Question

Show that the function$f:\square \to \square$ defined by$f\left(x\right)=x^2$ is not a homomorphism.

Short answer

The function$f:\square \to \square$ defined by$f\left(x\right)=x^2$ is not a homomorphism

Step-by-step solution

Definition of Isomorphism

Let G and H be the groups with the group operation denoted by +. Group G is isomorphicto a 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(a) + f(b) 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,

f(a)=f(b)

a²= b²

a = b

Hence, f is injective.

f is surjective

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

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

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

Hence, f is surjective.

f is not a homomorphism

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

role="math" localid="1651257752774" $$\begin{gathered} f(a+b)={(a+b)}^2 \\ =a^2+b^2+2ab \\ \ne f\left(a\right)+f\left(b\right) \end{gathered}$$

Thus, for all $a,b\in \square$,$f(a+b)\ne f\left(a\right)+f\left(b\right)$ and hence, f is not a homomorphism.

Therefore, the function $f:\square \to \square$defined by $f\left(x\right)=x^2$is not a homomorphism.