Question

Prove that the function$f:{ℂ}^{\ast }\to {ℝ}^{\ast \ast }$ given by$f(a+bi)=a^2+b^2$ is a surjective homomorphism of groups.

Short answer

It is proved that,$f$ is a homomorphism and $f$is surjective.

Step-by-step solution

To show f is a homomorphism

Definition of Group Homomorphism

Let$(G,\ast )$and$(G\text{'},\ast \text{'})$be any two groups. A function $f:G\to G\text{'}$is said to be a group homomorphism if$f(a\ast b)=f\left(a\right)\ast \text{'}f\left(b\right),\forall a,b\in G$.

Definition of Kernel of a Function

Let $f:G\to H$be a homomorphism of groups. Then the kernel of$f$is defined by the set $\{a\in G:f\left(a\right)=e_H\}$, where $e_H$is an identity element.

It is the mapping from elements in $G$onto an identity element in $H$by the homomorphism $f$.

Let $f:{ℂ}^{\ast }\to {ℝ}^{\ast \ast }$be the function defined as,$f(a+bi)=a^2+b^2$ .

We have to show that the function$f$ is a surjective homomorphism.

Let$a+bi,c+di\in {ℂ}^{\ast }$ .

Then, find $f\left((a+bi)(c+di)\right)$as:

$\begin{array}{c}f\left((a+bi)(c+di)\right)=f((ac-bd)+i(ad+bc))\\ ={(ac-bd)}^2+{(ad+bc)}^2\\ =a^2{c}^2-2acbd+b^2{d}^2+a^2{d}^2+2adbc+b^2{c}^2\\ =a^2{c}^2+b^2{d}^2+a^2{d}^2+b^2{c}^2\end{array}$

Further simplify as:

$\begin{array}{c}f\left((a+bi)(c+di)\right)=a^2(c^2+d^2)+b^2(c^2+d^2)\\ =(a^2+b^2)(c^2+d^2)\\ =f(a+ib)f(c+id)\end{array}$

.

Hence,$f$ is a homomorphism.

To show is surjective

Let$y\in {ℝ}^{\ast \ast }$.

Then, $\sqrt{y}\in {ℝ}^{\ast \ast }$and ${\left(\sqrt{y}\right)}^2=y$.

Take $a=\sqrt{y}$and $b=0$.

Then, $a+ib\in {ℂ}^{\ast }$.

.$\therefore f(a+ib)=a^2+b^2$

Simplify as:

$\begin{array}{c}f(a+ib)=a^2+b^2\\ ={\left(\sqrt{y}\right)}^2+0\\ =y\end{array}$

Hence,$f$ is surjective.

Thus,$f$ is a surjective homomorphism.