TITLE: Conjugation of $3$-cycles in $A_5$
QUESTION [6 upvotes]: 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!

REPLY [7 votes]: 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 $\tau(x_1x_2x_3)\tau^{-1}=(τx_1\ τx_2\ τx_3)=(y_1y_2y_3)$