Marvin J. Greenberg provided an elementary proof of the Kronecker–Weber theorem here (Amer. Math. Monthly, 81 (1974), no. 6, 601-607).
An argument in the lemma 4 was found to be wrong as noticed in *Correction to "An Elementary Proof of the Kronecker-Weber Theorem" * (Amer. Math. Monthly, 82 (1975), no. 8, 803):

Joe L. Mott informed me that the argument for the case $m>1$ in Lemma 4, Volume 81, (1974) 606, is incorrect. What is correct is that $V_i$ is the unique subgroup of $G$ of index $\lambda$, where $i$ is the smallest index such that $V_i \neq G$.

The lemma 4 is:

Let $K$ be an abelian extension of $\Bbb Q$ of degree $\lambda^m$, $\lambda$ an odd prime, in which $\lambda$ is the only ramified prime. Then $K/\Bbb Q$ is cyclic.

The case $m=1$ is dealt correctly. **However, I wasn't able to see where exactly Greenberg's proof fails for the case $m>1$** (the numbering and square brackets are mine) :

Returning to the case $m > 1$, we will show that $K/\Bbb Q$ is cyclic by showing $V_2$

[i.e. the second ramification group of $\lambda$ in $K$]is the unique subgroup of the Galois group $G = V_1$ of index $\lambda$. Let $H$ be any subgroup of index $\lambda$ in $G = \mathrm{Gal}(K/\Bbb Q)$, $K'$ its fixed field, $G' \cong G /H$ the Galois group of $K'$ over $\Bbb Q$, $V_j'$ the $j$-th ramification group of $K'$.(1) By restriction to $K'$, $V_j$ maps into $V_j'$. According to the sublemma

[i.e. the case $m=1$], $V_2'$ is trivial. Hence $V_2 \leq H$.(2) Applying this, in particular to the case where $H = V_j$ is the first ramification group which is not all of $G$, we see that $j = 2$ and $V_2$ has index $\lambda$. Hence $V_2$ is the unique subgroup of index $\lambda$.

I see no problem with (1) : let $f : G \to G' \cong G/H$ be the restriction to $K'$. Then $f(V_j) \subset V_j'$. Since $V_2' = 1$, we get $V_2 \subset \ker(f) = H$.

I see no problem with (2), since $V_{j-1} / V_j$ is non trivial, and embeds in $O_K/(\lambda)$ (fact 3 in Greenberg's paper), which has cardinality $\lambda$ because $\lambda$ is totally ramified in $K$ (see the very beginning of the proof of Lemma 4). So $V_j$ has indeed index $\lambda$ in $G$.

I tried to find a counter-example of an abelian extension $K/\Bbb Q$ whose degree and discriminant are both powers of an odd prime $\lambda$, such that the second ramification group $V_2$ of $\lambda$ (in $K$) is the whole Galois group, but it was without success. Maybe the following MAGMA code could help:

```
p := 3; r := 3;
a := RootOfUnity(p^r);
M := MinimalPolynomial(a + 1/a); #here K=NumberField(M) will be the subfield of Q(a) fixed by a subgroup of order p-1 in Gal(Q(a))
R<x> := PolynomialRing(RationalField());
K<a> := NumberField(M);
OK := RingOfIntegers(K); OK;
print " ";print " ";
Gal, _, Map := AutomorphismGroup(K); Gal;
P3 := Decomposition(OK, 3)[1][1]; P3;
print " ";print " ";
V2 := RamificationGroup(P3, 2); V2;
print "Cardinality of V2 is ", #V2;
```

Thank you very much for your help.