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Let $p$ be a prime. I'm trying to compute the subgroups of index $3$ in $\mathbb{Q}_p^\times$ to enumerate some cyclic extensions using CFT. I've essentially reduced the problem down to finding the index $3$ subgroups of $1+p\mathbb{Z}_p=U_1$.

My argument goes as follows. Pick an element $u\in U_1$. Then Hensel's lemma states that $X^3-u$ has a solution in $\mathbb{Z}_p$ if and only if $X^3-1$ has a solution modulo $p$. The latter always holds, so this shows that any element in $U_1$ has a cube root.

The problem is finding out what set those cube roots belong to. I would like to show that they belong to $U_1$ (or at least one root belongs), but this does not always hold. We know that any cube root of a unit, must have absolute value $1$, so if $a^3=u$, then $|a|=1$, so $a\in\mathbb{Z}_p^\times$.

Next there are two cases:

Case 1: $3\nmid p-1$. In this case we can use the isomorphism $\mathbb{Z}_p^\times/U_1\approx \mathbb{F}_p^\times$ to deduce that $a$ must represent a unit in this quotient group i.e. $a\in U_1$. This proves that there are no index $3$ subgroups of $U_1$.

Case 2: $3\mid p-1$. In this case the previous argument does not work, because we have nontrivial elements in the quotient group whose 3rd powers are trivial.

My question is if anyone has any ideas how to count the index $3$ subgroups of $1+p\mathbb{Z}_p$ when $3\mid p-1$? This is also equivalent to counting the index $3$ subgroups of the additive group of $\mathbb{Z}_p$.

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If $p\neq 2$ (true under your condition), so that $(1+p\mathbb{Z}_p,\times)\cong (\mathbb{Z}_p,+)$:

If $G\subset \mathbb{Z}_p$ is of index $3$ then $G\supset 3\mathbb{Z}_p$, but $3\mathbb{Z}_p=\mathbb{Z}_p$ unless $p=3$ (as $3$ is a unit of $\mathbb{Z}_p$). So no such subgroups. (note that for $p=3$ we do get (exactly one) such subgroup, namely $3\mathbb{Z}_3$)

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Just wanted to add that in the case $p=3$ this index 3 subgroup of $U_1$ is, of course, $1+9\mathbf{Z}_3$. Hensel lifting doesn't give cube roots of elements $u\in(1+3\mathbf{Z}_3)\setminus(1+9\mathbf{Z}_3)$, because the derivative of $x^3-u$ has a positive exponential value. – Jyrki Lahtonen Jun 11 '11 at 18:58

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