Multiplying two transpositions using cycle notation

Question

Let $(ab)$ and $(cd)$ be two disjoint cycles of $[1,n]$ for some positive integer $n$. I don't understand why $(ab)(cd)=(dac)(abd)$. Is there a way to actually derive this?

By "calculation", I end up with the disjoint cycles $(ab)$ and $(cd)$ again (for example $(ab)(cd)$ sends $a$ to $b$ and $b$ to $a$).

Is this a systematic result? Or is it deduced by trial and error?

Answer 1

$(dac)(abd)=(dc)(da)(ad)(ab)$ but since $(da)=(ad)=(da)^{-1}$ and $(ab),(cd)$ are disjoint we must have $(dac)(abd)=(dc)(ab)=(ab)(cd)$.

Answer 2

By direct calculation:

$$\begin{align} a&\stackrel{(cd)}{\mapsto}a\stackrel{(ab)}{\mapsto}b,\\ b&\stackrel{(cd)}{\mapsto}b\stackrel{(ab)}{\mapsto}a,\\ c&\stackrel{(cd)}{\mapsto}d\stackrel{(ab)}{\mapsto}d,\\ d&\stackrel{(cd)}{\mapsto}c\stackrel{(ab)}{\mapsto}c, \end{align}$$

whereas

$$\begin{align} a&\stackrel{(abd)}{\mapsto}b\stackrel{(dac)}{\mapsto}b,\\ b&\stackrel{(abd)}{\mapsto}d\stackrel{(dac)}{\mapsto}a,\\ c&\stackrel{(abd)}{\mapsto}c\stackrel{(dac)}{\mapsto}d,\\ d&\stackrel{(abd)}{\mapsto}a\stackrel{(dac)}{\mapsto}c,\\ \end{align}$$

so we can see that $(ac)(bd)$ and $(dac)(abd)$ map each element of $\overline{1, n}$ to the exact same element in each case; therefore, they are equal.