Commutativity of T(n) and R(n) as functions on lattices (Lang Introduction to Modular Forms)

Question

I am currently reading through Lang's Introduction to Modular Forms. In chapter II, he introduces the Hecke Operator as follows.

Let $\mathcal{L}$ be the free abelian group generated by the lattices in $\mathbb{C}$. We define the Hecke operator $T(n)$ for each positive integer $n$ to be the map $$ T(n):\mathcal{L}\rightarrow\mathcal{L}$$ such that $$T(n)L=\sum_{(L:L')=n}L'$$ He then defines another operator $R(n):\mathcal{L}\rightarrow\mathcal{L}$ to be such that $$R(n)L=nL$$ Note that the above is the sublattice of $nL\subset L$, and not the sum of $L$ by itself $n$ times in the free abelian group. He then says it is clear that the operators $R(n)$ and $T(m)$ commute with each other. I am having a hard time seeing why this is the case, my strategy is that it is sufficient to take some lattice $L$ and show that $R(n)T(m)L=T(m)R(n)L$. I have tried this as follows $$R(n)T(m)L=R(n)\sum_{(L:L')=m}L'=\sum_{(L:L')=m}R(n)L'=\sum_{(L:L')=m}nL'$$ compared wish $$T(m)R(n)L=T(m)nL=\sum_{(nL:L')=m}L'$$ I don't see how to bridge these equalities, or if there is an error in my understanding (I tried to do a ``change of variable'' but was unsuccessful). If anyone could help it would be much appreciated.

Answer 1

Note I figured out the solution. We wish to show that $$\sum_{(L':L)=m}nL'=\sum_{(L'':nL)=m}L''$$ This can be seen by showing that we have a correspondence between $L''\subset nL$ index $m$ sublattices of $nL$ and $nL'$ where $L'\subset L$ is an index $m$ sublattice of $L$.

Indeed if $L''\subset nL$ is an index $m$ sublattice, then we have that every $x\in L''$ can be written as $ny$ for some $y\in L$, and $L'=\{y\in L: ny\in L''\}\subset L$ will be the corresponding sublattice of $L$ of index $m$. From this construction it is clear that $L''=nL'$.

Conversely, given $L'\subset L$ of index $m$, then we will have that $nL'=L''\subset nL$ of index $m$, and this will give the correspondence between the two sums. Thus, we have that they commute.