Question

Let ${\mathbf{\mathbb{Q}}}^{*}$be the multiplicative group of nonzero rational numbers, ${\mathbf{\mathbb{Q}}}^{\mathbf{*}\mathbf{*}}$the subgroup of positive rationals, and $\mathbf{H}$the subgroup $\left\{1,-1\right\}$. Prove that ${\mathbf{\mathbb{Q}}}^{*}\mathbf{=}{\mathbf{\mathbb{Q}}}^{\mathbf{*}\mathbf{*}}\mathbf{\times }\mathbf{H}$.

Short answer

Answer:

It is proved that ${\mathrm{\mathbb{Q}}}^{*}={\mathrm{\mathbb{Q}}}^{**}\times H$.

Step-by-step solution

Find ℚ**∩H

Definition of subgroup:

Let $\left(G,*\right)$be a group under binary operation, say $*$. A non-empty subset $H$of $G$is said to be a subgroup of $G$, if $\left(H,*\right)$is itself a group.

In other words, if $e_G\in H$and $a,b\in H$then, $a{b}^{-1}\in H$.

Definition of Normal Subgroup:

A subgroup $N$of a group $G$is called a normal subgroup of $G$if $Na=aN,\forall a\in G$.

Let ${\mathrm{\mathbb{Q}}}^{*}$be the multiplicative group of rational numbers and the ${\mathrm{\mathbb{Q}}}^{**}$subgroup of positive rationals.

Let $H$be the subgroup of $\left\{1,-1\right\}$.

We have to prove that ${\mathrm{\mathbb{Q}}}^{*}={\mathrm{\mathbb{Q}}}^{**}\times H$.

Theorem 9.3can be stated as follows:

If $M$and $N$are normal subgroups of a group $G$such that $G=MN$and , then $G=M\times N$.

Now, ${\mathrm{\mathbb{Q}}}^{**}\cap H=\left\{1\right\}$, and both ${\mathrm{\mathbb{Q}}}^{**}$and $H$are normal subgroups of ${\mathrm{\mathbb{Q}}}^{*}$, since ${\mathrm{\mathbb{Q}}}^{*}$is abelian.

Prove ℚ*=ℚ**×H

For given any $a\in {\mathrm{\mathbb{Q}}}^{*}$, we have $a=a$.1 if is positive and $a=\left(-a\right)\left(-1\right)$if a is negative.

Thus, ${\mathrm{\mathbb{Q}}}^{*}={\mathrm{\mathbb{Q}}}^{**}H$.

Hence, byTheorem 9.3 ${\mathrm{\mathbb{Q}}}^{*}={\mathrm{\mathbb{Q}}}^{**}\times H$.