The contraction subgroup for $x$ in the locally compact group, $G$, $\mathrm{con}(x) = \left\{ g\in G \mid x^ngx^{-n} \to 1\text{ as }n\to\infty \right\}$, and the Levi subgroup is $\mathrm{lev}(x) = \left\{ g\in G \mid \{x^ngx^{-n}\}_{n\in\mathbb{Z}} \text{ has compact closure}\right\}$. The following will be shown. Let $G$ be a totally disconnected, locally compact group and $x\in G$. Let $y\in{\sf lev}(x)$. Then there are $x'\in G$ and a compact subgroup, $K\leq G$ such that:
-$K$ is normalised by $x'$ and $y$,
-$\mathrm{con}(x') = \mathrm{con}(x)$ and $\mathrm{lev}(x') = \mathrm{lev}(x)$ and
-the group $\langle x',y,K\rangle$ is abelian modulo $K$, and hence flat.
If no compact open subgroup of $G$ normalised by $x$ and no compact open subgroup of $\mathrm{lev}(x)$ normalised by $y$, then the flat-rank of $\langle x',y,K\rangle$ is equal to $2$.