Does there exist an example that the image of a valuation $\textrm{Im}(v)\setminus \{0\}$ is not a group?

Question

Consider a multiplicative valuation $v:K \rightarrow \Gamma\cup\{0\}$, where $\Gamma$ is a totally ordered abelian group and $K$ is a field. I saw the definition of its value group in some references, that a value group of a valuation $v$ is defined to be the subgroup generated by $\operatorname{Im}(v)\setminus\{0\}$. However, in other references, the value group is $\operatorname{Im}(v)\setminus\{0\}$. Is there an example where the image of a valuation $\operatorname{Im}(v)\setminus\{0\}$ is not a group? Or the two definitions are in fact the same?

Answer 1

Note that if $u \in K$ we have that $v(u) = 0$ if and only if $u = 0$. Then by definition $v$ is a homomorphism taking the group $K^\times = K \setminus \{0\}$ to the group $\Gamma$. By basic group theory, the image is a group.