Question

Show that the function $f:\square \to \square$given by $f\left(x\right)=3x$is an isomorphism of additive groups.

Short answer

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

Step-by-step solution

Definition of Isomorphism

Let G and H be the groups with the 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) $f(a+b)=f\left(a\right)+f\left(b\right)$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,

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

Hence, f is injective.

f is surjective

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

Thus, for any real number a in co-domain $\square$, there exists an element $\frac{b}{3}$in such that

role="math" localid="1651221830356" $$\begin{gathered} f\left(a\right)=3a \\ =3\left(\frac{b}{3}\right) \\ =b \end{gathered}$$

Hence, f is surjective.

f is a homomorphism

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

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

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

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