Understanding the cycles $(12)(13)(14)$ and $(13)(1234)(13)$

Question

I need help in understanding the decomposition in disjoint cycles for the permutations $(1 2) (1 3) (1 4)$ and $(1 3)(1 2 3 4)(1 3)$ which has been listed in the solutions as $(1 4 3 2)$ in both cases.

For the first cycle all I can think of us that we start with the first factor $(1 4)$ and we say that $1 \to 4$ and then $4 \to 1$, which to my mind would have constituted a cycle in and of itself, but given the answer I guess we then can say that $1 \to 3$ and then $3 \to 1$ and then again that $1 \to 2$ and $2 \to 1$

So then would we continue on with a cycle even if it appears that the cycle is completed (returned to the original number) if that original number is used in another factor that is involved in the product? Like we see the number $1$ is in the product above. In other words we usually know that a cycle is completed when we return to the original number, but if in the process of returning to that original number we come back to it more than once we only complete the cycle when we return to it for the last time and disregard all the intermediate returns to that number? I find it difficult to articulate precisely what I mean, but I hope you understand what I'm asking.

For the 2nd cycle I tried to solve it by starting with $1 \to 3$ then $3 \to 4$ and then $4 \to 1$, which, since it's the last factor, would complete the cycle but then we get $(1 3 4)$ then we have $(1 3)(1 3 4)$ which I was not able to factor. I then remembered the associative rule was valid for cycles and tried that by doing $[(1 3)(1 2 3 4)](1 3)$ which just led me to $(1 3 4)(1 3)$ which I got $(1 3 4)$ obviously not correct. I must be missing something here. Any help in understanding where I am going wrong is very much appreciated.

Answer 1

We proceed as follows:

$$\begin{align} 1&\xrightarrow{(14)}4\xrightarrow{(13)}4\xrightarrow{(12)}4\\ 4&\xrightarrow{(14)}1\xrightarrow{(13)}3\xrightarrow{(12)}3\\ 3&\xrightarrow{(14)}3\xrightarrow{(13)}1\xrightarrow{(12)}2\\ 2&\xrightarrow{(14)}2\xrightarrow{(13)}2\xrightarrow{(12)}1 \end{align}$$

So the permutation is $(4321)$ which is equivalent to $(1432)$.

The other permutation is done similarly.

Answer 2

I'm sharing this answer because your comment:

...$(134)(13) $ which I got $(134)$...

Is touching on a fun way to deal with elements of $S_n$, but you have to know that every element of $S_n$ can be written as a product of transpositions ($(ij)$ where $i\ne j$). See id* specifically for the property of transpositions that makes this so convenient.

So the first product can computed as follows, using only $(ijk)=(jki)$ and $(ij)(jk)=(ikj)$.

See: \begin{align} (1 2) (1 3) (1 4)& = (21) (1 3) (1 4)\\ &= (213)(14)\\ &= (321)(14)\\ &= (3214)\\ &=(1432) \end{align} and the following uses the fact that $(ij)(ij)=\text{id}.$ See:

\begin{align} (1 3)(1 2 3 4)(1 3)&= (1 3)(1 2 3)(3 4)(1 3)\\ &= (13)(312)(34)(13)\\ &=\underbrace{(13)(13)}_{\text{id*}}(12)(34)(13)\\ &=(12)(34)(13)\\ &=(12)(431)\\ &=(12)(143)\\ &=(2143)\\ &=(1432) \end{align}