Question

Prove that the function $h:{\square }_8\to {\square }_8$ defined by h(x) = 2x is a homomorphism that is neither injective nor surjective.

Short answer

The function $h:{\square }_8\to {\square }_8$defined by h(x) = 2x is a homomorphism that is neither injective nor surjective.

Step-by-step solution

Definition of Isomorphism

Let G and H be the groups with the group operation denoted by + . The group G isisomorphic to a group H if there is a function $f:G\to H$such that,

(i) f is injective

(ii) f is surjective

(iii) $f(a+b)=f\left(a\right)+f\left(b\right)$ for all $a,b\in G$

h is not injective

The elements in the group ${\square }_8$ are {0,1,2,3,4,5,6,7}.

Let $4,0\in {\square }_8$ such that, h(4) =h(0) implies,

h(4) = h(0)

2(4) = 2(0)

8 = 0 (since, 8=0mod8)

0 = 0

Here, h(4)=0 is not possible. Hence, h is not injective.

h is not surjective

Since, h is not injective and ${\square }_8$ is finite implies that h cannot be surjective.

Hence, h is not surjective.

h is a homomorphism

Let $a,b\in {\square }_8$then,

h(a+b) = 2(a+b)

= 2(a) + 2(b)

= h(a) + h(b)

Thus, for all $a,b\in {\square }_8$, h(a+b) = h(a) + h(b) and hence, h is a homomorphism.

Therefore, the function $h:{\square }_8\to {\square }_8$defined by h(x) = 2x is a homomorphism that is neither injective nor surjective.