Let $\Bbb Z[\zeta_p]$ denote the cyclotomic integers where $p$ is a prime and let $h_1 = h_1(p)$ denote its relative class number.
Question: Is it known whether there are infinitely many primes $p$ having the property that $h_1$ is not a power of 2?
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$\begingroup$ By $h_1$ you mean the "first factor" in $h(p)$, the negative part $h^-$, the factor not coming from the max'l real subfield of the cyclotomic field, right? If so, then I think this ought to be doable: I suspect you can prove from the formula for $h_1$ in terms of Bernoulli numbers (Iwasawa's "Lectures on p-adic L-functions" pg. 90) that if $p$ is irregular, then $p$ divides $h_1$. It works out for $p=37$ and $59$ but I haven't had a moment to check in general. Infinitude of the irregular primes then gives you that $h_1(p)$ is divisible by $p$ (hence not a power of $2$) for infinitely many $p$. $\endgroup$– user164898Commented Oct 9, 2022 at 19:01
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1$\begingroup$ Hmm, try this: suppose $p$ is irregular. Then $p$ divides the class number $h(p) = h^+h^-$, so $p$ divides either $h^-$ or $h^+$. A thm. of Kummer establishes that if $p$ divides $h^+$, then it also divides $h^-$ (see Thm.9 in Lozano-Robledo's "Bernoulli numbers, Hurwitz[...]" for a nice exposition). So if $p$ is irr., then $p$ divides $h^-$. There are infinitely many irrregular primes, so for infinitely many primes $p$, $h^-$ is divisible by $p$, hence fails to be a power of $p$. (Vandiver's conj. is that $p$ never divides $h^+$ when $p$ is irr., but you don't need it, due to Kummer's thm.) $\endgroup$– user164898Commented Oct 9, 2022 at 20:42
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1$\begingroup$ "$h_-$ is divisible by $p$, hence fails to be a power of $p$. Probably mistyped "...fails to be a power of $2$". $\endgroup$– Oscar LanziCommented Oct 9, 2022 at 21:42
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1$\begingroup$ @OscarLanzi Yes, that was indeed my typo! Sorry for the confusing typo. $\endgroup$– user164898Commented Oct 9, 2022 at 21:54
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1$\begingroup$ @JohnKlein I don't know anywhere where it's given explicitly in the literature, but I'm not the right person to ask, since I don't know the literature on class numbers as well as I'd like. But I am positive that the argument is not new at all, since it only uses ideas which were known to Kummer 130 years ago. $\endgroup$– user164898Commented Oct 9, 2022 at 21:58
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