how $e=\text{number of elements of $G'/G$}=2$?

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Consider $\Bbb Q_3$ and its quadratic extension $\Bbb Q_3(\sqrt{3})$. I want to find the ramification index $e$.

We have the valuation groups as follows:

For $\Bbb Q_3$, the valuation group $G=\{|a|_3: a \in \Bbb Q_3, \ a \neq 0 \}$ is generated by $3$,

For $\Bbb Q_3(\sqrt{3})$, the valuation group $G'=\{|b|: b \in \Bbb Q_3, \ b \neq 0 \}$ is generated by $\sqrt 3$.

Thus, $e=\text{number of elements of $G'/G$}= ?$

As $\Bbb Q_3(\sqrt 3)$ is ramified, so $e=2$.

But how $e=\text{number of elements of $G'/G$}=2$ ?

Can you explain it please?