Permutation representations of the orbits of the automorphism group of a finite module over discrete valuation ring

Abstract

Consider a discrete valuation ring R whose residue field is finite of cardinality at least 3. For a finite torsion module, we consider transitive subsets O under the action of the automorphism group of the module. We prove that the associated permutation representation on the complex vector space C[O] is multiplicity free. This is achieved by obtaining a complete description of the transitive subsets of O × O under the diagonal action of the automorphism group.

Appendix

Let R _λ denote the abelian group corresponding to the partition $(\lambda = \lambda _{1}^{\rho _1}>\lambda _{2}^{\rho _2}>{\lambda }_{3}^{{\rho } _3}>\cdots >{\lambda }_{k}^{{\rho } _k})$. In this section, we prove the existence of a certain combinatorial lattice structure among certain B-valued subsets of R _λ defined below in order to motivate Definition 1. This is a bigger lattice (additively closed for supremum) when compared to the lattice of characteristic subgroups of abelian p-groups for p>2.

DEFINITION 3 (B-val)

Let $B = \{0,1\} \subset \mathbb {N} \cup \{0\}$. A B-val $\underline a$ is an element of ${\bigcup }_{l \in \mathbb {N}} \{0,1\}^{l}$.

DEFINITION 4 (B-set)

Fix an odd prime p in this article. Let R _λ denote the abelian group corresponding to the partition $({\lambda } = {\lambda }_{1}^{\rho _1}>{\lambda }_{2}^{\rho _2}>{\lambda }_{3}^{\rho _3}>\cdots >{\lambda }_{k}^{\rho _k})$. Let (r ₁,r ₂,…,r _k ) ∈ {0,1,2,…,λ ₁} × {0,1,2,…,λ ₂} × ⋯ × {0,1,2,…,λ _k } and $\underline {a}=(a_{1},a_{2},\ldots ,a_{k}) \in \nolinebreak \{0,1\}^{k}$. Define a B-set denoted by $I(\underline r,\underline a)$ as the subset ${{\prod }^{k}_{i=1}}p^{r_{i}}(R^{\rho _{i}}_{\lambda _{i}})^{a_{i}}$ of the abelian group R _λ , where $p^{r}({R}^{{\rho }}_{\mu })^{0}=p^{r}(\mathbb {Z}/p^{\mu }\mathbb {Z})^{{\rho }}$ for 0≤r≤μ and $p^{r}(R^{\rho }_{\mu })^{1}=p^{r}\left ((\mathbb {Z}/p^{\mu }\mathbb {Z})^{\rho }\backslash p(\mathbb {Z}/p^{\mu }\mathbb {Z})^{\rho }\right )$ for 0≤r≤μ with $p^{\mu } (R^{\rho }_{\mu })^{1}=\{0\}$.

DEFINITION 5

Let $\mathcal {L}$ be a collection of subsets of R _λ . Define a lattice structure as follows. Let $L_{1},L_{2} \in \mathcal {L}$, then L ₁∧L ₂ is an element of $\mathcal {L}$ which is the biggest set in $\mathcal {L}$ which is contained in L ₁ and L ₂. Define L ₁∨L ₂ to be the smallest set in $\mathcal {L}$ that contains L ₁ + L ₂. We note that L ₁ + L ₂={l ₁ + l ₂ ∈ R _λ ∣l ₁ ∈ L ₁ ⊂ R _λ ,l ₂∈L ₂ ⊂ R _λ }.

Remark 2.

In the case, when the lattice $\mathcal {L}$ is the lattice of characteristic subgroups this definition is not needed as any subgroup L which contains two groups L ₁,L ₂ as subgroups also contains the group L ₁ + L ₂ generated by both of them, i.e. note the least upper bound (supremum) is additively closed instead of just closed under the set union.

Also we note that the lattice of characteristic subgroups has two-way distributivity (L ₁∨L ₂)∧L ₃=(L ₁∧L ₃)∨(L ₂∧L ₃),(L ₁∧L ₂)∨L ₃=(L ₁∨L ₃)∧(L ₂∨L ₃). This is because of the following identities. For any real numbers r,s,t, we have

$$\max(\min(r,s),t)=\min(\max(r,t),\max(s,t)), $$

$$\min(\max(r,s),t)= \max(\min(r,t),\min(s,t)). $$

The following two lemmas prove that the set of all such B-valued subsets is a lattice with respect to the Definition A3.

Lemma 4.

$$ I(\underline r,\underline a) \cap I(\underline s,\underline b) = \left\{ \begin{array}{c} I(\underline r \cup \underline s,\underline a + \underline b - \underline a\underline b), \\ \emptyset\ \text{ if there exists } i \text{ such that } r_{i}<s_{i}, a_{i}=1\\ \text{ or } s_{i}<r_{i}, b_{i}=1, \end{array} \right. $$

(36)

where $\underline a \underline b=(a_{1}b_{1},\ldots ,a_{k}b_{k})$ and $\underline r \cup \underline s=(\max (r_{i},s_{i}):i=1,2,\ldots ,k)$.

Proof.

Let i∈{1,2,3,…,k}. If r _i = s _i , then

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}*}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}*}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}*}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}}$.

If r _i ≠s _i , then assume without loss of generality that r _i <r _i +1≤s _i .

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=\emptyset$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=\emptyset$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}\cap p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}$.

This proves the lemma. □

Lemma 5 (Sum of two B-sets lemma).

Let $I(\underline r,a),I(\underline s,b)$ denote two B-subsets of R _λ with B-val $\underline a,\underline b$ . Then $I(\underline r,\underline a)+I(\underline s,\underline b)$ is a B-set. The B-val associated to the B-set $I(\underline r,\underline a)+I(\underline s,\underline b)$ is given by $\underline c = 1/2((a+b)sgn(r-s)-(a-b))sgn(r-s)$ . The B-set $I(\underline r,\underline a)+I(\underline s,\underline b) = I(\underline r \cap \underline s,\underline c)$ where $\underline r \cap \underline s=(\min (r_{i},s_{i}):i=1,2,\ldots ,k)$.

Proof.

We have that p is an odd prime. Let i∈{1,2,3,…,k}. If r _i = s _i , then

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}+ p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}},p$ is an odd prime,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}+ p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}+ p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}+ p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=p^{r_{i}=s_{i}}R_{\lambda _{i}}^{\rho _{i}}$.

If r _i ≠s _i , then assume without loss of generality that r _i <r _i +1≤s _i .

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}+p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}+p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}*}=p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}+p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}*}$,

• $p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}+p^{s_{i}}R_{\lambda _{i}}^{\rho _{i}}=p^{r_{i}}R_{\lambda _{i}}^{\rho _{i}}$.

Here s g n(r _i −s _i )=+1,0,−1 according as r _i >s _i ,r _i = s _i ,r _i <s _i respectively and s g n(r−s)=(s g n(r _i −s _i )∣i=1,2,…,k). Now the sum set equality

$$I(\underline{r},\underline{a})+I(\underline{s},\underline{b})=I(\underline{r}\cap \underline{s},\underline{c}). $$

follows. □