I need help with understand how to approach a question like this.
Given a mapping function Φ: Q -> Q defined by Φ(x) = 3x - 1.
I need to define a binary operation * such that Φ is an isomorphism mapping.
(Q,•) -> (Q,*)
Where • is the usual multiplication.
This is my attempt:
Given a,b in Q:
Φ(a•b) = Φ(a)*Φ(b)
3(ab)-1 = (3a-1)*(3b-1)
I know that I must get a term containing ab on the right hand side, so * must be based in part on multiplication. We begin by trying * = •
3(ab)-1 = 9ab - 3a - 3b + 1
The right hand side is too large. To fix this, we do the following:
3(a•b) - 1 = (1/3)(9a•b - 3a - 3b + 1 + 3a + 3b - 4) 3(a•b) - 1 = 3(a•b) - 1
Then the binary operation * follows from the steps performed above, starting from •:
a*b = (1/3)(a•b + 3a + 3b -4)
However, I know that an isomorphism mapping preserves the identity. Thus, if e is the identity for •, then Φ(e) should be the identity for *.
The identity for • is 1. Thus, Φ(1) = 3•1 -1 = 2, the identity for *.
Trying this:
a*e = a
(1/3)(a•2 + 3•a + 3•2 - 4) = a
This is not a true statement.
The correct answer is (1/3)(a•b + a + b -2) which does have identity 2.
(1/3)(a•2 + a + 2 -2) = a
How did the author arrive at this solution? How should I approach such problems? I have managed to correctly get some other results (particularly ones where the binary operation besides * is +).
As you say, you must have $$\Phi(ab)=\Phi(a)\ast\Phi(b).\tag{1}$$ But you want to find $x\ast y$. Let's solve $\Phi(a)=x$ and $\Phi(b)=y$. For $\Phi(a)=x$ you need $3a-1=x$, so $a=\frac13(x+1)$. Also $b=\frac13(y+1)$. From $(1)$ you get $$x\ast y=\Phi(ab)=3ab-1=\frac13(x+1)(y+1)-1$$ etc.