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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 $\langle\mathbb{Z}, *\rangle$

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I absolutely do not understand this. What is the solution doing? Why are we starting out with $m*n$? And why do we have $\phi(m-1) * \phi(n-1)$?

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3 Answers 3

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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.

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I am probably restating what you did. But couldn't we have NOT started out with m*n? Instead we compare $\phi(m * n)$ and $\phi(m) +\phi(n)$ So computing $\phi(m*n) = m*n +1$ and $\phi(n) + \phi(m) = n + m + 2$ Equating both equations, $m*n +1 = m + n + 2 \implies m*n = m + n + 1$? –  Hawk Oct 12 '12 at 4:37
    
Careful, the $\phi$ you've written is going the wrong way (i.e. from $(\mathbb{Z}, \ast)$ to $(\mathbb{Z}, +)$). But you are correct, you could have instead compared $\phi(m)\ast\phi(n)$ and $\phi(m + n)$, you would then know that $(m+1)\ast(n+1) = m+n+1$. –  Michael Albanese Oct 12 '12 at 9:19

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?

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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 $m\ast n=m+n-1$.

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Your solution makes more sense to me –  Hawk Oct 11 '12 at 2:46

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