TITLE: How can I write the product of two transpositions as product of 3-cycles?
QUESTION [5 upvotes]: Consider this equality in the group $S_n$ :
$$(ab)(cd)=(cad)(abc)$$
I can prove it for myself, but I get the impression that I should be able to see this very quickly. How could I get more intuition on this ?
$$
a \overset{(abc)}{\mapsto} b \overset{(cad)}{\mapsto} b \\  
b \overset{(abc)}{\mapsto} c \overset{(cad)}{\mapsto} a \\
c \overset{(abc)}{\mapsto} a \overset{(cad)}{\mapsto} d  \\
d \overset{(abc)}{\mapsto} d \overset{(cad)}{\mapsto} c \\$$

REPLY [7 votes]: Instead of tracing through where it maps elements, you can also derive this from basic cycle manipulations:
$$(cad)(abc) = (dca)(cab) = (dc)(ca)(ca)(ab) = (dc)(ab)$$
The two rules I'm using are

$(a_1\ldots a_n) = (a_2\ldots a_na_1)$
$(a_1\ldots a_n) = (a_1a_2)(a_2a_3)\cdots(a_{n-1}a_n)$