Thread: An Isomorphism Question!

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 and .

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.

Originally Posted by watchmath

By the given condition you have an isomorphism and .

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.

Should i use the funtion , and , then . Is that what u ment by the obvious choice?

We don't know what is the explicit formula for f and g . But now if you have a function from A to B and from B to C, how would you make a function from A to C (by using f and g)?

g o f?

Originally Posted by Lipticboven

g o f?

Exactly!!
Now your job is to verify that this function is a homomorphism, 1-1 and onto.

Originally Posted by watchmath

Exactly!!
Now your job is to verify that this function is a homomorphism, 1-1 and onto.

To prove it is a homomorhism would it be ok to say:
since(A,*) is isomorphic to (B,&) and (B,&) is isomorphic to (C,@) then,
f(a*b) = f(a) & f(b)
g(a&b) = g(a) @ g(b)

So

g(f(a*b)) = g(f(a) & f(b)) = g(f(a)) @ g(f(b))

Therefore (A,*) is isomorphic to (C,@).

Originally Posted by Lipticboven

To prove it is a homomorhism would it be ok to say:
since(A,*) is isomorphic to (B,&) and (B,&) is isomorphic to (C,@) then,
f(a*b) = f(a) & f(b)
g(a&b) = g(a) @ g(b)

So

g(f(a*b)) = g(f(a) & f(b)) = g(f(a)) @ g(f(b))

Therefore (A,*) is isomorphic to (C,@).

You just proved that g 0 f is a homomorphism. You need now to prove that this function is 1-1 and onto

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