Let $L=\mathbb Z^2$, $\gamma\in \mathrm M_2(\mathbb Z)$, $\det (\gamma)=n>0$. Then $\gamma$ acts on $L$ by left matrix multiplication. Is $[L:\gamma L]=n$?
I have tried proving this by Fundamental Theorem on Homomorphisms but have not made any progress. I think I can solve the case only when $n=1$.
What's more, it seems that when $L=\mathbb Z^k(k\in\mathbb Z_{\geq1})$ the claim is also true. How to prove the generalized proposition? The case $k=1$ is equivalent to $|\mathbb Z/n\mathbb Z|=n$.
Reduce $\gamma$ to Smith Normal Form using unimodular row and column operations. These do not change $|L:\gamma(L)|$ and they do not change $|\det(\gamma)|$, so you can assume that $\gamma$ is diagonal, and then the result is clear.