Intuition for expressing a cycle as a product of transpositions.

Question

I have recently learned the proof of:

$(a_1 \ a_2 \ \ldots \ a_n) = (a_1 \ a_k)(a_1 \ a_{k-1}) \ldots (a_1 \ a_2) = (a_k \ a_{k-1}) (a_k \ a_{k-2}) \ldots (ak \ a_1) = (a_1 \ a_2) (a_2 \ a_3 ) \ldots (a_{k-1} \ a_k)$

Now I can prove the above 3 equalities through induction but unfortunately induction, unlike a direct proof, doesn't quite tell me what is really going on. I'm struggling to understand this intuitively. I don't want to just memorize it.

Can anyone walk me through this with an intuitive explanation. Thank you.

Answer 1

$(a_1,a_2,\cdots, a_n) = (a_1,a_n)(a_1,a_{n-1})\cdots(a_2, a_1)$

Read the composition of cycles from right to left. How do they act on the element $a_1$? The right most transposition "sees" $a_1$ and takes it to $a_2.$ None of the rest of the transpositions act on $a_2.$

Consider the element $a_k,$ the first several transpositions do not act on $a_k.$ The first transposition that does is $(a_1,a_k).$ That takes $a_k$ to $a_1.$ The next transpositions takes that element to $a_{k+1}.$

In this chain of transpositions, $a_1$ is acting as the interchange for all of the other elements. Every element gets sent to $a_1$ and then sent from $a_1$ to its destination.

$(a_n,a_{n-1})\cdots (a_n,a_1)$ is the same idea, except $a_k$ is acting as the interchange.

$(a_1,a_2)(a_2,a_3)\cdots(a_{n-1},a_n)$ even though it is ultimately the same cycle, the activity that is going on is a little bit different. Consider the element $a_k,$ it glides along unaffected by the first several transpositions, until it hits $(a_{k+1},a_k)$ gets shifted and then again the rest of the transpositions do nothing.

Answer 2

The intuition you seem to need is how to read off from a permutation given as a product of cycles the mapping the permutation represents. If you consider a single cycle such as $(1 \; 2 \; 3)$, then it is the permutation that maps $1$ to $2$, $2$ to $3$ and $3$ to $1$. When you look at a product of two or more cycles that may overlap, then you have to decide whether to work from left to right or from right to left. If I work from left to right $(1 \; 2 \; 3)(2 \; 4)$ maps $1$ to $4$ (via $2$), $2$ to $3$, $3$ to $1$ and $4$ to $2$.

So now it should be intuitively clear that $(1 \; 2)(2 \;3)\ldots (47 \; 48) (48 \;1) $ (for example) is the cyclic permutation $(1\; 2\; \ldots \; 48)$. And, in general that if the $a_i$ are all distinct that $(a_1\;a_2)(a_2\;a_3)\ldots (a_{n-1}\;a_n)(a_n\;a_1)$ is the cyclic permutation $(a_1\; a_2\; \ldots \; a_n)$.