Isomorphism from $\langle Z, +\rangle$ onto $\langle Z, \ast\rangle$? I'm trying to do 3.16 in Fraleigh's algebra book.  Here it is:

The map $f: Z\to Z$ defined by$ f(n) = n + 1$ for $n$ in $Z$ is 1-1 and onto $Z$.  Give the definition of a binary operation $\ast$ on $Z$ such that $f$ is an isomorphism mapping $\langle Z, +\rangle$ onto $\langle Z, \ast\rangle$.

I know isomorphisms have to pass the homomorphism property:
$$f(mn) = f(m)f(n)$$
So:$ (mn) + 1 = (m + 1)(n + 1) = nm + n + m + 1$
So $0 = n + m$...
A bit lost at this point.
Any guidance would be great!
Thanks,
Mariogs
 A: You need to define the operation $*$ on $\mathbb Z$ in order to make $f$ and isomorphism.
So we need for $f$ to satisfy the homomorphism property: $$f(m+n) = m+n+1 = f(m)*f(n) = (m+1)*(n+1)$$
Now, what operation $*$ will give us the equality: $$m+n + 1 = (m+1)*(n+1)\;\;?$$
How about: $p*q = p + q -1$. Compute, now, $(m+1)*(n+1)$, and see if you obtain the desired $m+n +1$.
A: From your original question and also your comments, I think you are a little confused about what it means for $f$ to be a homomorphism.  You said that $f$ is a homomorphism if $$f(mn) = f(m)f(n).$$  But this is wrong.  That is what it means for $f$ to be a homomorphism from $\langle Z, \cdot\rangle$ to $\langle Z, \cdot\rangle$.
But here we want to define a new operation, called $\ast$, so that $f$ is a homomorphism from $\langle Z, +\rangle$ to $\langle Z, \ast\rangle$.  
The property for that to be true is that $$f(m+n) = f(m)\ast f(n).$$
Notice that there is no multiplication anywhere in this formula.  There is addition, and there is $\ast$, which is not multiplication. 
You know $f$, so you know everything about what this formula says except what $\ast$ means.  You can use algebra to figure out what $a\ast b$ must be in order for the formula to hold.
A: Hint: Multiplication isn't involved in the problem. The homomorphism property for $f:(\mathbb Z,+)\to(\mathbb Z,\ast)$ states that $f(m+n)=f(m)\ast f(n)$. Now apply the definition of $f$ and you should be back on track.
A: Define $\ast$ as $a \ast b = a+b-1$.
We then need to check $$\phi(a+b)=\phi(a)\ast\phi(b).$$
By definition, $$\phi(a+b)=a+b+1.$$
Also, by definition $$\phi(a)\ast \phi(b)=(a+1)\ast(b+1)=(a+1)+(b+1)-1=a+b+1.$$
Thus  $$\phi(a+b)=\phi(a)\ast \phi(b),$$ and $\phi$ is an isomorphism.
A: Their is a general procedure for transferring structures over bijections of sets. What do I mean by this? Well consider the following proposition:
Proposition Let $(G,\bullet)$ be a group and $S$ be a set, such that their is a bijection of sets, $$\phi:G\to S$$ (note that $\phi$ is not a group homomrphism, yet). (Note also that a in category theory you would say that their is a set isomorphism $\phi:UG\to S$, where $U$ is the forgetful functor to sets.) Then their is a unique group operation, $\star$ on $S$ that makes $\phi$ a group homomorphism.
sketch of proof: Lettuce start by  defining the group operation $\star$. Define: $$\textbf{1}:s_1\star  s_2=\phi(\phi^{-1}(s_1)\bullet\phi^{-1}(s_2))$$
We will now show that this product is associative.
$$(s_1\star s_2)\star s_3=\\
\phi[\phi^{-1}(s_1\star s_2)\bullet \phi^{-1}(s_3)]=\\
\phi[\phi^{-1}(\phi[\phi^{-1}(s_1)\bullet \phi^{-1}(s_2)])\bullet \phi^{-1}(s_3)]=\\
\phi[\phi^{-1}(s_1)\bullet [\phi^{-1}(s_2)\bullet \phi^{-1}(s_3)]]=\\
\phi[\phi^{-1}(s_1)\bullet \phi^{-1}(\phi([\phi^{-1}(s_2)\bullet \phi^{-1}(s_3)]])=\\
\phi(\phi^{-1}(s_1)\bullet\phi^{-1}(s_2\star s_3))=\\
s_1\star(s_2\star s_3).$$
I will leave the other axioms to you.
$$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\Box$$
Now what use is this proposition? Well in you situation, you have the following: A group, $(\mathbb{Z},+)$ and you are looking for another group operation on $\mathbb{Z}$ so that the bijection, $f(n)=n+1$ is a homomorphism. The second $\mathbb{Z}$ is just a set, without any group structure. So let us define $$n_1*n_2=f(f^{-1}(n_1)+f^{-1}(n_2)).$$ This is simply applying $\textbf{1}$ to our situation. Let us now work out what this must be.
$$n_1*n_2=f(f^{-1}(n_1)+f^{-1}(n_2))=\\f(n_1-1+n_2-1)=\\n_1+n_2-1.$$
You might say that we had to work hard to get this. But this proposition is quite powerful. What the answer does is to solve a more general problem. A few exercises that you could try (if you want) are as follows:


*

*Define a group operation, $*$ on $\mathbb{Z}$ such that $f(n)=n+5$ is a homomorphism from $(\mathbb{Z},+)$ to $(\mathbb{Z}, *)$ using this proposition.

*Define a group operation $*$ on $\mathbb{Z}/2\oplus\mathbb{Z}/4$ such that $f(m,n)=(m,n+1)$ is a homomorphsm from $(\mathbb{Z}/2\oplus\mathbb{Z}/4,+)$ to $(\mathbb{Z}/2\oplus\mathbb{Z}/4,*)$

*Consider the map, $f:\mathbb{Z}/3\to\{a,b,c\}$, such that $f(0)=a, f(1)=b, f(2)=c$. Use this proposition to to define a group operation $*$ on $\{a,b,c\}$ such that $f$ is a group homomorphism from $(\mathbb{Z}/3,+)$ to $(\{a,b,c\},*)$.

*(this one is harder) If you have a ring, $R$ and a set $S$, and a set bijection $f:R\to S$ Define a ring structure on $S$ such that $f$ is a ring homomorphism.  
Note also Propositions like the above are often called "transfer of structure theorems". So is the proposition indicated in exercise 4. I learned this proposition from Seth Warner's book "Modern Algebra". This is Theorem 6.3 on page 43 (of the copy of the book that I have). He calls this theorem the "Transplanting Theorem".
