\in Sl(2, \bar K)$ such that $g\lambda(t)g^{-1}$ is of the form $\mu(t)$. Then $\mu(t)$ is the instability 1-PS of $g \cdot f$. Let suppose $g \cdot f$ have the form: $$g\cdot f=X^Tg$$ for some nonnegative integer $T$ and some polynomial $g \in S^N(V)$ which is not divisible by $X$. Since $\mu(t)$ drives $g \cdot f$ to $0$, $T$ neccessarily satisfies $N/2 < T \leq N$. i.e. $f$ has a zero of multiplicity greater than $N/2$ on $\mathbb{P}^1_{\bar K}$ and hence a unique such zero.\
$~$Now, by using the fact that $K$ is seperably closed, by a suitable change of basis made over $K$, we can assume that $f$ can be factorized in the form: $$f= F_1 \cdot F_2 \cdots F_r$$ for some $0 \leq r \leq N$, with deg $F_1 \geq$ deg $F_2 \geq \cdots \geq$ deg $F_r$ and each $F_i$ of the form $(X^{p^{t_i}}- \alpha_i Y^{p^{t_i}})$ for some non-negative integer $t_i$.\
Note that ${t_1 \geq t_2 \geq ... \geq t_r \geq 0}$. Factorizing $f$ into product of linear polynomials over the field $K^{1/p^{t_1}}$, we get : $$f= (X - \alpha_1^{1/p^{t_1}}Y)^{p^{t_1}} \cdots (X - \alpha_r^{1/p^{t_r}}Y)^ {p^{t_r}}$$ Note that by Claim 1, $p^{t_1} = T$. By once again making a change 