# Find a permutation with the given square or cube

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 $x$.

### Progress

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.

• 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. – hardmath Oct 29 '14 at 2:46
• 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 – Abigail Oct 29 '14 at 2:51
• @hardmath Thanks for the comment on my solution; I had missed the "all possible solutions" part of the question. – angryavian Oct 29 '14 at 2:52

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