[proofplan]
We use the ramification filtration $G_0 \supseteq G_1 \supseteq \cdots \supseteq G_t = \{\mathrm{id}\}$ and the injective quotient maps $G_s/G_{s+1} \hookrightarrow U_L^{(s)}/U_L^{(s+1)}$. For $s \geq 1$, the target is isomorphic to $(k_L, +) \cong \mathbb{F}_{p^f}$, whose order is a power of $p$, so $|G_1|$ is a power of $p$. For $s = 0$, the target is $k_L^\times$ of order $p^f - 1$, coprime to $p$, so $|G_0/G_1|$ is coprime to $p$. This makes $G_1$ the unique Sylow $p$-subgroup of $G_0$.
[/proofplan]
[step:Show $|G_1|$ is a power of $p$ using the ramification filtration]
Since $\operatorname{Gal}(L/K)$ is finite, the filtration $G_0 \supseteq G_1 \supseteq G_2 \supseteq \cdots$ terminates: there exists $t \geq 1$ with $G_t = \{\mathrm{id}\}$. The order of $G_1$ factors as
\begin{align*}
|G_1| = |G_1/G_2| \cdot |G_2/G_3| \cdots |G_{t-1}/G_t| = \prod_{s=1}^{t-1} |G_s/G_{s+1}|.
\end{align*}
For each $s \geq 1$, the [Quotients of Ramification Groups](/theorems/???) gives an injective group homomorphism $G_s/G_{s+1} \hookrightarrow U_L^{(s)}/U_L^{(s+1)}$, and by the [Quotients of Higher Unit Groups](/theorems/???), $U_L^{(s)}/U_L^{(s+1)} \cong (k_L, +)$. Since $k_L \cong \mathbb{F}_{p^f}$ where $f = f_{L/K}$, the additive group $(k_L, +)$ has order $p^f$. Therefore $|G_s/G_{s+1}|$ divides $p^f$ for each $s \geq 1$, so $|G_s/G_{s+1}|$ is a power of $p$.
Since $|G_1|$ is a product of powers of $p$, $G_1$ is a $p$-group.
[guided]
The ramification filtration $G_0 \supseteq G_1 \supseteq G_2 \supseteq \cdots$ has the property that each quotient $G_s/G_{s+1}$ injects into $U_L^{(s)}/U_L^{(s+1)}$ via $\sigma \mapsto \sigma(\pi_L)/\pi_L$. For $s \geq 1$, the [Quotients of Higher Unit Groups](/theorems/???) identifies $U_L^{(s)}/U_L^{(s+1)}$ with the additive group of $k_L$.
Why does this force $p$-group structure? The residue field $k_L$ is a finite field of characteristic $p$ with $|k_L| = p^f$, so $(k_L, +) \cong (\mathbb{Z}/p\mathbb{Z})^f$ is an elementary abelian $p$-group. Any subgroup of a $p$-group is a $p$-group, and $G_s/G_{s+1}$ injects into $(k_L, +)$, so $|G_s/G_{s+1}|$ is a power of $p$.
The telescoping product $|G_1| = \prod_{s=1}^{t-1} |G_s/G_{s+1}|$ is then a product of powers of $p$, hence a power of $p$.
[/guided]
[/step]
[step:Show $|G_0/G_1|$ is coprime to $p$]
By the [Quotients of Ramification Groups](/theorems/???) with $s = 0$, there is an injective homomorphism $G_0/G_1 \hookrightarrow U_L^{(0)}/U_L^{(1)} = U_L/U_L^{(1)}$. By the [Quotients of Higher Unit Groups](/theorems/???), $U_L/U_L^{(1)} \cong k_L^\times$, which has order $|k_L| - 1 = p^f - 1$.
Since $\gcd(p, p^f - 1) = 1$ (as $p \mid p^f$ but $p \nmid p^f - 1$), the order $|G_0/G_1|$ divides $p^f - 1$ and is therefore coprime to $p$.
[/step]
[step:Conclude $G_1$ is the unique Sylow $p$-subgroup of $G_0$]
We have shown:
- $G_1$ is a $p$-group (its order is a power of $p$),
- $|G_0 : G_1| = |G_0/G_1|$ is coprime to $p$.
By definition, a Sylow $p$-subgroup of $G_0$ is a $p$-subgroup whose index is coprime to $p$. The subgroup $G_1$ satisfies both conditions, so $G_1$ is a Sylow $p$-subgroup of $G_0$.
Moreover, $G_1$ is normal in $G_0$ (it is the kernel of the homomorphism $\phi_0: G_0 \to U_L/U_L^{(1)}$ from the [Quotients of Ramification Groups](/theorems/???)). A normal Sylow $p$-subgroup is unique (by the Sylow theorems, all Sylow $p$-subgroups are conjugate, and normality means the conjugacy class has exactly one element). Therefore $G_1$ is the unique Sylow $p$-subgroup of $G_0$.
[guided]
Let us verify the Sylow $p$-subgroup claim in detail.
We write $|G_0| = |G_1| \cdot |G_0/G_1|$. We have shown $|G_1|$ is a power of $p$ and $|G_0/G_1|$ divides $p^f - 1$, which is coprime to $p$. So $|G_0| = p^a \cdot m$ with $p^a = |G_1|$ and $\gcd(m, p) = 1$ where $m = |G_0/G_1|$. By Sylow's theorem, a Sylow $p$-subgroup of $G_0$ is a subgroup of order $p^a$. Since $|G_1| = p^a$ and $G_1 \leq G_0$, the subgroup $G_1$ is a Sylow $p$-subgroup.
For uniqueness: the [Quotients of Ramification Groups](/theorems/???) gives a group homomorphism
\begin{align*}
\phi_0 : G_0 \to U_L^{(0)}/U_L^{(1)} \cong k_L^\times
\end{align*}
defined by $\sigma \mapsto \sigma(\pi_L)/\pi_L \pmod{U_L^{(1)}}$, and $G_1 = \ker(\phi_0)$. As the kernel of a group homomorphism, $G_1$ is normal in $G_0$.
By the Sylow theorems, all Sylow $p$-subgroups of $G_0$ are conjugate. Since $G_1 \trianglelefteq G_0$, it is fixed by conjugation, meaning it is its own conjugacy class, which has exactly one element. Therefore $G_1$ is the unique Sylow $p$-subgroup of $G_0$.
Uniqueness has the following consequence: any group automorphism of $G_0$ must send $G_1$ to itself. In particular, $G_1$ is characteristic in $G_0$, meaning it is preserved by all automorphisms of $G_0$, not just inner ones.
[/guided]
[/step]
