Simple permutation multiplication of two 4-cycles from $S_4$, specifically, $ (1324)(1423)$.

Question

For some reason I am confused on how to multiply these together with $4$ elements. I can do it with $2$ elements but theres a little gap in my logic and I am sure one of you can help me with a simple clarification of where the elements get sent to. Thanks!

Answer 1

Let $\sigma=(1324), \tau=(1423)$.

Then from right to left, we have

$$\begin{align} 1&\stackrel{\tau}{\mapsto}4\stackrel{\sigma}{\mapsto}1,\\ 2&\stackrel{\tau}{\mapsto}3\stackrel{\sigma}{\mapsto}2,\\ 3&\stackrel{\tau}{\mapsto}1\stackrel{\sigma}{\mapsto}3,\\ 4&\stackrel{\tau}{\mapsto}2\stackrel{\sigma}{\mapsto}4. \end{align}$$

Thus $(1324)(1423)={\rm id},$ the identity permutation.

Answer 2

I'll add an explanation on how the determine very simply the inverse of a cycle (and thus see instantly that in your case, the cycles are inverses of each other):

If $\gamma=(a_1\,a_2\,\dots\,a_{n-1}\,a_n)$ is any cycle, imagine it consists of $n$ points on a circle, one after the other counterclockwise. The inverse cycle $\gamma^{-1}$ consists of the same points, one after the other, but clockwise, so that we get, always starting with $a_1$: $$ \gamma^{-1}=(a_1\,a_n\,a_{n-1}\,\dots\,a_2).$$