Doubling on the integers is not an isomorphism

Example

Let $G = (\mathbb{Z}, +)$ be the integers with addition. Define $d : G \to G$ by $d(u) = 2u$. Then $d(u + v) = 2u + 2v = d(u) + d(v)$, so $d$ is a homomorphism. If $2u = 0$, then $u = 0$. Thus the kernel is $\{0\}$, and $d$ is injective. The image of $d$ is the set of even integers. Thus $1$ is not in the image, and $d$ is not surjective. Thus $d$ is not an isomorphism.

Remarks

The image $2\mathbb{Z}$ is isomorphic to $\mathbb{Z}$. Thus $d$ is an isomorphism from $\mathbb{Z}$ onto $2\mathbb{Z}$, but not onto $\mathbb{Z}$.