Valuations with the same valuation ring Given a field $k$, a valuation map $v$ is a surjective group homomorphism $k^\times\twoheadrightarrow H$, where $H$ is an ordered group (called the valuation group), such that $v(a+b)\ge \operatorname{min}(v(a),v(b))$. I must prove that if I have another valuation $v':k\twoheadrightarrow H'$, with the property that $v(a)\ge 0\iff v'(a)\ge 0$, then there is an order-preserving isomorphism $f:H\to H'$ such that $v'=f\circ v$.
Firstly, observe that, with $a,b\in k$, $v(a)\ge v(b)$ if and only if $v'(a)\ge v'(b)$: $$v(a)\ge v(b)\iff v(ab^{-1})\ge 0\iff v'(ab^{-1})\ge 0 \iff v'(a)\ge v'(b).$$
From $v'=f\circ v$ we see that on an element $h\in H$ (using that $h=v(a_h)$ for some $a_h\in k$) we must set $f(h):=v'(a_h)$. This shows that if such $f$ is a preserving order isomorphism, it will be the unique one. The observation above guarantees that $f(h)$ doesn't depend on the choice of $a_h$, and that $f$ is order-preserving.
If $h,l\in H$, in order to prove that $f$ is an homomorphism: $$f(h+l)=f(v(a_{h})+v(a_{l}))=f(v(a_ha_l))=v'(a_ha_l)=v'(a_{h})+v'(a_{l})=f(h)+f(l).$$ Plus, $f$ is bijective because the map $f':H'\to H$ defined $h'\mapsto v(a_{h'})$, for $h'\in H'$ and $a_{h'}$ defined analogously to $a_h$, is the inverse of $f$.
Does this outline of the proof make sense? I'm a bit suspicious because I don't find where one should use that $v(a+b)\ge \operatorname{min}(v(a),v(b))$; did I use this hypothesis without seeing it?
 A: Your proof is fine and it is indeed sufficient to consider only the multiplicative structure on $k^\times$, i.e., you need not be worried that you did not use the min-property for addition at all.
Here's the same result from a different perspective:
Any group homomorphism $f\colon G\to H$ from a group $G$ to an ordered group $H$ straightforwardly induces an order on $G/\ker f$ by us declaring those elements of $G/\ker f$ positive/negative that are mapped to positive/negative elements of $H$. With this, the induced monomorphism $\overline f\colon G/\ker f\to H$ is a monomorphism of ordered groups and hence an isomorphism of ordered groups from $G/\ker f$ to the image $\operatorname{im}\overline f$. In particular, if $f$ is onto, we obtain a isomorphism between ordered groups.
Now in the OP situation, we are given that the two valuations $v$ and $v'$ have the same kernel $$\begin{align}N:=\ker v&=\{\,a\in k^\times\mid v(a)\ge0\land v(a^{-1})\ge 0\,\}\\&=\{\,a\in k^\times\mid v'(a)\ge0\land v'(a^{-1})\ge 0\,\}=\ker v'\end{align}$$ and that they induce the same order on the quotient $k^\times/N$. From the above, this is isomorphic as ordered group to both $H$ and $H'$, i.e., $H$ and $H'$ are isomorphic as ordered group, as was to be shown.
