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.... 



Permutation covering of a $G$-lattice
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.)


