The Klein group, the centralizer of a permutation, and its relation with the alternating group

The following reflexion came out of my irrepressible need to understand why the 3 double transpositions in $\mathfrak S_4$, together with the identity, formed a group $V$. Of course, one might just say: “they are stable under multiplication, as one sees by computing the 4·3/2 = 6 different products”, but who makes this computation anyway? And since I wanted not only to understand this, but to explain it to Lean, I needed an argument that could actually be done, for real. So here is an argument that requires no computation, besides the one that says that there are 3 double transpositions.

Prop. — The subgroup of $\mathfrak S_4$ generated by the 3 double transpositions is the unique 2-sylow subgroup of $\mathfrak A_4$. In particular, it has order 4 and consists of these 3 double transpositions and of the identity.
Proof. — Let $V$ be the subset of $\mathfrak S_4$ consisting of these 3 double transpositions and of the identity. Let $S$ be a 2-sylow subgroup in $\mathfrak A_4$.
We first prove $S \subseteq V$. The subgroup $S$ has order 4. Let $g\in S$. The order of $g$ divides 4, so its cycles have lengths 1, 2 or 4. If there were one cycle of length 4, then $g$ would be that cycle, hence of odd sign. Consequently, either $g=1$, or $g$ has a cycle of length 2, and then there must be a second because $g$ is even. Consequently, $S\subseteq V$, as claimed.
Since $4=\operatorname{\rm Card}(S)=\operatorname{\rm Card}(V)$, this shows that $S=V$, hence $S=\langle V\rangle$.

At this point, we still need to understand why there are 3 double transpositions. More generally, I wanted to prove that the number of permutations in $\mathfrak S_n$ of given orbit type. The orbit type a permutation $g$ is a multiset of strictly positive integers with sum $n$ given by the cardinalities of the orbits of $g$ on $\{1,\dots,n\}$. We write it as $1^{n_1} 2^{n_2}\dots r^{n_r}$, meaning that $g$ has $n_1$ orbits of length $1$ (fixed points), $n_2$ orbits of cardinality $2$, etc., so that $n= \sum n_i i$. Let $\mathscr O_g$ be the set of orbits of $g$. The action of $g$ on a given orbit $c$ coincides with a circular permutation with order the length $\ell(c)$ of this orbit; when it is nontrivial, such a permutation will be called a cycle of $g$. The supports of these cycles are pairwise disjoint, so that these cycles commute, and their product is exactly $g$. In fact, this is the only way of writing $g$ as a product of cycles with pairwise disjoint supports. (By convention, the identity is not a cycle.)

Theorem. — There are exactly $$N(1^{n_1}\dots r^{n_r}) = \frac{n!}{1^{n_1}\dots r^{n_r} n_1!\dots n_r!}$$ permutations with orbit type $1^{n_1} 2^{n_2}\dots r^{n_r}$.

A standard proof of this result goes as follows. Write the decomposition of such a permutation $g$ into cycles with disjoint supports as $g=(\cdot)\dots (\cdot)(\cdot,\cdot)\dots(\cdot,\cdot,\dots)$, leaving blank spaces for the values of the cycles (and, contradicting our convention, allowing for cycles of length 1…). There are $n!$ ways to fill these spaces with the $n$ distinct integers between $1$ and $n$, but some of them will give rise to the same permutation. Indeed, the entries in a cycle of length $s$ only count up to a circular permutation, so that we need to divide the result by $1^{n_1}\dots r^{n_r}$. Moreoveer, we can switch the order of the cycles of given length, hence we also need to divide that result by $n_s!$ (number of ways of switching the various cycles of length $s$), for all possible length $s$.

This is reasonably convincing but one could wish for something more precise, both in the proof, and in the statement. In fact, in the preceding formula, the numerator $n!$ is the order of $\mathfrak S_n$. Since all permutations with a given orbit type are conjugate by $\mathfrak S_n$, the left hand side appears as the cardinality of the orbit of a permutation $g$ of that orbit type, so that the denominator has to be equal the cardinality of the stabilizer of this permutation under the action by conjugation. Therefore, a more precise proof of this formula could run by elucidating the structure of this centralizer. This may also be interesting once one wishes to relativize the result to the alternating group $\mathfrak A_n$ in order to obtain a formula for the cardinality of the various conjugacy classes in $\mathfrak A_n$.

Let us fix a permutation $g\in\mathfrak S_n$ with orbit type $1^{n_1}\dots r^{n_r}$. The stabilizer of $g$ under the action by conjugation is its centralizer $Z_g$, the subgroup of all $k\in\mathfrak S_n$ which commute with $g$.

We first define a morphism of groups $$\tau \colon Z_g \to \mathfrak S_{n_1}\times \mathfrak S_{n_2}\times\dots \mathfrak S_{n_r}.$$ Let $\mathscr O_g$ be the set of orbits of $g$; this is a set with cardinality $n_1+n_2+\dots+n_r$. Restricted to one orbit, the action of $g$ coincides with that of a circular permutation on (which fixes the complementary subset); these circular permuations have disjoint supports, hence they commute pairwise and their product is equal to $g$. For $c\in\mathscr O_g$, we write $\ell(c)$ for its cardinality of its support, this is also the order of the cycle of $g$ defined by this orbit. If $k\in Z_g$, then $kgk^{-1}=g$. Consequently, the action of $k$ permutes the orbits of $g$, respecting their cardinalities. This defines the desired group morphism $\tau$ from $Z_g$ to a product of permutation groups $\mathfrak S_{n_1}\times \dots \mathfrak S_{n_r}$.

This morphism $\tau$ is surjective.
Indeed, given permutations $\sigma_1$ of the set of fixed points of $g$, $\sigma_2$ of the set of orbits of length 2, etc., we construct $k_\sigma\in Z_g$ such that $\tau(k_\tau)=(\sigma_1,\dots,\sigma_r)$. We fix a point $a_c$ in each orbit $c$ and decide that $k_\sigma(a_c)=a_{\sigma_i(c)}$ if $c$ has length $i$. The formula $k_\sigma g=g_\sigma$ imposes $k_\sigma (g^n a_c)=g^n a_{\sigma_i(c)}$ for all $n\in\mathbf Z$, and it remains to check that this formula gives a well defined element in $Z_g$. In fact, this formula defines a group theoretic section of $\tau$.

What is the kernel of this morphism $\tau$?
If $\tau(k)=1$, then $k$ fixes every orbit $c\in\mathscr O_g$. Since $kg=gk$, we obtain that on each orbit $c$, $k$ coincides with some power of the corresponding cycle, which has order $\ell(c)$. We thus obtain an isomorphism $$\ker(\tau) \simeq \prod_{c\in\mathscr C_g} (\mathbf Z/\ell(c)\mathbf Z).$$

To compute the cardinality of $Z_g$, it is now sufficient to compute those of $\operatorname{\rm im}(\tau)$ and $\ker(\tau)$, and this gives the formula $$\operatorname{\rm Card} (Z_g) = \operatorname{\rm Card} (\ker(\tau)) \operatorname{\rm Card} (\operatorname{\rm im}(\tau)) = 1^{n_1}\dots r^{n_r} n_1! \dots n_r!,$$ as was to be shown.

One of the interest of this argument is that it can be pushed forward to understand the structure of the conjugacy classes in the alternating group $\mathfrak A_n$. The case $n\leq 1$ is uninteresting, hence we assume $n\geq 2$. Then $\mathfrak A_n$ has index 2 in $\mathfrak S_n$, and the formulas $$\operatorname  {\rm Card}((g)_{\mathfrak A_n}) = \frac{{\rm Card}({\mathfrak A_n})}{{\rm Card}(Z_g \cap {\mathfrak A_n})} \quad\text{\rm and}\quad \operatorname  {\rm Card}((g)_{\mathfrak S_n}) = \frac{{\rm Card}({\mathfrak S_n})}{{\rm Card}(Z_g)}$$ for the cardinalities of the conjugacy classes $(g)_{\mathfrak A_n}$ and $(g)_{\mathfrak S_n}$ imply that both are equal if and only if $Z_g$ is not contained in $\mathfrak A_n$; otherwise, the conjugacy class $(g)_{\mathfrak S_n}$ is the disjoint union of $(g)_{\mathfrak A_n}$ and of a conjugacy class $(g')_{\mathfrak A_n}$ of a permutation $g'$ which is conjugate to $g$ in $\mathfrak S_n$ but not in $\mathfrak A_n$, and both have the same cardinality.

Examples of this phenomenon are classical. For example, the 5-cycles in $\mathfrak S_5$ are conjugate, but they constitute two distinct conjugacy classes under $\mathfrak A_5$. Even more elementary, the 3-cycles $(1\,2\,3)$ and $(1\,3\,2)$ are conjugate in $\mathfrak S_3$, but they are not conjugate in $\mathfrak A_3$ since that group is commutative!

So let us use our description of $Z_g$ to give a full description of this phenomenon.

As a first step, when is $\ker(\tau)$ contained in $\mathfrak A_n$? We have seen that $\ker(\tau)$ is generated by the cycles $c\in\mathscr C_g$. Consequently, $\ker(\tau)$ is contained in $\mathfrak A_n$ if and only if all of them are contained in $\mathfrak A_n$, which means that their lengths are odd.

We assume that this condition holds, so that $\ker(\tau)\subseteq \mathfrak A_n$, and now work on the image of $\tau$. Its surjectivity was proved by the means of an explicit section $\sigma\mapsto k_\sigma$. Given the preceding condition that $\ker(\tau)\subseteq \mathfrak A_n$, a necessary and sufficient condition for the inclusion $Z_g\subseteq \mathfrak A_n$ will be that the image of this section consists of even permutations. This section is a morphism of groups, so it suffices to understand the sign of $k_\sigma$ when $\sigma$ consists of a cycle $(c_1,\dots,c_s)$ in $\mathfrak S_{n_i}$ and is trivial on the other factors. Then $\ell(c_1)=\dots=\ell(c_s)$, by definition of $\sigma$. The formula $k_\sigma(g^n a_c)=g^n a_{\sigma(c)}$ shows that the non trivial cycles of $k_\sigma$ are of the form $(g^n a_{c_1},\dots, g^n a_{c_s})$; they all have the same length, $s$, and there are $\ell(c_1)$ of them. Consequently, the sign of $k_\sigma$ is equal to $(-1)^{(s-1)\ell(c_1)}=(-1)^{s-1}$ since $\ell(c_1)$ is odd. This proves that the sign of $k_\sigma$ is equal to the sign of $\sigma$. In addition to the condition that the orbits of $g$ have odd cardinalities, a necessary and sufficient condition for the image of $\sigma\mapsto k_\sigma$ to be contained in $\mathfrak A_n$ is thus that all symmetric groups $\mathfrak S_{n_i}$ coincide with their alternating groups, that is, $n_i\leq 1$ for all $i$. We can now conclude:

Theorem. — Let $1^{n_1}\dots r^{n_r}$ be a partition of $n$.

1. If $n_i=0$ for even $i$, and $n_i\leq 1$ for all $i$, then there exist two permutations in $\mathfrak S_n$ with orbit type $1^{n_1}\dots r^{n_r}$ which are not conjugate in $\mathfrak A_n$.
2. Otherwise, any two permutations in $\mathfrak S_n$ with that orbit type are conjugate in $\mathfrak A_n$.

We can check the initial example of two 5-cycles in $\mathfrak S_5$ which are not conjugate in $\mathfrak A_5$. Their orbit type is $5^1$: the only length that appears is 5, hence odd, and it has multiplicity $1$. In fact, this is the only orbit type in $\mathfrak S_5$ where this phenomenon appears!