Chapter 6, Problem 1E

To determine

To find: An isomorphism from to group of integers under addition to the group of even integers under addition.

Explanation of Solution

Given information:

Concept used:

Isomorphism: - A homomorphism $\varphi$ from G into $\overline{G}$ is said to be an isomorphism if $\varphi$ is one-one.

Isomorphic groups: - When an isomorphism of G onto $G^{*}$ exist then the two groups $G,G^{*}$ are called isomorphic.

One-one function: - A function which maps distinct elements from the domain to the distinct elements of the co-domain is said to be one-one function.

Onto function: - A function $f:X\to Y$ is said to be onto if for every element (let y ) in the co-domain Y of f , there exist at least one element (let x ) in the domain, that is X such that $f\left(x\right)=y$ .

Calculation:

Construct a function $f:\mathbb{Z}\to \langle 2\rangle \text{ by }f\left(x\right)=2x\text{ such that }x\in \mathbb{Z}$

Let us consider two elements as $a,b\text{ such that }a,b\in \mathbb{Z}$

Let $f\left(a\right)=f\left(b\right)$

Then from the defined mapping $f\left(a\right)=2a\text{ and }f\left(b\right)=2b$

So,

  $\begin{array}{l}2a=2b\\ \Rightarrow a=b\left(\text{dividing both side by 2}\right)\end{array}$

Hence it is clear that the function is one-one.

Now consider $x\in \langle 2\rangle$ , then there exist an integer c such that

  $\begin{array}{c}f\left(c\right)=2c\\ =x\end{array}$

So, f is onto.

Now

  $\begin{array}{c}f\left(a+b\right)=2\left(a+b\right)\\ =2a+2b\\ =f\left(a\right)+f\left(b\right)\end{array}$

Hence f is a group isomorphism and hence $\mathbb{Z}\approx \langle 2\rangle$ .