Showing product of disjoint cycle

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $c$.