Question

Describe the quotient group ${\mathrm{ℝ}}^{\ast }/{\mathrm{ℝ}}^{\ast \ast }$, where ${\mathrm{ℝ}}^{\ast }$and ${\mathrm{ℝ}}^{\ast \ast }$ are as in Exercise 22.

Short answer

The quotient group${\mathrm{ℝ}}^{\ast }/{\mathrm{ℝ}}^{\ast \ast }$ is $\{1,-1\}$.

Step-by-step solution

First isomorphic theorem

Theorem 8.20

Let $\mathit{f}\mathbf{:}\mathit{G}\mathbf{\to }\mathit{H}$ be a surjective homomorphism of a group with kernel K. Then quotient group $\mathit{G}\mathbf{/}\mathit{K}$ is isomorphic to H.

Describing the quotient group ℝ∗/ℝ∗∗

As given in the question

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

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

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

So, it can be defined by

$\varphi \left(h\right)=\left\{\begin{array}{cc}1& h>1\\ -1& h<1\end{array}\right.$

Therefore, from first theorem of isomorphism, the quotient group can be given as

$N=\varphi \left(h\right)\Rightarrow N=\{1,-1\}.$