0
$\begingroup$

Given an integer $n$, we can determine the structure of the multiplicative group of integers modulo $n$ ($U(\mathbb{Z}/n\mathbb{Z})$) by the factorization of $n$. Hence we can easily find all the cyclic subgroups $\mathbb{Z}/m\mathbb{Z}$ of $U(\mathbb{Z}/n\mathbb{Z})$.

Conversely, given an integer $m$, how to find the smallest integer $n$ such that $U(\mathbb{Z}/n\mathbb{Z})$ has a subgroup homomorphic to $\mathbb{Z}/m\mathbb{Z}$? Or given a finite abelian group $A$, how to find the smallest integer $n$ such that $U(\mathbb{Z}/n\mathbb{Z})$ has a subgroup homomorphic to $A$?

Improve this question
$\endgroup$
2
  • $\begingroup$ Smallest $n$ such that the units mod $n$ has a subgroup isomorphic to the integers mod $m$ is probably not much different from smallest prime congruent to 1 mod $m$, which is tabulated (with some links) at oeis.org/A034694 $\endgroup$
    – Gerry Myerson
    Commented Dec 31, 2013 at 15:41
  • 1
    $\begingroup$ As @GerryMyerson says it won't be too different from the least prime that is $1\pmod m$, but it can be different. For example if $m=144$ then $n=17 \times 19=323$ works, whereas the least prime that is $1\pmod {144}$ is $433$. A simple idea would be to factor $m$ into a product of coprime numbers, take the least prime that is one mod each coprime factor, multiply these together and minimize over all such factorizations of $m$. $\endgroup$
    – Lucia
    Commented Dec 31, 2013 at 16:21

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .