Question

Show that the function $g:\square **\to \square **$ given by$g\left(x\right)=\sqrt{x}$ is an isomorphism.

Short answer

It is proved that the function $g:\square **\to \square **$defined by $g\left(x\right)=\sqrt{x}$is an isomorphism.

Step-by-step solution

Definition of Isomorphism

Let G and H be the groups with a group operation denoted by **. Group G is isomorphic to group H if there is a function $f:G\to H$such that:

(i) f is injective

(ii) f is surjective

(iii) $f\left(ab\right)=f\left(a\right)f\left(b\right)$for all a, b$\in$G

f is injective

Let $a,b\in \square **$such that $g\left(a\right)=g\left(b\right)$implies,

$$\begin{gathered} g\left(a\right)=g\left(b\right) \\ \sqrt{a}=\sqrt{b} \\ {a}^{\frac{1}{2}}=b^{\frac{1}{2}} \\ a=b \end{gathered}$$

Hence, g is injective.

f is surjective

Let $a\in \square **$in co-domain then,$g\left(a\right)=b$such that $\sqrt{a}=b$implies $a=b^2$.

Thus, for any real number a in $\square **$, there exists an element$b^2$in $\square **$such that,

$$\begin{gathered} g\left(a\right)=a^{\frac{1}{2}} \\ =b^{2\left(\frac{1}{2}\right)} \\ =b \end{gathered}$$

Hence, g is surjective.

f is a homomorphism

Let $a,b\in \square **$then,

$$\begin{gathered} g\left(ab\right)=\sqrt{ab} \\ =\sqrt{a}.\sqrt{b} \\ =g\left(a\right).g\left(b\right) \end{gathered}$$

Thus, for all a,b$\in \square **$, $g\left(ab\right)=g\left(a\right)g\left(b\right)$and hence, g is a homomorphism.

Therefore, the function$g:\square **\to \square **$defined by$g\left(x\right)=\sqrt{x}$is an isomorphism.