TITLE: Show that not all 5 cycles in $A_5$ are conjugate in $A_5$
QUESTION [7 upvotes]: This is a practice question (not assignment). One thing I am wondering is what this question exactly means. From my understanding, the conjugate of $a \in A_5$ is $gag^{-1} \in A_5$ where $g \in A_5$. I was trying to look at a similar question here on MSE and someone mentioned $(12345) \in A_5$ is not conjuagte in $A_5$. So does that means there exists a $g \in A_5$ such that $g(12345)g^{-1} \notin A_5$ (I am guessing this is incorrect since $A_5$ is a group). I am also away of the property that $g(12345)g^{-1} = (g(1)g(2)g(3)g(4)g(5))$.

REPLY [3 votes]: Spelling/fleshing out the counting argument by Cameron Buie from the comments under his answer (which is also the underlying idea in Dune's +1 answer).
The size of a conjugacy class $[g]$ of the element $g$ in a finite group $G$ is
$$
|[g]|=\frac{|G|}{|C_G(g)|}.
$$
Here $C_G(g)=\{x\in G\mid xg=gx\}$ is the centralizer of $g$ in $G$.
Our knowledge about conjugacy classes in $S_5$ tells us that a 5-cycles has 24 conjugates in $S_5$. Let $\sigma\in S_5$ be a fixed 5-cycle. The above formula tells us that $C_{S_5}(\sigma)$ is of order five. As any element commutes with its powers, we can conclude that
$$C_{S_5}(\sigma)=\langle \sigma\rangle.$$
As this subgroup is contained in $A_5$, we can conclude that 
$$
C_{A_5}(\sigma)=C_{S_5}(\sigma).
$$
Another application of the formula above then tells us that the conjugacy class of any 5-cycle in $A_5$ has exactly 12 elements. In other words, the $S_5$ conjugacy class splits in two.

More detailed information can be obtained by studying the normalizer of a Sylow $5$-subgroup in $S_5$. There are six such Sylow subgroups, so their normalizers have order $20$. If we specify $\sigma=(12345), P_5=\langle\sigma\rangle$ we see that the 4-cycle
$\beta=(2354)$ satisfies the relation
$$
\beta\sigma\beta^{-1}=[1\mapsto\beta(2)\mapsto\beta(3)\mapsto\beta(4)\mapsto\beta(5)\mapsto1]=(13524)=\sigma^2.
$$
Therefore $N_{S_5}(P_5)=\langle\sigma,\beta$ as the latter clearly normalizes $P_5$ and it has the correct order $20$.
The problem comes from the fact that $\beta\notin A_5$. Any element $\tau\in S_5$
with the property $\tau\sigma\tau^{-1}=\sigma^2$ has the property that $\tau\beta^{-1}$ must commute with $\sigma$. In other words we must have $\tau=\beta\sigma^k$ for some $k$. But all these permutations are odd and thus not available in $A_5$.