Admittedly, my question is a bit provocative. Let me be precise.
The fundamental theorem of Galois theory states three things (for a fixed finite normal separable extension $M/K$):
(1) The maps $L\mapsto \mathrm{Gal}(M/L)$ and $H\mapsto \mathrm{Fix}(H)$ determine an anti-isomorphism between the poset of intermediate fields of $M/K$ and the poset of subgroups of $\mathrm{Gal}(M/K)$.
(2) For each intermediate field $L$, $|\mathrm{Gal}(M/L)| = [M:L]$ and for each subgroup $H$ of $\mathrm{Gal}(M/K)$, $[M:\mathrm{Fix}(H)]=|H|$.
(3) $L$ is a normal extension of $K$ if and only if $\mathrm{Gal}(M/L)$ is a normal subgroup of $\mathrm{Gal}(M/K)$.
The main ingredient of Galois' proof of the Abel-Ruffini theorem is the following theorem that I will denote by $(*)$: a polynomial $f$ over $\mathbb Q$ is solvable if and only if $\mathrm{Gal}(f)$ is solvable.
Certainly the proof of $(*)$ uses (3) of the fundamental theorem of Galois theory to translate between the solvability of $f$ and $\mathrm{Gal}(f)$.
Question: Does the proof of $(*)$ use (1) somewhere? Or can one prove $(*)$ without proving (1)?