# In $S_{5}$ show there are $5$ elements $\rho$ with $\rho \sigma \rho^{-1}=\tau$ for given $\sigma$ and $\tau$

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 $$5$$.
• Would then showing with $\tau = (13524) = (25241) = (52413) = (24135) = (41352)$, with all its own corresponding $\rho$ be sufficient? – Mathbeginner May 23 at 9:09
• @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. – Arthur May 23 at 9:10
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 $$5$$.