# A table for irreducible integral representation of finite cyclic groups

Is there such a table where the irreducible integral representations of finite cyclic groups are listed?

Edited:

Thanks for Todd Leason's comment.Acutally,i want to know all inequivalent indecomposable integral representations of cyclic groups. I think this is related to determining conjugacy classes of finite cyclic subgroups of $GL(n,Z)$. I want to know such subgroups of $GL(n,Z)$ explicitely.Is there such a list?

For example,cyclic subgroups of $GL(3,Z)$ are determined explicitely in:

ON THE FINITE SUBGROUPS OF GL (3, Z)

• table? A theorem, perhaps... – Dima Pasechnik May 2 '15 at 16:42
• Note that for integral representations there is an alternative notion for irreducibility, called Z-irreducibility, that leads to more interesting results. In particular, the Z-irreducible modules of cyclic groups are described in Curtis, Reiner: Representation Theory of Finite Groups, Theorem (74.3). – Todd Leason May 3 '15 at 14:05
• Concerning the edit, it's worth emphasizing that there is a big difference between "irreducible" and "indecomposable" in this kind of representation theory. – Jim Humphreys May 9 '15 at 15:10

For any representation $M$ over the integers, $pM$ is a submodule for any prime $p$. Thus, if $M$ is irreducible, then for any prime, either $pM=0$ or $pM=M$. For a finitely generated $\mathbb{Z}$-module $M\neq pM$ for at least one $p$. If $pM=qM=0$ for two distinct primes, then $M=(p,q)M=0$. Thus any non-zero irreducible has a unique $p$ such that $pM=0$, and for all others, we have $qM=M$. That is, $M$ is an irreducible representation over $\mathbb{F}_p$.

These are easily determined for a cyclic group of order $n$. There's one for each $n$th root of unity $\zeta$ in $\bar{\mathbb{F}}_p$; let $q=p^\ell$ be minimal such that $\zeta\in \mathbb{F}_q\subset \bar{\mathbb{F}}_p$, take the obvious 1-dimensional representation over $\mathbb{F}_q$ with a generator acting by $\zeta$, and restrict scalars to $\mathbb{F}_p$.

For cyclic groups $C_p$ of prime order $p$ the irreducible integral representations are known (I don't know if there are results for cyclic groups of composite order but it's likely since the result for prime orders is very old):

Let $\zeta \in \mathbb{C}$ be a $p$-th root of unity and let $h_p$ be the class number of $\mathbb{Q}(\zeta)$. Up to isomorphism there are exactly $1+2h_p$ irreducible $\mathbb{Z}C_p$-modules.

If $B_1,...,B_h \subseteq \mathbb{Q}(\zeta)$ are fractional ideals that represent the elements of the class group of $\mathbb{Q}(\zeta)$ und $b_i \in B_i \setminus (1-\zeta)B_i$, then the irreducible $\mathbb{Z}C_p$-modules are: $$\mathbb{Z},\,\,B_i,\,\,B_i\oplus \mathbb{Z}\qquad(i=1,...,h)$$ If $\sigma$ is a generator of $C_p$ then the action is given by $\sigma \cdot b = \zeta b, \, \sigma\cdot (b,1) = (\zeta b+b_i,1)$.

Reference: Curtis, Reiner: Representation Theory of Finite Groups and Associative Algebras, Theorem (74.3) and Exercise 4 in § 74.9.

Note: These irreducible representations depend on the class number of cyclotomic fields. Since this class number is not known in general there is no explicit list in general.

• You mean indecomposable, rather than irreducible. And you are implicitly talking about the $\mathbb{Z}$-free ones. – Alex B. May 4 '15 at 10:57
• @Alex B.: I'm talking about Z-irreducible modules in the sense of C-R as should be clear from the reference. Z-irreducible modules are of course indecomposable. The OP explained to be interested in (cyclic) subgroups of GL_n(Z). Therefore only representations arising from Z-free modules are of interest here. – Todd Leason May 7 '15 at 10:55
• Yes, I hadn't seen the edit. – Alex B. May 8 '15 at 11:04

It is known that if a finite cyclic group $G$ has order divisible by $p^{3}$ for some prime $p$, then there are infinitely many non-isomorphic indecomposable $\mathbb{Z}G$-modules ( which are torsion free as $\mathbb{Z}$-modules). One reference for this and related theorems is a 1963 paper of Alfredo Jones, which is available with Open Access on Project Euclid (https://projecteuclid.org/euclid.mmj/1028998908).

• But there are only finitely many of $\mathbb{Z}$-rank $\le n$ for fixed $n$. Do you know a result on their number ? – Todd Leason May 3 '15 at 22:25
• @ToddLeason : I don't myself, but others may. – Geoff Robinson May 3 '15 at 23:17