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Given $a,b\in S_n$ such that their commutator has at least $n-4$ fixed points, is there an element $z\in S_n$ such that $a^z=a^{-1}$, and $b^z=b^{-1}$? Here $a^z:=z^{-1}az$.

Magma says that the answer is yes up to $n\leq 11$ (computed by M. Zieve). The answer is also yes if $a$ and $b$ commute via the argument below. It would be great to have an answer in case a,b do not commute, even under the assumption that $\langle a,b\rangle$ contains the alternating group $A_n$. If $\langle a,b\rangle$ contains $A_n$ with $n\geq 4$, then any $z$ as above is unique.

The commutative case: If $a^b=a$, the orbits (and cycles) of $a$ are permuted by $b$. Say $b$ permutes under conjugation cyclicly the orbits $A_1,A_2,\ldots,A_r$ of $a$. Let $a_i\in S_n$ be the cycle induced by $a$ on $A_i$, for $i=1,\ldots,r$. Choose $c\in S_n$ to be element of order $r$ such that $b=c$ on $A_i$ for $i=1,...,r-1$. (This forces $a_r^c=a_1$). Since $b=c$ outside $A_1$ and $a_1^{bc^{-1}}=a_1$, we have $b=a_1^e c$ for some integer $e$.

Choose $v$ to be an involution on $A_1$ such that $a_1^v=a_1^{-1}$ on $A_1$, and trivial outside $A_1$. Choose $w$ to be the involution in the symmetric group on $1,\ldots,r$ such that $(1,\ldots,r)^w=(r,\ldots,1)$ and $1^w=2$. Define $u$ to be a permutation such that if $(i,j)$, $i\leq j$, is an orbit of $w$, then $\alpha^u = \alpha^{c^{j-i}}$ for every $\alpha\in A_i$, $i=1,\ldots,r$. This way $c^u =c^{-1}$. Let $z:=vv^c\cdots v^{c^{r-1}}u$, so that $c^z=c^{-1}$, and $a^z=a^{-1}$. (Moreover, $a_i^z=a_{i^w}^{-1}$ for $i=1,\ldots,r$). It follows that $b^z=a_{1^w}^{-e}c^{-1} = a_2^{-e}c^{-1}=c^{-1}a_1^{-e} = b^{-1}$.

Some might consider this proof (in the commutative case) to be simpler in the language of wreath products.

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    $\begingroup$ After doing lots of random experiments I am starting to believe that this might be true. If so, then I am afraid that it might not be easy to prove! $\endgroup$ – Derek Holt Apr 14 '17 at 19:11
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    $\begingroup$ Note also that it is very easy to find examples in which $[a,b]$ has $n-5$ fixed points and there is no element $z$ inverting both $a$ and $b$. $\endgroup$ – Derek Holt Apr 18 '17 at 15:16
  • $\begingroup$ Totally naive comment, but would the 4 vs 5 thing have anything to do with when A_n is simple? $\endgroup$ – Sam Hopkins Apr 18 '17 at 15:19
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    $\begingroup$ @SamHopkins I don't think so. I think a reason why many fixed points matter is as follows: Suppose $a^{-1}b^{-1}ab=x$. If $z$ inverts $a$ and $b$, then $x^z=a^{-z}b^{-z}a^zb^z=aba^{-1}b^{-1}$, so $x^{zab}=x$. Thus the $z$ we are looking for is in the coset $C_G(x)b^{-1}a^{-1}$. Note that $C_G(x)$ is $S_{n-k}$ times a subgroup of $S_k$, where $k\le4$. I believe the reason for the existence of $z$ is the hugeness of $C_G(x)$. $\endgroup$ – Peter Mueller Apr 18 '17 at 15:44
  • $\begingroup$ Have you tried using Rutilo Moreno, Luis Manuel Rivera, Blocks in cycles and $k$-commuting permutations, SpringerPlus, December 2016, 5:1949? (A preprint also appears as arXiv:1306.5708.) Corollary 1 characterizes when the commutator of two permutations has at least $n-k$ fixed points (the authors say that these two permutations $k$-commute). $\endgroup$ – darij grinberg Jun 16 '18 at 21:55

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