After reading the article of the cake cutting problem in the September 2023 issue of science news. I was thinking about using a divide and conquer strategy by having the cake eaters divide themselves into two groups then using the known "cut and choose" two person strategy and repeat. This cuts the cake in $\log(n)$ (where $n$ is the number of cake eaters) cut and choose steps. That only leaves splitting into two groups and preforming the actual cutting and choosing. I figured that each cake eater can choose which of the two groups they want to be in $1$ step each for a total of $n$ steps each cake eater split step. That comes to $n \times \log(n)$ steps needed by the already considered steps. But Procaaccia proved that envy-free cake cutting takes at least $n^2$ steps in "Thou shalt covet thy neighbor's cake" (Ariel D. Procaccia. 2009. Thou shalt covet thy neighbor's cake. In Proceedings of the 21st International Joint Conference on Artificial Intelligence (IJCAI'09). Morgan Kaufmann Publishers Inc., San Francisco, CA, USA, 239–244.) So that leaves $(n^2-n)\over \log(n)$ (To multiply to $n^2$ each of the $\log(n)$ steps need to include $n^2\over \log(n)$ steps, we've already accounted for $n$ of these steps so $(n^2-n)\over \log(n)$) steps unaccounted for. And the only place I can see for those unaccounted step to be happing is in the $\log(n)$ cutting and dividing steps.
Where am I going wrong? Is my algorithm not actually envy-free? I'm I being too optimistic in assuming that each cake eater can choose between the groups in constant time? Does the fact that we are dealing with groups rather than individuals change the "cut" and/or "choose" step(s)?