# On the symmetric group of 2^n elements

Consider the set $X_1^n=\{1,2,...,2^n\}$. Then define $X_2^n$ to be the set of two element subsets of $X_1^n$. I will construct $X_i$ by induction on $i$. $X_i^n$ is the set of two element subsets of $X_{i-1}^n$ so that those two elements are disjoint, i.e. $X_i^n=\{\{x,y\}| x,y\in S_{i-1}, x\cap y=\phi\}$. Here by intersection I mean that, we have a map $\phi_i$ from $X_i^n$ to subsets of $X_1^n$ which is defined by induction as follows:- $\phi_{i+1}(\{x,y\})=\phi_i(x)\cup \phi_i(y)$. So by $x\cap y$ I meant $\phi_{i-1}(x)\cap \phi_{i-1}(y)$. For example, an element of $X_3^n$ will look something like this:- $\{\{1,2\},\{3,4\}\}$. An element of $X_4^n$ will look like $\{\{\{1,3\},\{2,5\}\},\{\{4,7\},\{6,8\}\}\}$. But $\{\{1,2\},\{2,3\}\}$ is not an element of $X_3^n$ because $\{1,2\}\cap \{2,3\}\ne \phi$. Note that, for $i>n+1$, $X_i^n=\phi$.

Edit:- A way to imagine the set $X_k^n$ is to consider the set of un-ordered binary trees of height $k-1$ and $2^{k-1}$ nodes, whose leaves are numbered by a subset of $\{1,2,...,2^n\}$ so that the leaves have distinct numbers.

For further clarification, as Amritanshu Prasad mentioned this in comment, $\{\{\{1,5\},\{2,3\}\}, \{\{1,3\},\{2,4\}\}\}$ is not an element of $X_3^n$ because $\{1,5,2,3\}\cap \{1,3,2,4\}\ne \phi$.

I am interested in the following directed system of finite dimensional Hilbert spaces:- $H_i^n$ is the Hilbert space which has an orthonormal basis $\{e_s\}_{s\in X_{n+2-i}^n}$. Here $1\le i\le n+1$. We have an isometry $\psi_i:H_i^n\to H_{i+1}^n\otimes H_{i+1}^n$ given by $\psi_i(e_{\{x,y\}})=\frac{e_x\otimes e_y+e_y\otimes e_x}{\sqrt {2}}$. This gives the following directed system of Hilbert spaces:- $$H_1^n\to (H_2^n)^{\otimes 2}\to (H_3^n)^{\otimes 4}\to .....\to (H_{n+1}^n)^{\otimes 2^n}$$ where the map $(H_i^n)^{\otimes 2^{i-1}}\to (H_{i+1}^n)^{\otimes 2^{i}}$ is given by $\psi_i^{\otimes 2^{i-1}}$. A nice property of this directed system is that the image of $\sum\limits_{x\in X_{n+1}^n} e_x\in H_1^n$ in $(H_{n+1}^n)^{\otimes 2^{n}}$ under the map $H_1^n\to (H_2^n)^{\otimes 2}\to ....\to (H_{n+1}^n)^{\otimes 2^{n}}$ is $\sum\limits_{\sigma\in S_{2^n}} e_{\sigma(1)}\otimes e_{\sigma(2)}\otimes ....\otimes e_{\sigma(2^n)}$ upto multiplication by a constant which implies that image of $\sum\limits_{x\in X_{n+1}^n} e_x\in H_1^n$ under the map $H_1^n\to (H_2^n)^{\otimes 2}\to ....\to (H_{i}^n)^{\otimes 2^{i-1}}$ lies in $Sym^{2^{i-1}}(H_i^n)$ but its not of the form $v^{\otimes 2^{i-1}}$ for some $v\in H_i^n$.

Suppose, $H^n$ is the Hilbert space which is the direct limit of the above directed system. I am interested in some specific type of isometries from $H^n$ to $H^{n+1}$ and the directed system $$H^1\to H^2\to H^3\to ...$$ arising from these maps.

I am working on something related to this and I think this type of things about the permutation group $S_{2^n}$ have been studied before. So any reference would be really helpful.

• Edited the first paragraph to make it clear.
– DLN
Apr 10, 2016 at 21:46
• It looks like $X_k^n$ is empty whenever $2^{k-1} > n$. Am I mistaken? Apr 11, 2016 at 3:19
• Just checking if I understand correctly: is $\{\{\{1,3\},\{2,5\}\},\{\{1,5\},\{2,3\}\}\}$ also an element of $X^4_3$? Apr 11, 2016 at 4:32
• @Amritanshu Prasad, No. because $\{1,3,2,5\}\cap \{1,5,2,3\}\ne \phi$
– DLN
Apr 11, 2016 at 5:13
• @S. Carnahan, for any $x\in X_k^n$, $\phi_k(x)$ has $2^{k-1}$ elements and $\phi_k(x)$ is a subset of $X_1^n$. So $X_k^n=\phi$ for $k>n+1$.
– DLN
Apr 11, 2016 at 5:18

These spaces are related to $2$-Sylow subgroups of $S_n$. For example, if $n=2^k$, then $X^n_{k+1}$ is the set of $2$-Sylow subgroups of $S_n$. To see this, note that $S_n$ acts transitively on $X^n_{k+1}$, and that the stabilizer of any fixed element of $X^n_k$ is a $2$-Sylow subgroup of $S_n$. I don't really have a reference for this, but it is not too hard to work out the order of the automorphism group, and see that it is $2^{v_2(n!)}$ using the fact that $$v_2(n!) = \lfloor n/2 \rfloor + \lfloor n/4 \rfloor + \dotsb$$
• Nice observation. I think you meant $X_{k+1}^k$ is the set of 2-sylow subgroups of $S_n$.
• Also, I guess we can do the same thing for $p$-sylow subgroups of $S_{p^n}$ if instead of taking $2$ element subsets, we take $p$-element subsets.