φ maps S onto S' in a one-to-one fashion, meaning that each element u in S' has a unique corresponding element a in S such that φ(a) = u.
Since a is unique, the inverse map φ-1(u) = a from S' to S is well defined and unique. This shows φ-1 is a one-to-one map from S' to S.
Since a is an arbitrary element of S and φ is an onto map from S to S' then φ-1(φ(a)) = a maps S' onto S.
Let φ(a) = u and φ(b) = v, then since φ is an isomorphism φ(ab) = uv, ie, φ(φ-1(u)φ-1(v))=uv, and finally applying φ-1 to both sides gives us φ-1(u)φ-1(v) = φ-1(uv).
Comments
φ maps S onto S' in a one-to-one fashion, meaning that each element u in S' has a unique corresponding element a in S such that φ(a) = u.
Since a is unique, the inverse map φ-1(u) = a from S' to S is well defined and unique. This shows φ-1 is a one-to-one map from S' to S.
Since a is an arbitrary element of S and φ is an onto map from S to S' then φ-1(φ(a)) = a maps S' onto S.
Let φ(a) = u and φ(b) = v, then since φ is an isomorphism φ(ab) = uv, ie, φ(φ-1(u)φ-1(v))=uv, and finally applying φ-1 to both sides gives us φ-1(u)φ-1(v) = φ-1(uv).
first of all, if Ï is an isomorphism, it is bijective, so it has an inverse FUNCTION Ï^-1.
this function Ï^-1 is also bijective, because it has the inverse Ï
(a function is bijective if and only if it has a (two-sided) inverse. you might try proving this, it's instructive).
so all we need to do is show that Ï^-1 is also a homomorphism.
so let x,y be any two elements of S'.
since Ï is surjective, x = Ï(u) for some u in S, and y = Ï(v) for some v in S.
moreover, since Ï is injective, we know that Ï^-1(x) = u, and Ï^-1(y) = v.
thus:
Ï^-1(x*'y) = Ï^-1(Ï(u)*'Ï(v))
= Ï^-1(Ï(u*v)) = (Ï^-1oÏ)(u*v) = u*v
= Ï^-1(x)*Ï^-1(y), so Ï^-1 is indeed a homomorphism, and being bijective,
is an isomorphism.
Fish. The answer...is fish.