Consider the group of $mn\times mn$ permutation matrices $\mathfrak{S}\_{mn}$ and partition each such matrix $P$ into $n^2$ blocks of $m\times m$ matrices $Q\_{i,j}$. Now, transpose each $Q\_{i,j}$ (independently) to form a new $mn\times mn$ matrix denoted $P^t$ (with an abuse of notation). Let's construct the set $U\_{mn}:=\{P^t\in\mathfrak{S}\_{mn}:\, P\in\mathfrak{S}\_{m,n}\}$.


It's clear that if $m=1$ then $U\_{mn}=\mathfrak{S}\_{mn}$; the same if $n=1$.



> 
> **Question 1.** For which $m$ and $n$, does $U\_{mn}$ form a group?
> 
> 
> UPDATE. Negative answer shown below.
> 
> 
> 


A cute note: $\frac{\#\mathfrak{S}\_{2n}}{\# U\_{2n}}=C\_n$ is the [Catalan number](https://en.wikipedia.org/wiki/Catalan_number). Is $\# U\_{2n}$ a subgroup?
**Answer.** No. See below. Another counterexample to the converse of [Lagrange's theorem](https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)): $\# U\_{2n}$ divides $\#\mathfrak{S}\_{2n}$ but $U\_{2n}$ is not a subgroup of $\mathfrak{S}\_{2n}$.



> 
> **Question 2.** View $P\in U\_{2n}$ as $P\leftrightarrow\sigma$ as a $1$-line permutation $\sigma=(\sigma\_1,\dots,\sigma\_{2n})$. In this way, what is an equivalent (to the above "transpose") characterization of $P$ in the language of $\sigma$?
> 
> 
> UPDTE. I am Still waiting for an answer.
> 
> 
> 


If you're interested in enumeration then [visit here](https://mathoverflow.net/questions/258625/transposing-unrimmed-permutations).



does this set of permutations form a group? And more
For $m, n > 1$ it's never a group.


Let $\sigma \in \mathfrak{S}\_n$, $\sigma$ switches $1, m + 1$, and in fact $\sigma = \sigma^t$. Clearly, $\sigma \in U\_{mn}$. Similarly, let $\tau$ switch $1, 2$; similarly, $\tau = \tau^t$. Then $\sigma \tau \sigma$ switches $2, m+1$. This clearly is not in $U\_{mn}$. 


