Scalar multiple of one lattice contained in another I believe my question boils down to the following question:

Given lattices $L$ and $L'$ in $k^{n}$, does there exist $\lambda \in k^{\times}$ so that $\lambda L' \subseteq L$ and $\lambda L' \not\subseteq \mathfrak{m}L$?

A fuller background:
Let $k$ be a locally compact ultrametric field, $A$ be the valuation ring of $k$, $\mathfrak{m}$ the maximal ideal of $A$.  Let $K_{L}$ be the subgroup of $\rm{GL}_{n}(k)$ which sends $L$ onto $L$.  Let $L$ and $L'$ be two lattices in $k^{n}$ and suppose $K_{L} \subseteq K_{L'}$.  The claim is that there is a $\lambda \in k^{\times}$ so that $L' = \lambda L$.  To this end, it is first observed that $K_{L'} = K_{\lambda L'}$.  Following this, it is stated that we can assume the above statement.
Let me know if I have left out any necessary information.
 A: By the characterization of locally compact, ultra-metric fields, you know that your ideal, $\mathfrak{m}=(\pi)$ for some local uniformizing parameter, $\pi$, so this is really a question of whether or not $\lambda L'\subseteq L$ and $\lambda L'\not\subseteq \pi L$, but then the real question just boils down to the valuation, $v_\pi(\lambda)$, since $uL=L$ for any unit, $u$.
In particular, if $\lambda L'\subseteq \pi L$ then $\lambda\pi^{-1}L'\subseteq \pi^{-1}\pi L=L$, and so you can, by induction, find a $\lambda$ with exactly the right valuation so that the inclusions you want follow for some $\lambda$ simply by using the well-ordering principle on the set of $\{n\in\Bbb Z:\pi^n L'\subseteq L\}$ which is a bounded from below set, so has a minimal element.
Proof of boundedness from below:
If you multiply by too many $\pi^{-1}$ factors, you are causing your set to scale up (i.e. become bigger) and--by considerations on the Haar measure, $\mu$ of the lattices (which are compact, open sets in $k^n$)--you can see that if $|\pi|=c^{-1}<1$ that
$$0<\mu(\pi^{-n}L)=c^n\mu(L)<\infty$$
the last inequality since $L$ is compact, and the first because it is open.
Then for some $n\in\Bbb Z$ we have for $n>N$, we have $\mu(\pi^{-n}L)>\mu(L')$ (since $0<\mu(L')<\infty$ due to the compactness and openness of $L'$) so that it is impossible for $\pi^{-n}L\subseteq L'$, proving the boundedness from below of the set from below. In particular, if $\mu(L')=m'$ and $\mu(L)=m$ then you can compute for certain from $c^n\mu(L)>\mu(L')=m$ that $n\log c+\log(m)>\log m'$ i.e. any $N< \log_c {m'\over m}$ fails.
