## ChrisS 4 years ago 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 $\phi(x*y)=\phi(x)*\phi(y)$

1. anonymous

i think this is not one to one

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. ChrisS

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

6. anonymous

yw

7. ChrisS

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 $<M_{2}(\mathbb{R}),*>\ with <\mathbb{R},*> where\ \phi(A)\ is\ the\ determinant\ of\ matrix\ A$ This one is also not an isomorphism because multiplication of real numbers is commutative, where Matrix multiplication is not.

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. ChrisS

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 :)

14. anonymous

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