Problem: find a permutation such that

  1. $x^2 = (1\;3\;4\;5\;7)$, $x\in S_7$
  2. $x^3 = (1\;3\;4\;5\;7)$, $x\in S_7$

Must find all possible solutions for $x$.


I have solved for the first part, $x=(57134)$ but I am not sure on how to prove that there is only one $x$ value.

  • $\begingroup$ Surely you are posing two separate problems, since we cannot have $x^2=x^3$ unless $x = x^2 = x^3$ is the identity map. $\endgroup$ – hardmath Oct 29 '14 at 2:46
  • $\begingroup$ I have solved for the first part, x=(57134) but I am not sure on how to prove that there is only one x value $\endgroup$ – Abigail Oct 29 '14 at 2:51
  • $\begingroup$ @hardmath Thanks for the comment on my solution; I had missed the "all possible solutions" part of the question. $\endgroup$ – angryavian Oct 29 '14 at 2:52

Note that in cycle notation for permutations, $(1~3~4~5~7) = (5~7~1~3~4)$.

Part 1: The idea here would be to create a cycle that moved 1 to 3 after two moves, 3 to 4 after two moves, 4 to 5 after two moves, 5 to 7 after two moves, and 7 to one after two moves.

Then (arbitrarily and without loss of generality), 1 is first in our cycle notation, followed by something, then 3, then something, then 4. So we have $(1~a_1~3~a_2~4)$ as our cycle. Now we try to figure out what goes in place of $a_1$ and $a_2$. After 1 iteration of this permutation, 4 maps to 1. On the second iteration, it should map to five, so that $x^2(4) = 5$. Note that I'm viewing $x$ as a function with $x^2 = x \circ x$. We now have $x = (1~5~3~a_2~4)$. Next, we need $x^2(5) = 7$. The way to do that is to set $a_2 = 7$. Thus we see, and can confirm, that $(1~5~3~7~4)^2 = (1~3~4~5~7)$. There's a pair left over in $S_7$; we haven't moved 2 or 6. Since $(2~6)^2$ is the identity, we can set $x = (1~5~3~7~4)(2~6)$ as well.

Hopefully this helps; the teacher in me won't let me do the whole problem (or the actual proof) for you. :)


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