=\phi(y) then\ x^{2}=y^{2}$$ $$Then\ \sqrt{x^{2}}=\sqrt{y^{2}}$$ So x=y Thus the map is one-to-one. Now I need to show that it is onto and I need to show $$\phi(x*y)=\phi(x)*\phi(y)$$

Question

Determine whether the given map $$\phi$$ is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? $$<\mathbb{Q},*>\ with\ <\mathbb{Q},*>\ where\ \phi(x)=x^{2}\ for\ x \in \mathbb{Q}$$ I have the one-to-one requirement $$If\ \phi(x)=\phi(y) then\ x^{2}=y^{2}$$ $$Then\ \sqrt{x^{2}}=\sqrt{y^{2}}$$ So x=y Thus the map is one-to-one. Now I need to show that it is onto and I need to show $$\phi(x*y)=\phi(x)*\phi(y)$$

anonymous · Answer

i think this is not one to one

anonymous · Answer

for example 
$$\phi(2)=\phi(-2)$$

anonymous · Answer

and the mistake is in this line 
$$x^2=y^2\iff x=y$$ should be 
$$x^2=y^2\iff x = \pm y$$

anonymous · Answer

also it is not onto since nothing gets mapped to negative rationals

chriss · Answer

right... I see I was totally off on this now.  Thanks!

anonymous · Answer

yw

chriss · Answer

Hopefully you won't mind verifying one more for me... Same question but different binary structures... I think I have the answer but would like to run it by you if I could.. for this one the binary structure is $$\ with <\mathbb{R},*> where\ \phi(A)\ is\ the\ determinant\ of\ matrix\ A$$ This one is also not an isomorphism because multiplication of real numbers is commutative, where Matrix multiplication is not.

anonymous · Answer

no it is not an isomorphism because the set of two by two matrices is the same as 
$$\mathbb{R^4}$$

anonymous · Answer

it is a homomorphism however, because the determinant of the product is the product of the determinants

anonymous · Answer

that is 
$$\phi(AB)=\phi(A)\phi(B)$$

anonymous · Answer

i am assuming the " * " in both cases means multiplication yes?

anonymous · Answer

in any case it is certainly not injective because an infinite number of matrices can have the same determinant

chriss · Answer

awesome, thanks again.  We haven't got to homomorphisms yet, but I will note this for when we do.  

Yes, I didn't see a symbol for using a dot for multiplication.  I know it's confusing since * can be a generic binary operation symbol too but I thought it would make sense in the context of the question.

I do appreciate your help though.  I'll add a medal to one of the other questions you have answered since I can't give you two here lol.

Thanks a bunch :)

anonymous · Answer

no problem, i have medals to spare ( and i can't seem to redeem them at the package store)