Let $\sigma = (12345)$ and $\tau = (13524)$, find an element $\rho$ such that $\rho \sigma \rho^{-1}=\tau$ and then show there are exactly $5$ such elements.

Now I computed $\rho$ using $\rho \sigma \rho^{-1} = \left(\rho(1) \rho(2) \rho(3) \rho(4) \rho(5)\right) = (13524),$ thus $\rho(1)=1, \rho(2)=3, \rho(3)=5, \rho(4)= 2, \rho(5)=4$, which results in $\rho = (2354)$.

Now how can I show there are exactly $5$ such elements? I imagine another $\rho$ could be found by just shifting the permutation, such as $\rho_{2} = (3542)$, but this would give me $4$ elements in stead of $5$.


Your "shifted" $\rho$ is actually the same permutation. However, note that $(13524)$ is also the same permutation as $(35241)$, and using that as your $\tau$ will actually give you a different $\rho$.

As for showing that these are the only $5$ you can get, consider splitting into cases depending on, say, what $\rho(1)$ is, and show that this forces the value of $\rho$ on $2, 3, 4$ and $5$.

  • $\begingroup$ Would then showing with $\tau = (13524) = (25241) = (52413) = (24135) = (41352)$, with all its own corresponding $\rho$ be sufficient? $\endgroup$ – Mathbeginner May 23 at 9:09
  • $\begingroup$ @Mathbeginner That will give you five different $\rho$, which is half of hat you are asked. You are also asked to show that there are no other possible $\rho$ (they say "exactly 5"), and that's what my second paragraph is about. $\endgroup$ – Arthur May 23 at 9:10
  • $\begingroup$ thanks! That makes sense. $\endgroup$ – Mathbeginner May 25 at 10:02

Consider $S_5$ acts on the set of $5$-cycles (by conjugation). The stabilizer is of order $5!/4! = 5$, where $5!=|S_5|$ and $4!$ is the number of $5$-cycles. Actually the stabilzer is exactly the centralizer of $\sigma$, by the definition of action. What you are going to find is the size of a coset of the stabilizer, so there are $5$.


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