Question

Prove that the function $f:\square *\to \square **$defined by f(x) = |x| is a surjective homomorphism that is not injective.

Short answer

The function $f:\square *\to \square **$ defined by$f\left(x\right)=\left|x\right|$ is a surjective homomorphism that is not injective.

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 a group H if there is a function $f:G\to H$such that,

(i) f is injective

(ii) f is surjective

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

f is not injective

Let $-2\in \square *$such that,

$$\begin{gathered} f(-2)=\left|-2\right| \\ =\left|2\right| \\ =2 \end{gathered}$$

Here, $f(-2)=2$ is not possible. Hence, f is not injective.

f is surjective

If $x\in \square **$ then there exists x > 0 such that x = |x| and then, f(x) = x .

Hence, f is surjective.

f is a homomorphism

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

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

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 $f\left(x\right)=\left|x\right|$ is a surjective homomorphism that is not injective.