Question

Prove that the function $f:\square *\to \square *$defined by $f\left(x\right)=x^3$is an isomorphism.

Short answer

The function $f:\square *\to \square *$defined by$f\left(x\right)=x^3$ is an isomorphism.

Step-by-step solution

Definition of Isomorphism

Let G and H be groups with a group operation denoted by. Group G is isomorphic to group Hif 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 $f\left(a\right)=f\left(b\right)$implies,

$$\begin{gathered} f\left(a\right)=f\left(b\right) \\ {a}^3=b^3 \\ a=b \end{gathered}$$

Hence, f is injective.

f is surjective

Let $a\in \square *$in co-domain then $f\left(a\right)=b$, such a³= b that implies.$a=\sqrt[3]{b}$

Thus, for any real number a in $\square *$, there exists an element $\sqrt[3]{b}$in $\square *$such that,

role="math" localid="1651228677821" $$\begin{gathered} f\left(a\right)=a^3 \\ =b^{3\left(\frac{1}{3}\right)} \\ =b \end{gathered}$$

Hence, f is surjective.

f is a homomorphism

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

$$\begin{gathered} f\left(ab\right)={\left(ab\right)}^3=a^3{b}^3 \\ =f\left(a\right)f\left(b\right) \end{gathered}$$

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

Therefore, the function $f:\square *\to \square *$defined by$f\left(x\right)=x^3$ is an isomorphism.