Let $L/K$ be a Galois extension of number fields with $p$-power degree ($p$ prime) in which exactly one prime, say $\mathfrak{q}$ ramifies. Let $H_L$ be the $p$-Hilbert class field for $L$ and suppose that $[H_L:L] > 1$ so that $p$ divides the class number of $L$. Now, if $G =$ Gal($H_L/K)$, we have one of the inertia subgroups $T$ for the single ramified prime $\mathfrak{q}$ in $H_L/K$. Since $G$ is a $p$-group, there is a normal subgroup $G_1$ containing $T$ in $G$ which is index $p$ in $G$. Hence, if $F$ is the fixed field of $G_1$, we have that $F/K$ is unramified since it contains the fixed field for $T$, and Abelian since it is degree $p$. Hence $p$ must divide the class number of $K$.
I feel like the proof is done, but it is also mentioned that since $G_1$ is normal, it also contains all the other inertia subgroups of $\mathfrak{q}$ in $H_L/k$ (since they are all conjugate). My question is, why do we have to worry about all the other inertia groups? Shouldn't even just one inertia group $T$ contained in $G_1$ give us that $F/K$ is unramified? Is there something important from Galois/algebraic number theory that I'm forgetting?