Question

Let ${\mathrm{ℝ}}^{\ast }$ be the multiplicative group of nonzero real numbers and let N be the subgroup $\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{\}}$. Prove that ${ℝ}^{\ast }/N$ is isomorphic to the multiplicative group ${ℝ}^{\ast \ast }$ of positive real numbers.

Short answer

It is proved that ${\mathrm{ℝ}}^{\ast }/\mathrm{N}$ is isomorphic to ${\mathrm{ℝ}}^{\ast \ast }$.

Step-by-step solution

As given in the question

${\mathrm{ℝ}}^{\ast }$= multiplicative group of nonzero numbers.

${\mathrm{ℝ}}^{\ast \ast }$= multiplicative group of positive real numbers.

$\mathbf{N}\mathbf{=}\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{\}}$

Proving that ℝ∗/N  is isomorphic to ℝ∗∗

Let, $\varphi :{ℝ}^{\ast }\to {ℝ}^{\ast \ast }$

So, it can be defined by

$\varphi \left(h\right)=\left|h\right|$

First, we will show that $\Phi$ is a homomorphism.

Taking two arbitrary $a,b\in {ℝ}^{\ast }$

$\begin{array}{c}\varphi \left(ab\right)=\left|a\right|\left|b\right|\\ =\varphi \left(a\right)\varphi \left(b\right)\end{array}$

Which shows that $\varphi$ is a homomorphism.

If $h\in \ker \varphi$, then $\varphi \left(h\right)=\left|h\right|=1$

Since h is the identity element of ${ℝ}^{\ast }$, it can be either 1 or -1. Hence = N. So $\varphi$ is surjective too.

Therefore, from the first theorem of isomorphism and the above results, we can conclude that ${ℝ}^{\ast }/N$ is isomorphic to ${\mathrm{ℝ}}^{\ast \ast }$.