Timeline for How to find $n$ such that the group of units $U(\mathbb{Z}/n\mathbb{Z})$ has a given abelian subgroup?
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| Dec 31, 2013 at 16:21 | comment | added | Lucia | 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 $m$. | |
| Dec 31, 2013 at 15:41 | comment | added | Gerry Myerson | 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 | |
| Dec 31, 2013 at 13:35 | history | asked | XMDu | CC BY-SA 3.0 |