Verify that the map $h: \mathbb{Z}_{18} \rightarrow \mathbb{Z}_3$, given by $h(x)=2x$, is a group homomorphism and find its kernel.

Question

Verify that the map $h: \mathbb{Z}_{18} \rightarrow \mathbb{Z}_3$, given by $h(x)=2x$, is a group homomorphism and find its kernel.

In order to verify it is a group homomorphism let $x,y \in \mathbb{Z}_{18}$ then I want to show that $h(xy)=h(x)h(y)$

$h(xy)=2xy$

Yet notice that

$h(x)h(y)=(2x)(2y)=xy$

I'm not sure how to show this group homomorphism is operation preserving.

The kernel would be all $x \in \mathbb{Z}_{18}$ such that $h(x)=e$ so the set $\left\{2,5,8 \right\}$ would be the kernel of this homomorphism.

Answer 1

Here the group operation is addition and not multiplication. $\Bbb{Z}_18$ and $\Bbb{Z}_3$ are additive groups, not multiplicative.

If you want to show that $h:G\rightarrow G'$ is a homomorphism then you need to prove $h(x\circ y)=h(x)\cdot h(y)$ (where $\circ$ is the operation for $G$ and $\cdot$ is the operation of $G'$)

$\Bbb{Z}_{18}$ is a group under the operation of addition mod $18$ and $\Bbb{Z}_3$ is a group under the operation of addition mod 3.