Suppose that $f_{1},...,f_{k}:\{0,1\}^{r}\rightarrow\{0,1\}^{r}$ are bijective functions.

For all $n\geq r$, let $G_{f_{1},...,f_{k};r}=\subseteq S(\{0,1\}^{n})$ be the subgroup generated by

i. the functions $f_{i}\times Id_{n-r}$ where $Id_{n-r}:\{0,1\}^{n-r}\rightarrow\{0,1\}^{n-r}$ is the indetity function, and

ii. the mappings $\pi_{\sigma}$ for permutations $\sigma\in S_{n}$ where $\pi_{\sigma}(x_{1},...,x_{n})=(x_{\sigma(1)},...,x_{\sigma(n)})$.

Let $t_{f_{1},...,f_{k};n}$ be the largest natural number $m$ such that the symmetry group $S_{m}$ embeds in the group $G_{f_{1},...,f_{k};r}$. We say that $(f_{1},...,f_{k})$ is of exponential growth if there is some $\alpha>1$ and $c>0$ where $t_{f_{1},...,f_{k};n}>c\cdot\alpha^{n}$ for all $n\geq r$. We say that $(f_{1},...,f_{k})$ is of polynomial growth if there is some polynomial $p$ where $t_{f_{1},...,f_{k};n}<p(n)$ for all $n\geq r$. We say that $(f_{1},...,f_{k})$ is of intermediate growth if it is not of polynomial nor exponential growth. Does there exist a collection of bijective functions $(f_{1},...,f_{k})$ of intermediate growth?

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.