Let $n = 2k \gt 0$ be an even integer, and $S_n$ the symmetric group on $\{1,..,n\}$.

Let $\mu$ be the permutation s.t. $\mu(x) = n-x+1$.

Question: What is the centralizer of $\mu$ in $S_n$ ? That is, what is the subgroup $Z(\mu)$ of all permutations in $S_n$ that commute with $\mu$ ?

Note that $\mu$ is the product of $k$ disjoint transpositions, and thus an involution (self-inverse), $\mu = (1\;n) (2\;n-1) \ldots (k\;k+1)$. Because they are disjoint, these transpositions commute, and it follows that $\mu$ commutes with any product of a subset of these.

But such products do not generally account for the entire centralizer $Z(\mu)$. For example with $n=6$, the permutation $(1\;3) (4\;6)$ commutes with $\mu$ but is not expressible as a product of the disjoint transpositions above that form $\mu$.

If possible an explicit characterization of $Z(\mu)$ elements would be preferred, but a construction of generators (valid for all $n$) would also be useful. The application I have in mind is the enumeration of diagonal latin squares; see also a related question I posted previously.

The idiosyncratic term used there, $\mu$-derangement, may be justified by its definition being equivalent to both a permutation $g$ and $\mu g$ being derangements (permutations without fixed points). Computational experience suggests that $\mu$-derangements which share certain easily computed invariants result in equal numbers of "normalized" latin squares, and that the broadest extent of such equivalence may derive from conjugation by permutations in $Z(\mu)$.


2 Answers 2


The commutativity condition $\mu\pi=\pi\mu$ can be written as $\mu\pi\mu^{-1}=\pi$. Thus a permutation $\pi$ commutes with $\mu$ if and only if relabeling the cycle structure of $\pi$ according to $\mu$ leaves $\pi$ invariant. That means that for a given cycle of $\pi$, either the cycle has to have even length and pairs of elements at maximal distance from each other in the cycle must be mapped to each other by $\mu$, or the cycle has to be matched with another cycle of the same length and $\mu$ has to map the elements of the cycles to each other in order. (In particular, $\pi$ must have an even number of cycles of every odd length; note the similarity to the condition for a permutation to be the square of some permutation, which is that there must be an even number of cycles of every even length.)

There are $2^kk!$ such permutations, and they are in bijection with $\mathbb Z_2^k\times S_k$. The permutation in the centralizer of $\mu$ corresponding to $(v,\rho)\in\mathbb Z_2^k\times S_k$ is generated from the cycle structure of $\rho$ by making binary choices according to the elements of $v$. For each cycle $\gamma$ of $\rho$, the element $v_i$ corresponding to the least element $i$ in $\gamma$ determines whether $\gamma$ gets cloned to form a pair of cycles mapped to each other under relabeling with $\mu$, or whether $\gamma$ gets doubled in length to form a cycle mapped to itself under relabeling with $\mu$. The elements $v_j$ corresponding to the remaining elements $j\ne i$ in $\gamma$ determine whether $j$ or $n+1-j$ takes the place of $j$ in the cloned or doubled cycle(s).


The centralizer of $\mu$ is a example of a wreath product: Specifically, it is isomorphic to $C_{2} \wr S_{k},$ so its order is indeed $k! 2^{k}.$ The "base group" is elementary Abelian group of order $2^{k},$ and is generated by the $k$-disjoint transpositions appearing in $\mu,$ and the $S_{k}$ permutes these $k$ generating transpositions around as if they were individual points. But it should be noted that the centralizer is definitely not a direct product- it is a particular kind of semi-direct product, but the $S_{k}$ induces non-trivial automorphisms of the base group.

  • $\begingroup$ Thanks, that's a nice observation. It reminds me of another Question I asked about a semi-direct product (one of fixed size). $\endgroup$
    – hardmath
    Jun 11, 2014 at 23:03

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