Permutation covering of a $G$-lattice

Question

Let $G$ be a finite group. By a $G$-lattice we mean a finitely generated free abelian group $L$ with an action of $G$. We say that $L$ is a permutation $G$-lattice if $L$ has a ${{\mathbf{Z}}}$-basis permuted by $G$. By a permutation covering of a $G$-lattice $L$ we mean a surjective morphism of $G$-lattices $P\to L$, where $P$ is a permutation $G$-lattice.

Let $G_1=C_p$, a cyclic group of prime order $p$. Consider the standard permutation $G_1$-lattice $P_1={{\mathbf{Z}}}^p$. Set $L_1=P_1/{{\mathbf{Z}}}$, where ${{\mathbf{Z}}}$ is embedded in $P_1={{\mathbf{Z}}}^p$ diagonally. We have a permutation covering $P_1\to L_1$ of rank $p$.

Let $G_2=C_p$, and define $P_2={{\mathbf{Z}}}^p$ and $L_2=P_2/{{\mathbf{Z}}}$ similarly. Now take $$ G=G_1\times G_2,\qquad P=P_1\otimes_{{\mathbf{Z}}} P_2, \qquad L=L_1\otimes_{{\mathbf{Z}}} L_2. $$ Then we have a $G$-lattice $L$ of rank $(p-1)^2$ and a permutation covering $P\to L$ of rank $p^2$.

Question. Does there exist a permutation covering of $L$ of smaller rank, say of rank $p^2-p$?

For $p=2$ the answer is YES: the 1-dimensional lattice $L$ clearly has a permutation covering of rank 2.

This must be an easy question....

Answer 1

The answer for $p>2$ is NO:

Suppose $M \to L$ is a permutation covering with kernel $K\leq M$, and the rank of $M$ smaller than $p^2$. The permutation lattice $M$ decomposes into a direct sum
$$ M = M_1 \oplus \dotsb \oplus M_r, $$ where each summand corresponds to an orbit of the underlying $G$-set. These orbits can have sizes $1$ or $p$, as summands of rank $p^2$ are forbidden. But summands of rank $1$ are in the kernel $K$, since $L$ contains no elements that are fixed by $G$. Thus we can omit these from the beginning, and assume that each $M_i$ has rank $p$.

Each $M_i$ has a (unique) submodule $S_i$ of rank $1$ on which $G$ acts trivially. Again, we must have $S_i \leq K$, as no element of $L$ is fixed by $G$. So each $M_i$ contributes at most rank $p-1$ to $L$. Thus we must have $r\geq p-1$. Since we assume that $M$ has rank $<p^2$, we have $r=p-1$.
Now both sublattices $$ S_1 \oplus \dotsb \oplus S_{p-1} \subseteq K $$ are pure sublattices of $M$, and by comparing ranks we can conclude that actually $$ S_1 \oplus \dotsb \oplus S_{p-1} = K.$$ Thus we would have $$ L \cong (M_1/S_1) \oplus \dotsb \oplus (M_{p-1}/S_{p-1}). $$ But I claim that $L$ is indecomposable. (Edited later, in view of comments, and since the first proof was somewhat unclear:) The $P$ of the question is clearly the group ring $\mathbf{Z}G$, and $L = \mathbf{Z}G/I$, where $I$ is some ideal of $\mathbf{Z}G$. It is, however, well known that the group ring of a $p$-group over a field of characteristic $p$ or over a local ring with residue field of characteristic $p$ is local. Thus $L\otimes \mathbf{F}_p$ or $L\otimes \mathbf{Z}_p$ (localisation at $p$) has a unique maximal submodule, and is indecomposable. Therefore, $L$ must also be indecomposable. So when $p-1>1$, we get a contradiction.
(On the other hand, $L \otimes \mathbf{Q}$ has a decomposition into $p-1$ irreducible $\mathbf{Q}G$-modules. An alternative proof would be to compute matrices of the projections to these subspaces in terms of a $\mathbf{Z}$-basis of $L$, and see that they do not have integer entries.)

Answer 2

If we take as base ring $\mathbb{Q}$ in place of $\mathbb{Z}$, the answer is Yes.

You surely have good reasons for taking the integers as base ring. But in view of your more general question Minimal rank of a permutation resolution of a $G$-lattice I believe that over $\mathbb{Q}$ (any other field those char. doesn't divide the group order will also do) there is a fair chance to solve the general question. Therefore the subsequent consideration might be of interest to you.

Let $G = C_p \times C_p= \langle \sigma,\tau\rangle$ and for $H \le G$ let $e_H = \sum_{h \in H} h \in \mathbb{Q}[G]$. Define a map of $\mathbb{Q}[G]$-modules by
$$\phi_H: \mathbb{Q}[G/H] \to \mathbb{Q}[G],\,\,gH \mapsto g\cdot e_H.$$ Let $H_j = \langle \sigma\tau^j\rangle\,\,(j=0,...,p-1)$ and $H_p=\langle\tau\rangle$ and $$\phi: \bigoplus_{i=1}^{p-1}\mathbb{Q}[G/H_j] \to \mathbb{Q}[G]\twoheadrightarrow L.$$ Note: The left hand side is a $\mathbb{Q}$-vector space of dimension $(p-1)p$.

Claim: $\phi$ is an epimorphism.

Since $H_i \cap H_j = 1$, we find $\sum_{j=0}^p e_{H_j}= p + e_G$. Moreover, $L = \mathbb{Q}[G]/\langle e_{H_0},e_{H_p}\rangle$ and $e_G = e_{H_0}e_{H_p}$. Hence $e_{H_0},e_{H_p},e_G$ vanish in $L$. Therefore $\phi(\sum_{j=1}^{p-1}H_j) = p \in L$ and $\phi(\frac{1}{p} \sum_{j=1}^{p-1}gH_j) = \bar{g} \in L$ for all $g \in G$. This shows the surjectivity of $\phi$.