Show that not all 5 cycles in $A_5$ are conjugate in $A_5$ 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 $g(12345)g^{-1} = (g(1)g(2)g(3)g(4)g(5))$. 
 A: Let me suggest a correction on the approach you're talking about in the comments:


*

*Start with a $5$-cycle $(a\:b\:c\:d\:e).$

*Show that there exists another $5$-cycle $(f\:g\:h\:i\:j)$ such that for all $r\in A_5,$ we have $$\bigl(r(a)\:r(b)\:r(c)\:r(d)\:r(e)\bigr)\ne (f\:g\:h\:i\:j)$$


This is a rather cumbersome approach, though, especially since there are $5$ equivalent but distinct ways to represent a given $5$-cycle with this notation. We could also guess and check, but that is even more cumbersome.
If you know that the order of a conjugacy class must divide the order of the group, then we can proceed much more simply. How many $5$-cycles are there? What is the order of $A_5$?
A: The conjugacy map on a 5-cycle $(a,b,c,d,e)$ (or any cycle) by $\sigma$ is obtained on replacing each element of the cycle by $\sigma$ applied to that element. So select some $\sigma$ in $S_5$ but not in $A_5$, for example the 2-cycle $(a,b)$, and conjugate a given 5-cycle by that. 
Specific example: The two cycles $(12345)$ and $(21345)$ cannot be conjugate in $A_5$. (The second cycle can of course be rewritten as $(13452)$ if you want it to start with the lowest entry as is usually done.)
ADDED: as SteveD has noted in a comment, this needs a further check. The cycle (21345) may be written in 5 ways, each of which has to be checked to see whether the conjugating map from 12345 to it is in $A_5$. It turns out non e of them are:
$$12345 \to 21345 : (12)\\ 12345 \to 13452 : (2345) \\
12345 \to 34521: (135)(24) \\ 12345 \to 45213 : (14)(253) \\ 12345 \to 52134 : (1543).$$
In each case the cycle form of the conjugating map lies in $S_5-A_5$, i.e. the conjugating map is an odd permutation. This means there is no conjugation of $(12345)$ to $(21345)$ by an element of $A_5.$
A: Here is a quite general lemma, you might be interested in.

Let $G$ be a finite group acting transitively on a set $\Omega$ and let $N \trianglelefteq G$ be a normal subgroup. Then the number of $N$-orbits of $\Omega$ equals $|G : N G_\omega|$ where $G_\omega$ is the stabilizer of an arbitrary $\omega \in \Omega$ in $G$. Furthermore they all have the same length.

Consider now $G = S_n$ and let $\Omega$ be the set of all permutations of $S_n$ with the same given cycle type (e.g. all cycles of length 5). It is well known that $S_n$ acts transitively on $\Omega$ by conjugation, so the above lemma tells us that $\Omega$ splits into $|S_n : A_n C_\sigma| \in \{1,2\}$ $A_n$-orbits where $C_\sigma$ is the centralizer of some permutation $\sigma \in \Omega$. In particular there are elements in $\Omega$ which are not conjugated in $A_n$ if and only if the centralizer of one (and hence all) permutation of $\Omega$ is contained in $A_n$.
From the proof of that lemma we see that in this case $\sigma$ and $\sigma^\tau$ are not conjugated in $A_n$ for any $\sigma \in \Omega$ and $\tau \in S_n \setminus A_n$.
A: 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 $A_5$.
