Let $G$ be a finite group and $\phi:M\to N$ be a surjective homomorphism of $G$-lattices (i.e. finitely generated $\mathbb{Z}[G]$-modules, free as $\mathbb Z$-modules), with kernel $K$. For every $n\geq 1$, let $M\_n:=\phi^{-1}(nN)$. Since $nN\cong N$, we have an exact sequence $0\to K\to M\_n \to N\to 0$. I was wondering if it is possible to determine the class of this extension in terms of $n$ and the class of $0\to K\to M\to N\to 0$ inside the finite abelian group $Ext^1\_G(N,K)$?


For example, is it true that $[M\_n]=a+bn$, for some $a,b\in Ext^1\_G(N,K)$?



Extensions of lattices
It is what you think it is:



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> **Lemma.** *The class of the short exact sequence $0 \to K \to M\_n \to N \to 0$ in $\operatorname{Ext}^1\_G(N,K)$ is $n$ times the class of the sequence $0 \to K \to M \to N \to 0$.*
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*Proof.* By construction, the class in $\operatorname{Ext}^1\_R(C,A)$ of a short exact sequence $0 \to A \to B \to C \to 0$ of $R$-modules is the image of $\operatorname{id}\_C$ under the boundary map $\operatorname{Hom}(C,C) \to \operatorname{Ext}^1\_R(C,A)$ coming from the long exact sequence
$$0 \to \operatorname{Hom}(-,A) \to \operatorname{Hom}(-,B) \to \operatorname{Hom}(-,C) \to \operatorname{Ext}^1\_R(-,A) \to \ldots.$$
In our situation, we have a commutative diagram with exact rows
$$\begin{array}{ccccccccc}0 & \to & K & \to & M\_n & \to & N & \to & 0 \\ & & |\!| & & \downarrow & & \downarrow{\scriptsize n}\!\!\!\! & & \\ 0 & \to & K & \to & M & \to & N & \to & 0.\!\end{array}$$
The long exact sequences give a commutative diagram
$$\begin{array}{ccccccccccc}0 & \to & \operatorname{Hom}(N,K) & \to & \operatorname{Hom}(N,M\_n) & \to & \operatorname{Hom}(N,N) & \to & \operatorname{Ext}^1\_G(N,K) & \to & \ldots \\ & & |\!| & & \downarrow & & \downarrow{\scriptsize n}\!\!\!\!\!\!\! & & |\!| & \\ 0 & \to & \operatorname{Hom}(N,K) & \to & \operatorname{Hom}(N,M) & \to & \operatorname{Hom}(N,N) & \to & \operatorname{Ext}^1\_G(N,K) & \to & \ldots,\!\!\!\end{array}$$
hence the image of $\operatorname{id}\_N$ in the first row maps to $n$ times the image of $\operatorname{id}\_N$ in the second row. $\square$


