Conjugation of $3$-cycles in $A_5$ 
How do you show that any two 3-cycles are conjugates in $A_5$. 

I know we have to take $2$ $3$-cycles say $a$ and $b$ in $A_5$ then we have to show there exists a $c\in A_5$ such that 
$a = c b c^{-1}$. But I don't know where to begin. Help me please!
 A: Let $(x_1x_2x_3)$ and $(y_1y_2y_3)$ represent the two $3$-cycles. Assume that $x_4,x_5$, respectively $y_4,y_5$ are the remaining two symbols in the five-element set $S_5$ operates on. There is a permutation $\tau$ sending $x_i$ to $\tau x_i=y_i$ for $i=1,...,5$. This will be either a permutation with positive sign or with negative sign. But if it's of negative sign, i.e. an element of $S_5\setminus A_5$, then we can compose $\tau$ with a transposition $(x_4x_5)$ to get an element in $A_5$. Now $\tau(x_1x_2x_3)\tau^{-1}=(τx_1\ τx_2\ τx_3)=(y_1y_2y_3)$
A: You could check it by hand! If you have never done such a thing, it can be instructional to concretely calculate with permutations. Take a particular $3$ cycle, for instance $(123)$, and caculate all $c(123)c^{-1}$ with $c$ running in $A_5$. That would no more than 60 calculations, which shouldn't take more than an hour if you carry them out entirely by hand, maybe even less than half an hour.
You can speed up the process if you know (or prove) that for any cycle $c=(a_1\:a_2\:\cdots a_p)\in S_n$ and $\sigma\in S_n$,
$$\sigma c\sigma^{-1}=\big(\sigma(a_1)\:\sigma(a_2)\:\cdots \sigma(a_p)\big)$$
Actually, once you know this formula, you can create, for every distinct $a,b,c\in\lbrace 1,2,3,4,5\rbrace$, a permutation $\sigma=\sigma_{a,b,c}\in S_5$ with $\sigma(123)\sigma^{-1}=(abc)$, just take
$$\sigma_{a,b,c}=\begin{pmatrix}
1&2&3&4&5\\
a&b&c&d&e
\end{pmatrix}$$
or 
$$\sigma_{a,b,c}'=\begin{pmatrix}
1&2&3&4&5\\
a&b&c&e&d
\end{pmatrix}$$
where $d,e$ are the two elements in $\lbrace 1,2,3,4,5\rbrace\smallsetminus\lbrace a,b,c\rbrace$
Now, at first sight, this doesn't solve the question for conjugacy classes in $A_5$. After all, the two choices I gave for $\sigma$ lie in $S_5$, and you want conjugacy in $A_5$. But if you take a closer look at how the two choices of conjugating element are related, you'll find that this actually answers your question. If you want I can add details tomorow, but you should be try to work this out by yourself.

More generally, the conjugacy classes of elements in $A_n$ are the same as those in $S_n$, except for those elements of $A_n$ that are products of even cycles of distinct lengths (fixed points being cycles of length one) : the conjugacy classes of such elements in $S_n$ split into two conjugacy classes in $A_n$.
For instance the $5$-cycles are conjugate in $S_5$ and $S_6$, yet they aren't in $A_5$ and $A_6$, for a $5$-cycle has a cycle decomposition as a $5$-cycle in $A_5$, and a $5$-cycle in $A_6$ has cycle structure a $5$-cycle plus a $1$-cycle. $3$-cycles are conjugate in $A_5$ because their cycle structure is one $3$-cycle plus two $1$-cycles
