Showing product of disjoint cycle I am trying to show the product of two disjoint cycles such that they have nothing in common for $A_n$ for $n\ge 3$.
So I have the two cycles $(ab)(cd)$. I have read here: http://people.math.gatech.edu/~ecroot/3cycle.pdf, that this will be (dac)(abd). I am having trouble seeing this can someone please explain.
Note I found that proof online and I have no relationship with it.
 A: Trace what happens to each element:


*

*in $(ab)(cd)$ the element $d$ goes to $c$ which then is not changed;

*in $(dac)(abd)$ the element $d$ goes to $a$ which then goes to $c$.


So $d$ ends up at $c$ in both cases.  Check $a,b,c$ in the same way.  And any element $x$ other than $a,b,c,d$ is not affected by either permutation, so ends up as $x$ itself in both cases.
A: The cycle $(abd)$ sends $a$ to $b$, $b$ to $d$, and $d$ to $a$, leaving $c$ alone. The cycle $(dac)$ sends $d$ to $a$, $a$ to $c$, and $c$ to $d$, leaving $b$ alone. What happens when you apply first $(abd)$ and then $(dac)$?
$$\begin{align*}
&a\mapsto b\mapsto b\\
&b\mapsto d\mapsto a\\
&c\mapsto c\mapsto d\\
&d\mapsto a\mapsto c
\end{align*}$$
Notice that the net effect is the same as that of $(ab)(cd)$: $a$ goes to $b$, $b$ goes to $a$, $c$ goes to $d$, and $d$ goes to $c$.
A: So you want to see that $$\sigma = (ab)(cd) = (dac)(abd)= \tau$$ You can simply try to see what happens to each "number". 
Note that
$$
\sigma(a) = b\\
\sigma(b) = a\\
\sigma(c) = d\\
\sigma(d) = c\\
\\
\tau(a) = b\\
\tau(b) = a\\
\dots
$$
For example, $\tau$ applied to $d$ first sends $d$ to $a$ and then $a$ to $c$.
Since $\sigma(x) = \tau(x)$ for $X$ being $a$, $b$, $c$, or $d$, they are equal.
