Finding permutations off given action

28 Views Asked by At

Let $\mathfrak{A}=\{A,\dots,Z\}$ and $\pi_i$ be involutions on $\mathfrak{A}$ without fixed points, $i=1,2,3$. Given are all triples $\pi_1(a)\pi_2(a)\pi_3(a)$ for $a\in\mathfrak{A}$:

APN BOY CCQ DTT ESF FXM GQE HEO IHR JRB KLU LZG MYJ

NJI OFA PAV QGC RWW SMK TKX UBH VVP WIZ XDL YND ZUS

Task: Find $\pi_1,\pi_2$ and $\pi_3$.


What I've got so far:

1) We have $\pi_i\pi_j(a_j)=a_i$ for all $i,j\in\{1,2,3\}$ since $\pi_i^2=\text{id}_\mathfrak{A}$. With this I was able to find all $\pi_i\pi_j$. I will not list them here as this takes up too much time and space.

2) My goal now is to find either one of the $\pi_i$'s since knowing one implies knowing the others.

3) If $a$ is a fixed point of $\pi_i\pi_j$, then $\pi_j(a)$ is a also a fixed point of $\pi_i\pi_j$ and $(a,\pi_j(a))$ is a cycle of $\pi_j$.

4) Since $\pi_1\pi_2$ has fixed points C and V and $\pi_2$ doesn't have fixed points we can conclude $\pi_2(C)=V$ and $\pi_2(V)=C$. Simmilar considering $\pi_3\pi_2$ we can obtain $\pi_2(T)=W$ and $\pi_2(W)=T$.

So in total $\pi_2=(CV)(TW)\cdots$ is what I have. Is there a smart way to get more?