Question

If $\mathit{f}\mathbf{:}\mathbf{\mathbb{Z}}\mathbf{\to }\mathbf{\mathbb{Z}}$is an isomorphism, prove that $\mathit{f}$is the identity map. [Hint: What are $\mathit{f}\left(1\right)\mathbf{,}\mathit{f}\left(1+1\right)\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{?}$ ]

Short answer

Hence proved that $f$is an identity map.

Step-by-step solution

Property of Rings

If any ring$f$has elements such that,$a,b\in R$ , then, addition and multiplication of the function of its elements is respectively given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Isomorphism

We have an isomorphism as:$f:\mathrm{\mathbb{Z}}\to \mathrm{\mathbb{Z}}$

In this case, we get:

$$\begin{gathered} f\left(1\right)=f\left(1+0\right) \\ =f\left(1\right)+f\left(0\right) \\ f\left(0\right)=0................\left\{1\right\} \end{gathered}$$

Since, the function is surjective due to isomorphism, we have:

$$\begin{gathered} f\left(n\right)=f\left(n\right)f\left(1\right) \\ f\left(1\right)=1..............\left\{2\right\} \end{gathered}$$

Also, for some positive integer, $n$,$\forall n\in \mathrm{\mathbb{Z}}$ again we have:

$$\begin{gathered} f\left(0\right)=f\left(n+\left(-n\right)\right) \\ 0=f\left(n\right)+f\left(-n\right) \\ 0=n+f\left(-n\right) \\ f\left(-n\right)=-n....................\left\{3\right\} \end{gathered}$$

From, equations 1, 2, and 3, we have:

Hence proved, $f$is an identity map.