In the paper 


P.J. Webb: Bounding the ranks of ZG-lattices by their restrictions to elementary abelian groups. J. Pure Appl. Algebra 23 (3) (1982), 311-318. 


the author writes in the introduction without giving any reference or explanation ($G$ denotes a finite group): 



> 
> If $M$ is any $\mathbb{Z}G$-lattice, that is a $\mathbb{Z}G$-module which is a free abelian group of finite rank, we may write $M=\text{core}(M)\oplus \text{proj}(M)$ where $\text{proj}(M)$ is a projective module and $\text{core}(M)$ is a submodule of $M$ with no projective summands. In general the submodule $\text{core}(M)$ is not uniquely determined by these requirements, but all the possible choices for $\text{core}(M)$ will have the same ranks as an abelian group. 
> 
> 
> 


I managed to show that each lattice can be decomposed in the described way, but I'm not able to prove that all $\text{core}(M)$ have the same rank. 


**Question:** Why is the rank of $\text{core}(M)$ uniquely determined ? 



Uniqueness of the rank of the core of a lattice 
Even more is true: Two possible cores of $M$ lie in the same genus, i.e. they become isomorphic in the localization at any prime of $\mathbb Z$. ~~However, I couldn't find an explicit proof of this anywhere in literature~~ **Edit:** It's Proposition 5.4 in Gruenberg: "Relation Modules of Finite Groups" (though unfortunately there's apparently no online version of this book besides google books).


Assume you have two decompositions $M=C\_1\oplus P\_1 = C\_2 \oplus P\_2$ with $P\_1$ and $P\_2$ projective. Assume furthermore that one cannot split off any further non-zero projective summands from $C\_1$ and $C\_2$.


By a theorem of Swan (see [Swan: Projective modules over finite groups](https://web.archive.org/web/20200721135250/https://projecteuclid.org/euclid.bams/1183523366)) $\mathbb Q\otimes P\_1$ and $\mathbb Q\otimes P\_2$ are free, and therefore so is $\mathbb Z\_{(p)} \otimes P\_1$ and $\mathbb Z\_{(p)}\otimes P\_2$ ($\mathbb Z\_{(p)}$ denotes the localization of $\mathbb Z$ at $p$; this simply uses the fact that since the columns of the decomposition matrix of a finite group are linearly independent, the isomorphism type of the $\mathbb Q$-span of a projective $\mathbb Z\_{(p)}$-module determines the isomorphism type of the projective module; alternatively see Theorem (77.14) on p. 548 in Curtis, Reiner: Representation theory of finite groups and associative algebras).


So now apply cancellation for $\mathbb Z\_{(p)}G$-lattices (see [Corollary 1 in this paper of Reiner](https://web.archive.org/web/20200612153355/https://projecteuclid.org/euclid.bams/1183524252)) to find that unless $P\_1$ and $P\_2$ have the same $\mathbb Z$-rank (in which case we would be done), either $C\_1$ or $C\_2$ (say, w.l.o.g. $C\_1$) has a non-zero free summand in every localization. So if this is the case, then take for each prime $p\_1,\ldots,p\_n$ that divides $G$ an epimorphism
$$
 \phi\_i: \mathbb Z\_{(p\_i)} \otimes C\_1 \twoheadrightarrow \mathbb Z\_{(p\_i)}G \subset \mathbb QG
$$
By multiplying each $\phi\_i$ with a sufficiently large integer relatively prime to $p\_i$ we may assume $\phi\_i(C\_1) \subseteq \mathbb Z G $ for each $i$. Now we get a map
$$
 \phi := \sum\_{i=1}^n \left(\prod\_{j\neq i} p\_j \right)\cdot \phi\_i|\_{C\_1}
$$
$\phi$ is a homomorphism from $C\_1$ to $\mathbb Z G$, which is an epimorphism in the localization at each $p\_i$ (by construction: the factor $\prod\_{j\neq i} p\_j$ is a unit in $\mathbb Z\_{(p\_i)}$ and in the radical of each $\mathbb Z\_{(p\_j)}$. Hence the image of $\phi$ in the localization at $p\_i$ is the same as the image of $\phi\_i$). The image of $\phi$ in the localization at any other prime is projective anyways, since if $p$ doesn't divide the order of $G$, then any $\mathbb Z\_{(p)}G$-lattice is projective. So now we have a homomorphism from $C\_1$ onto the image of $\phi$, and the image of $\phi$ is projective in each localization. But then by Theorem (78.1) on p. 550 in Curtis&Reiner the image of $\phi$ is projective, and therefore the image of $\phi$ is a projective summand of $C\_1$. This contradicts the assumption that we couldn't split off any further projective summands from $C\_1$.


