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I am looking to find real quadratic fields whose Hilbert class field is abelian over $\Bbb Q$. Then I learned about genus numbers and genus field of the number field. It is enough to find a number field whose class number is equal to the genus number. In YOSHIOMI FURUTA article
I found the formula, but I have a problem with computing the order of the quotient. So I was wondering, is there a way to find genus number using Sage just like the class number.

Any help is very appreciated

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    $\begingroup$ The genus number of quadratic number fields over $\mathbf{Q}$ was computed by Hasse (J. Math. Soc. Japan 3, 1951); he shows the genus number of $\mathbf{Q}(\sqrt{d})$ is $2^{r-1}$, where $r$ is the number of distinct prime divisors of $d$. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Aug 27, 2021 at 8:18
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    $\begingroup$ For example, you can take $d = 2,3,5,7,10,...$ for your problem, i.e. the class and genus number of e.g. $\mathbf{Q}(\sqrt{10})$ are equal (both are $2$ in this case). $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Aug 27, 2021 at 8:34

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I may as well promote my comment to an answer, so this is closed. In a short paper, H. Hasse [1] computed the genus field and genus number of every real quadratic field. In particular, he shows that, given a real quadratic field $\Omega = \mathbb{Q}(\sqrt{d})$, the genus number of $\Omega$ is $2^{r-1}$, where $r$ is the number of distinct prime divisors of (the square-free) $d$. Thus, for example, the genus number of $\mathbb{Q}(\sqrt{210})$ is $8$, as $210 = 2 \cdot 3 \cdot 5 \cdot 7$.

Thus the genus number is (rather trivially) computable in Sage, for example by the command 2^(len(set(d.factor()))-1), where $d$ is as above.

[1] Hasse, Helmut, Zur Geschlechtertheorie in quadratischen Zahlkörpern, J. Math. Soc. Japan 3, 45-51 (1951). ZBL0043.04002.

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