Suppose that (A,*), (B,&) and (C,@) are sets with prescribed binary operations.
Prove that if (A,*) is isomorphic to (B,&) and (B,&) is isomorphic to (C,@), then (A,*) is isomorphic to (C,@).
By the given condition you have an isomorphism $\displaystyle f:A\to B$ and $\displaystyle g:B\to C$.
What is the obvious choice if you want to make an isomorphism from A to C?
Claim this obvious choice is an isomorphism and prove it, meaning you need to show that this map is a 1-1, onto, homomorphism.
and to prove it is one-to-one would it be ok to say:
Since (A,*) is isomorphic to (B,&) it means,
if f(a) = f(b) then a = b
Similarly since (B,&) is isomorphic to (C,@) it means,
if g(a) = g(b) then a = b
Therefore if g(f(a)) = g(f(b) it means f(a) = f(b), which in turn means a = b therefore it shows (A,*) and (C,@) are one to one.
I am a bit stuck proving it is onto thou
Yest, thats OK!
For onto, take arbitrary element c in C. You show that this is onto if you can find x in A such that g 0 f (x)= g(f(x))=c.
You need to use the onto-ness of f and g.
Since g is onto there is an element b in B such that g(b)=c.
So -----> g(f(x))=c <---------- what you want
--------> g(b)=c <------------------------ what you have
So now how to produce x so that from what you have you get what you want