By Higman's theorem we know that if $G$ is abelian and finite then $\mathbb{Z}G$ has only trivial units iff G is has exponent $1,2,3,4, \text{or}\ 6$ But what if $G=C_{m}$ where $m \ge 7$. In this case has someone calculated what the $\mathcal{U}({\mathbb{Z}C_m})$ looks like?
Asked
Active
Viewed 27 times
1
-
It seems to that the analogues of the cyclotomic units $(1-c^k)/(1-c)$, $\gcd(k,m)=1$, at least will be there. I don't know whether they generate the group of units? IIRC they generated a finite index subgroup of the units in the case $\Bbb{Z}[\zeta_m]$, but I'm too ignorant :-(. – Jyrki Lahtonen May 24 '15 at 13:52
-
Did you look at Higman's paper? See http://www.maths.ed.ac.uk/~aar/papers/higman.pdf. He proves the theorem you cite (see his Theorem 6 with $G$ being $C_m$) as a consequence of studying more general cases. You definitely want to understand the structure of unit groups in cyclotomic rings of integers, as Jyrki indicates, and their ranks can be computed from the Dirichlet unit theorem. – KCd May 24 '15 at 14:27
-
E.g. if $G$ is cyclic of prime order $m=p$, then $\mathbb ZG \cong \mathbb Z \times \mathbb Z[\zeta_p]$. – Thomas Poguntke May 24 '15 at 14:29