Characterization of transpositions in symmetric group?

Question

For a symmetric group on an arbitrary (indeed potentially infinite) set, is it true that $x$ is a transposition if and only if $x^2=1$, $x \not= 1$, and $y^2=1$ and $C(x) \subseteq C(y)$ implies $y=x$ or $y=1$? Here $C(x) = \{y | xy=yx \}$ is the centralizer of $x$. If this is wrong, can you find a similar characterization of transpositions?

(This is used in Wilfrid Hodges's Model Theory Section 5.4 Example 2 of Interpreting a Set in its Symmetric Group on page 224, though the permutation group question stands on its own).

I believe I have proved that for such an $x = \Pi_{i \in I}(a_ib_i)$, permutation $\sigma$ commutes with $x$ $\iff x \sigma x = \sigma \iff \sigma$ sends all $\{a_i,b_i\}$ setwise to some $\{a_j,b_j\}$ (seen straightforwardly by plugging $a_i$ into both sides). With this, however, I believe that for $x^2=1,x\not=1$ and $y^2=1$ such that $C(x) \subseteq C(y)$, actually always $y=x$ or $1$, so the above characterization in the book is wrong? (If so, does anyone have another reference for the above interpretation, or know such an interpretation actually doesn't exist?)

Thank you for any help as always.

UPDATE: The characterization of transpositions ($x$ transposition $\iff \forall y ((xx^y)^2=1 \vee (xx^y)^3=1$) is valid for $S(\Omega)$ with $|\Omega|=5$ and $|\Omega| \geq 7$ including infinite $\Omega$ (therefore Hodges has a mistake in their transposition definition and the exercise asking to show it works for 6 which can't be due to the exceptional outer automorphism). I'll quickly add that they are trivially definable for $|\Omega|=1,2,3$, and for $|\Omega|=4$ they are exactly the elements of order 2 where it is possible to multiply on the left by an element of order 4 and get an element of order 3.

Answer 1

Here is an alternative way to characterize the transpositions, which works in $S_X$ whenever $|X|>6$: $\sigma$ is a transposition if and only if $\sigma$ has order $2$, and for every conjugate $\sigma' = \rho\sigma\rho^{-1}$, $\sigma\sigma'$ has order $1$, $2$, or $3$.

Using this characterization, I sketched the interpretation in $S_X$ of the group action of $S_X$ on $X$ in this answer.

In $S_n$ for $n<6$, the set of transpositions is an automorphism-invariant set (since every automorphism of $S_n$ is inner), and hence definable. But the outer automorphism of $S_6$ does not preserve the set of transpositions, so the set of transpositions is not even definable in $S_6$.

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Note that Hodges only claims that his characterization works for symmetric groups on infinite sets. But as established in the comments, the characterization is wrong even in this context. Let's figure out which permutations his condition actually defines.

Let $A$ be a set and $S_A$ the symmetric group on $A$. For $\sigma\in S_A$, let $\DeclareMathOperator{\supp}{supp}\supp(\sigma) = \{a\in A\mid \sigma(a)\neq a\}$. If $\sigma^2 = 1$, then $\supp(\sigma)$ is partitioned into orbits of size $2$, and $\sigma$ transposes the elements of each of these orbits. Let $O_\sigma$ be the set of these orbits, and note that $\sigma$ is uniquely determined by $O_\sigma$.

Now for $\rho\in S_A$, $\rho\in C(\sigma)$ if and only if $\sigma = \rho\sigma\rho^{-1}$ if and only if $O_\sigma = O_{\rho\sigma\rho^{-1}}$ if and only if $O_\sigma = \rho(O_\sigma) = \{\{\rho(a),\rho(b)\}\mid \{a,b\}\in O_\sigma\}$. In other words, $\rho$ commutes with $\sigma$ if and only if $\rho$ permutes the non-trivial orbits of $\sigma$ (and acts arbitrarily on the trivial orbits).

Claim: If $\sigma^2 = \tau^2 = 1$, then $C(\sigma)\leq C(\tau)$ if and only if (a) $\tau = 1$, (b) $\tau = \sigma$, or (c) $\supp(\sigma) = A\setminus \{a,b\}$ for some $a,b\in A$, and $\tau = (a\,b)$ or $\tau = \sigma (a\,b)$.

Proof: If $\tau = 1$, then $C(\sigma)\leq C(\tau) = S_A$. If $\tau = \sigma$, then $C(\sigma) = C(\tau)$. And if (c) holds, then for all $\rho$, if $\rho(O_\sigma) = O_\sigma$, then $\rho(\{a,b\}) = \{a,b\}$, so $\rho(O_\tau) = O_\tau$. Thus $C(\sigma)\leq C(\tau)$.

Conversely, assume $\sigma^2 = \tau^2 = 1$ and $C(\sigma)\leq C(\tau)$. Let $a\in \supp(\sigma)\cap \supp(\tau)$, and suppose $\{a,b\}\in O_\sigma$. Then $(a\,b)\in C(\sigma)$, so $(a\,b)\in C(\tau)$, and thus $\{a,b\}\in O_\tau$. In particular, $\sigma(a) = \tau(a)$. Thus $\sigma$ and $\tau$ agree on $\supp(\sigma)\cap \supp(\tau)$.

Now if $\supp(\sigma)\cap \supp(\tau)$ is non-empty, I claim that $\supp(\sigma)\subseteq \supp(\tau)$. Indeed, pick some $a\in \supp(\sigma)\cap \supp(\tau)$. Now let $b\in \supp(\sigma)$. Let $a' = \sigma(a) = \tau(a)$, and let $b' = \sigma(b)$. Then $(a\,b)(a'\,b')$ swaps the $\sigma$-orbits $\{a,a'\}$ and $\{b,b'\}$, so it is in $C(\sigma)$ and hence in $C(\tau)$. Thus $\{b,b'\}$ is also a $\tau$-orbit, and $b\in \supp(\tau)$.

Case 1: $\supp(\tau)\subseteq \supp(\sigma)$. If $\supp(\tau) = \varnothing$, then $\tau = 1$ and (a) holds. If $\supp(\tau) \neq \varnothing$, then by the claim, $\supp(\sigma)\subseteq \supp(\tau)$, so they are equal. Then $\sigma = \tau$, since they agree on their supports, and (b) holds.

Case 2: $\supp(\tau)\not\subseteq \supp(\sigma)$. Let $a\in \supp(\tau)\setminus \supp(\sigma)$, and let $b = \tau(a)$. Note that $b\notin \supp(\sigma)$, otherwise we would have $\sigma(b) = \tau(b) = a$ and $a\in \supp(\sigma)$. Suppose for contradiction that there is some $c\in A\setminus \supp(\sigma)$ with $c\neq a$ and $c\neq b$. Then $(a\,c)\in C(\sigma)$, since neither $a$ nor $c$ are in $\supp(\sigma)$, but $(a\,c)\notin C(\tau)$, contradiction. Thus $A\setminus \supp(\sigma) = \{a,b\}$.

Now if $\supp(\tau)$ is disjoint from $\supp(\sigma)$, we must have $\tau = (a\,b)$. Otherwise, by the claim, $\supp(\sigma)\subseteq \supp(\tau)$, and since $\tau$ moves $a$ to $b$ and agrees with $\sigma$ on $\supp(\sigma)$, $\tau = \sigma(a\,b)$. In either case, (c) holds. $\square$

This shows that the condition in Hodges actually characterizes those elements of $S_A$ of order $2$ whose suppose is not of the form $S_a\setminus \{a,b\}$ for some $a,b\in A$.