Abstract Algebra Isomorphism

[lesaltersvi]

Prove that if φ : S →S’ is an isomorphism of <S, *> with <S’ , *’> then φ^-1 is an isomorphism of <S’,*’> with <S,*>?

 

[daon2]

You haven't specified the structure. Is it a group isomorphism?

Where are you having difficulties?

 

[HallsofIvy]

That's pretty much the definition of "isomorphism". A "homomorphism" from one algebraic structure to another is a function that "preserves" the operations. A "homomorphism" is an "isomorphism" if and only if it is both "one-to-one" and "onto" which, in turn, means that the function is invertible. Two structures being "isomorphic" is an equivalence relation.

 

[lesaltersvi]

follow up

This is what I have now how does it look
( for all a and b are elements of S)
Φ(a *b) = φ(a) *’ φ(b)
Because of onto { x,y} is a subset of S’ and a and b are elements of S
Then φ(a) =x and φ(b) =y and c is an element of S φ(c)= x *’y

Because of 1-1
Φ(a*b) = φ(a) *’ φ(b)
= x*’y = φ(c)
= a*b=c
Then φ^-1(x*’y) = φ^-1(φ(c))
= c
=a*b
= φ^-1(x)* φ^-1(y)