EDIT. Here is the part of the answer that has been rewritten:
We give below a short proof of the Fundamental Theorem of Galois Theory (FTGT) for finite degree extensions. We derive the FTGT from two statements, denoted (a) and (b). These two statements, and the way they are proved here, go back at least to Emil Artin (precise references are given below).
The derivation of the FTGT from (a) and (b) takes about four lines, but I haven't been able to find these four lines in the literature, and all the proofs of the FTGT I have seen so far are much more complicated. So, if you find either a mistake in these four lines, or a trace of them the literature, please let me know.
The argument is essentially taken from Chapter II (link) of Emil Artin's Notre Dame Lectures [A]. More precisely, statement (a) below is implicitly contained in the proof Theorem 10 page 31 of [A], in which the uniqueness up to isomorphism of the splitting field of a polynomial is verified. Artin's proof shows in fact that, when the roots of the polynomial are distinct, the number of automorphisms of the splitting extension coincides with the degree of the extension. Statement (b) below is proved as Theorem 14 page 42 of [A]. The proof given here (using Artin's argument) was written with Keith Conrad's help.
Theorem. Let $E/F$ be an extension of fields, let $a_1,\dots,a_n$ be distinct generators of $E/F$ such that the product of the $X-a_i$ is in $F[X]$. Then
the group $G$ of automorphisms of $E/F$ is finite,
there is a bijective correspondence between the sub-extensions $S/F$ of $E/F$ and the subgroups $H$ of $G$, and we have
$$
S\leftrightarrow H\iff H=\text{Aut}(E/S)\iff S=E^H
$$
$$
\implies[E:S]=|H|,
$$
where $E^H$ is the fixed subfield of $H$, where $[E:S]$ is the degree (that is the dimension) of $E$ over $S$, and where $|H|$ is the order of $H$.
PROOF
We claim:
(a) If $S/F$ is a sub-extension of $E/F$, then $[E:S]=|\text{Aut}(E/S)|$.
(b) If $H$ is a subgroup of $G$, then $|H|=[E:E^H]$.
Proof that (a) and (b) imply the theorem. Let $S/F$ be a sub-extension of $E/F$ and put $H:=\text{Aut}(E/S)$. Then we have trivially $S\subset E^H$, and (a) and (b) imply
$$
[E:S]=[E:E^H].
$$
Conversely let $H$ be a subgroup of $G$ and set $\overline H:=\text{Aut}(E/E^H)$. Then we have trivially $H\subset\overline H$, and (a) and (b) imply $|H|=|\overline H|$.
Proof of (a). Let $1\le i\le n$. Put $K:=S(a_1,\dots,a_{i-1})$ and $L:=K(a_i)$. It suffices to check that any $F$-embedding $\phi$ of $K$ in $E$ has exactly $[L:K]$ extensions to an $F$-embedding $\Phi$ of $L$ in $E$; or, equivalently, that the polynomial $p\in\phi(K)[X]$ which is the image under $\phi$ of the minimal polynomial of $a_i$ over $K$ has $[L:K]$ distinct roots in $E$. But this is clear since $p$ divides the product of the $X-a_j$.
Proof of (b). In view of (a) it is enough to check $|H|\ge[E:E^H]$. Let $k$ be an integer larger than $|H|$, and pick a
$$
b=(b_1,\dots,b_k)\in E^k.
$$
We must show that the $b_i$ are linearly dependent over $E^H$, or equivalently that $b^\perp\cap(E^H)^k$ is nonzero, where $\bullet^\perp$ denotes the vectors orthogonal to $\bullet$ in $E^k$ with respect to the dot product on $E^k$. Any element of $b^\perp\cap (E^H)^k$ is necessarily orthogonal to $hb$ for any $h\in H$, so
$$
b^\perp\cap(E^H)^k=(Hb)^\perp\cap(E^H)^k,
$$
where $Hb$ is the $H$-orbit of $b$. We will show $(Hb)^\perp\cap(E^H)^k$ is nonzero. Since the span of $Hb$ in $E^k$ has $E$-dimension at most $|H| < k$, $(Hb)^\perp$ is nonzero. Choose a nonzero vector $x$ in $(Hb)^\perp$ such that $x_i=0$ for the largest number of $i$ as possible among all nonzero vectors in $(Hb)^\perp$. Some coordinate $x_j$ is nonzero in $E$, so by scaling we can assume $x_j=1$ for some $j$. Since the subspace $(Hb)^\perp$ in $E^k$ is stable under the action of $H$, for any $h$ in $H$ we have $hx\in(Hb)^\perp$, so $hx-x\in(Hb)^\perp$. Since $x_j=1$, the $j$-th coordinate of $hx-x$ is $0$, so $hx-x=0$ by the choice of $x$. Since this holds for all $h$ in $H$, $x$ is in $(E^H)^k$.
[A] Emil Artin, Galois Theory, Lectures Delivered at the University of Notre Dame, Chapter II, available here.
PDF version: http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Selected_Texts/st.pdf
Here is the part of the answer that has not been rewritten:
Although I'm very interested in the history of Galois Theory, I know almost nothing about it. Here are a few things I believe. Thank you for correcting me if I'm wrong. My main source is
http://www-history.mcs.st-and.ac.uk/history/Projects/Brunk/Chapters/Ch3.html
Artin was the first mathematician to formulate Galois Theory in terms of a lattice anti-isomorphism.
The first publication of this formulation was van der Waerden's "Moderne Algebra", in 1930.
The first publications of this formulation by Artin himself were "Foundations of Galois Theory" (1938) and "Galois Theory" (1942).
Artin himself doesn't seem to have ever explicitly claimed this discovery.
Assuming all this is true, my (probably naive) question is:
Why does somebody who makes such a revolutionary discovery wait so many years before publishing it?
I also hope this is not completely unrelated to the question.
I don't think Dedekind's complaint can be that a proof in an axiomatic system must be invalid if the objects in the system have not been constructed from simpler objects. This obviously leads to an infinite regress. Today we typically take "set" to be undefined, but Dedekind would surely have been familiar with treatments of Euclidean geometry in which "point" is an undefined term.
I suspect that Dedekind was complaining that people were not working in a full axiomatic system for the reals, but rather were using only the elementary axioms and did not realize that the completeness axiom was needed. Note that in the quote, the argument he criticizes uses only elementary properties such as commutativity and associativity.
Without completeness, we don't have any way of proving the existence of square roots, but the textbook treatments he complains about presumably don't admit that the theorem $\sqrt{2}\sqrt{3}=\sqrt{6}$ is conditional on the unproved existence of these roots.
Why couldn't it have been proved by Euclid?
Euclid couldn't have proved it to modern standards of rigor because Euclid couldn't have proved that the circles in your diagram intersected. This is equivalent to lacking the completeness property of the reals.
Best Answer
Gauss, in his Disquisitiones, used ad hoc notation for the class number when he needed it. He did not use h. Dirichlet used h for the class number in 1838 when he proved the class number formula for binary quadratic forms. I somewhat doubt that he was thinking of "Hauptform" in this connection - back then, the group structure was not as omnipresent as it is today, and the result that $Q^h$ is the principal form was known (and written additively), but did not play any role. Kummer, 10 years later, used H for the class number of the field of p-th roots of unity, and h for the class number of a subfield generated by Gaussiam periods (and "proved" that $h \mid H$); in the introduction he quotes Dirichlet's work on forms at length.