Question

Prove that the function $g:\square \to \square *$defined by g(x) =2^x is an injective homomorphism that is not surjective.

Short answer

The function $g:\square \to \square *$defined by g(x) = 2^xis an injective homomorphism that is not surjective.

Step-by-step solution

Definition of Isomorphism 

Let G and H be the groups with a group operation denoted by $*$. Group G isisomorphic 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$a,b\in G$

g is injective

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

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

Hence, g is injective.

g is not surjective

Since,$2^x>0$ for all $x\in \square$so that the image of g is only $\square **$.

Hence, g is not surjective.

g is a homomorphism

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

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

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

Therefore, the function $g:\square \to \square *$defined by $g\left(x\right)=2^x$is an injective homomorphism that is not surjective.