of basis over the field $K^{1/p^{t_1}}$, sending $$(X - \alpha_1^{1/p^{t_1}}Y) \rightarrow X'$$ $$Y \rightarrow Y'$$

and calling the resulting polynomial $f'$ (which is a translate of $f$ by an element in $Sl(2,K^{1/p^{t_1}})$), we see that $f'$ has the form $$f'= X'^{p^{t_1}}(X'-\beta_1 Y')^{p^{t_2}} \cdots (X'-\beta_r Y')^{p^{t_r}}$$ with all the $\beta_i$’s distinct. Note that $\beta_1,...,\beta_r$ belong to $K^{1/p^{t_1}}$. Since $f$ has a unique root of multiplicity $> N/2$, we see that the factor occuring in the above factorization with the highest power is neccessarily unique. i.e. $t_1$ is unique.\
$~$ Claim 2: The 1-PS $\mu(t)$ is an instability 1-PS of $f'$.\
Proof of claim 2 : The proof of the claim is quite obvious. We only sketch it briefly. Note that $\nu(f',\mu)$ = $t_1/(\mid\mid \mu \mid\mid)$. Suppose there exists another 1-PS $\mu'(t)$ such that $\nu(f',\mu') >
\nu(f',\mu)$. Since all 1-PS’s of $G$ are conjugates over ${\bar K}$, there exists an element $h \in G(\bar
{K})$ which conjugates $\mu$ into $\mu'$. Then $\mu(t)$ will be the instability 1-PS of $hf'$. It is easy to see that the highest power of $X'$ occuring in $f'$ is greater than or equal to the highest power of $X'$ occuring in $hf'$. Hence we see that $m(f',\mu) 