Hw3 Problem 2

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### Problem 2

The map $\phi: \mathbb{Z} \rightarrow \mathbb{Z}$ defined by $\phi(n) = n+1$ 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 $\langle\mathbb{Z}, *\rangle$.
(b) $\langle\mathbb{Z}, *\rangle$ onto $\langle\mathbb{Z}, +\rangle$.
In each case, give the identity element of $\langle\mathbb{Z}, *\rangle$.

Solution

(a) It is already stated that $\phi$ defines a function that is one to one and onto, meaning that all we have to show is the homomorphism property on $\langle\mathbb{Z}, +\rangle$ and $\langle\mathbb{Z}, *\rangle$. To satisfy the homomorphism property, where $x,y \in \mathbb{Z}$,

(1)
\begin{align} \phi(x+y) = \phi(x) * \phi(y) \end{align}
(2)
\begin{equation} (x+y)+1 = (x+1)*(y+1). \end{equation}

Now, by letting $*$ be define by $x*y = x+y-1$.

(3)
\begin{equation} (x+y)+1 = [(x+1)+(y+1)]-1 \end{equation}
(4)
\begin{equation} (x+y)+1 = (x+y)+1 \end{equation}

Now, finding the identity

(5)
\begin{equation} e*x = x*e = x \end{equation}
(6)
\begin{equation} 1*x = x*1= (1+x)-1 = (x+1)-1= x \end{equation}

Hence, $e=1$.

(b) It is already stated that $\phi$ defines a function that is one to one and onto, meaning that all we have to show is the homomorphism property on $\langle\mathbb{Z}, *\rangle$ and $\langle\mathbb{Z}, +\rangle$. To satisfy the homomorphism property, where $x,y \in \mathbb{Z}$,

(7)
\begin{align} \phi(x*y) = \phi(x) + \phi(y) \end{align}
(8)
\begin{equation} (x*y)+1 = (x+1)+(y+1). \end{equation}

Now, by letting $*$ be define by $x*y = x+y+1$.

(9)
\begin{equation} (x+y+1)+1 = [(x+1)+(y+1)] \end{equation}
(10)
\begin{equation} (x+y)+2 = (x+y)+2 \end{equation}

Now, finding the identity

(11)
\begin{equation} e*x = x*e = x \end{equation}
(12)
\begin{equation} -1*x = x*-1= ((-1)+x)+1 = (x+(-1))+1= x \end{equation}

Hence, $e=-1$.

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