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,@).

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- October 9th 2008, 08:37 AMLipticbovenAn 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,@). - October 9th 2008, 08:44 AMwatchmath
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. - October 9th 2008, 08:47 AMLipticboven
- October 9th 2008, 08:59 AMwatchmath
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)?

- October 9th 2008, 09:11 AMLipticboven
g o f?

- October 9th 2008, 09:17 AMwatchmath
- October 9th 2008, 09:34 AMLipticboven
- October 9th 2008, 09:37 AMwatchmath
- October 9th 2008, 09:43 AMLipticboven
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 (Angry) - October 9th 2008, 10:18 AMwatchmath
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 :D