Chapter 3.6, Problem 1TFE

Label each of the following statements as either true or false.

Every homomorphism is an isomorphism.

To determine

Whether the statement, “Every homomorphism is an isomorphism” is true or false.

Answer to Problem 1TFE

Solution:

The statement, “Every homomorphism is an isomorphism” is false.

Explanation of Solution

Formula used:

1) Homomorphism: Let $G$ be a group with respect to $\ast$ and $G\text{'}$ be the group with respect to $\ast \text{'}$. A mapping $\varphi :G\to G\text{'}$ is homomorphism from $G$ to $G\text{'}$ if

$\varphi (x\ast y)=\varphi (x)\ast \text{'}\varphi (y)$ for all x, y in $G$.

2) Isomorphism: Let $G$ be a group with respect $\ast$ and $G\text{'}$ be a group with respect to $\ast \text{'}$. A mapping $\varphi :G\to G\text{'}$ is an isomorphism from $G$ to $G\text{'}$ if

1. $\varphi$ is one-to-one correspondence from $G$ to $G\text{'}$

(bijective) and

2. $\varphi (x\ast y)=\varphi (x)\ast \text{'}\varphi (y)$ for all $x$ and $y$ in $G$

(homomorphism).

Explanation:

Consider the statement, “Every homomorphism is an isomorphism.”

Counter example:

Let $\mathbb{Z}$ be a group with operation multiplication.

Mapping $\varphi :\mathbb{Z}\to \mathbb{Z}$ by $\varphi (x)=x^2$.

First to check whether $\varphi$ is a homomorphism.

$\varphi (x\cdot y)$

$={(x\cdot y)}^2$

$=x^2\cdot y^2$

$=\varphi (x)\cdot \varphi (y)$

Thus, $\varphi$ is a homomorphism.

To check whether $\varphi$ is isomorphism.

$\begin{array}{l}\varphi (-x)={(}^{-}=x^2\\ \\ \varphi (x)={(}^x=x^2\end{array}$

As $x^2=x^2\Rightarrow \varphi (-x)=\varphi (x)$ but $-x\ne x$.

Therefore, $\varphi$ is not one-to-one correspondence.

So, $\varphi$ is not an isomorphism.

Thus $\varphi$ is a homomorphism but not an isomorphism.

Hence, the statement “Every homomorphism is an isomorphism” is false.