Basic Isomorphism Question The map $\phi: \mathbb{Z} \to \mathbb{Z}$ defined by $\phi (n) = n + 1$ for $n \in \mathbb{Z}$ is one to one and onto $\mathbb{Z}$. Give the definition of a binary operation $*$ on $\mathbb{Z}$ such that $\phi$ is an isomorphism mapping 
a) $\langle\mathbb{Z},+\rangle$ onto $\langle\mathbb{Z}, *\rangle$

I absolutely do not understand this. What is the solution doing? Why are we starting out with $m*n$? And why do we have $\phi(m-1) * \phi(n-1)$?
 A: The problem is stated very badly. A better way to state it would be:

Let $+$ be the usual addition of integers. The map $\phi: \mathbb{Z} \to \mathbb{Z}$ defined by $\phi (n) = n + 1$ for $n \in \mathbb{Z}$ is one-to-one and onto $\mathbb{Z}$. Give the definition of a binary operation $*$ on $\mathbb Z$ such that $\phi$ is an isomorphism from the group $\langle\mathbb Z,+\rangle$ to the group $\langle\mathbb Z,*\rangle$.

Does the solution now make more sense?
A: You're trying to define an operation $\ast$ on $\mathbb Z$ such that $\phi$ is an isomorphism. We want $\ast$ to be preserved by $\phi$ in particular that $$\phi(m+n)=\phi(m)\ast\phi(n)=(m+1)\ast(n+1).$$
The solution is starting with $m\ast n$ so instead we get
$$m\ast n=\phi(m-1)\ast \phi(n-1).$$
Now if $\phi$ is an isomorphism then we'd have to have $$\phi(m-1)\ast \phi(n-1)=\phi((m-1)+(n-2))=\phi(m+n-2)$$ and finally note that 
$$\phi(m+n-2)=m+n-2+1=m+n-1.$$
Thereby we must have that $m\ast n=m+n-1$.
A: The question asks you to define a binary operation $\ast$ on $\mathbb{Z}$ such that $(\mathbb{Z}, \ast)$ is a group and $\phi : (\mathbb{Z}, +) \to (\mathbb{Z}, \ast)$ is an isomorphism.
Suppose such a binary operation exists. Given $m, n \in \mathbb{Z}$ (thinking of this at the target of $\phi$), how do we define $m\ast n$? Well, by definition of $\phi$, we have 
$$m\ast n = \phi(m - 1)\ast\phi(n - 1).$$ 
Now, if $\phi$ is an isomorphism from $(\mathbb{Z}, +)$ to $(\mathbb{Z}, \ast)$, we have $\phi(a)\ast\phi(b) = \phi(a + b)$ for all $a, b \in \mathbb{Z}$. In particular, we have 
$$\phi(m-1)\ast\phi(n-1) = \phi((m-1) + (n-1)) = \phi(m + n - 2).$$
Combining the two equations we have 
$$m\ast n = \phi(m + n - 2),$$
but we can simplify this by applying the definition of $\phi$:
$$m\ast n = \phi(m + n - 2) = (m + n - 2) + 1 = m + n - 1.$$
Therefore, if $\ast$ with the claimed properties exists, then $m\ast n$ must be defined as $m + n - 1$. It is then easy to check that by defining $\ast$ in this way, $(\mathbb{Z}, \ast)$ is indeed a group.
