Potential incongruence between two seemingly equivalent methods for computing compositions of permutations?

Question

I am learning about permutations right now and the text has described two methods (one more formal, the other less so) for computing compositions of permutations in disjoint cycle notation. I am doing a problem that involves finding $p^2$ for $p: S_8 \rightarrow S_8$

• where $p = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 4 & 1 & 3 & 2 & 7 & 6 & 8 & 5 \end{pmatrix}$

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Through the course of writing this question out I've come to see that potentially my 2 answers for the different methods are actually the same answer. I remembered that there is a proposition that states that:

Given $n \geq 2$, any $\sigma \in S_n$ can be expressed as a composition of $2$-cycles.

Given this, would my answers for method $1$ and method $2$ be the same permutation, just represented differently? Namely, does: $p^2 = (124)(587) = (12)(41)(58)(75)$? It sure seems like it. In which case, I apologize for such a long wall of text for next to no purpose. Though I'll leave it up for posterity. Maybe it'll help someone else working through these kinds of problems one day. My work is given below.

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So I have computed this composition for each of the methods, which I will describe briefly before carrying out the calculation, and got different results. I believe I know which one is correct, but I am confused as to why I'm getting different results for methods that were given as equivalent in the text.

1. The first is the "formal" method where we have for each $x \in S_8, p{^2}(x) = p \circ p(x) = p(p(x))$ and so carrying out this calculation I got:

   • $1 \rightarrow p(p(1)) = p(4) = 2 \rightarrow p(p(2)) = p(1) = 4 \rightarrow p(p(4) = p(2) = 1$
     • So our disjoint cycle is $(124)$
   • $3$ is fixed
   • $5 \rightarrow p(p(5)) = p(7) = 8 \rightarrow p(p(8)) = p(5) = 7 \rightarrow p(p(7)) = p(8) = 5$
     • So our disjoint cycle is $(587)$
   • $6$ is fixed.
   • Thus we have $p^2 = (124)(587)$

2. The next, less "formal" method involves writing out the composition in terms of the disjoint cycles of each permutation, in this case, $(142)(578)(142)(578)$, and starting on the right side (which makes sense since our convention is such that we carry out the composition of functions starting with the function on the right side first) go until you find the first cycle with the first number you want to start with, say $1$, and "follow" it until you go all the way to the left then start over with that number you left off with after having reach the farthest left point (it seems, from experience, that there are $n$ images to follow before starting over at the right for a composition of $n$ permutations in disjoint cycle notation) and continue the cycle until you run through all the numbers. To this end, I got:

   • Starting with the number $1: 1 \rightarrow 4 \rightarrow 2, 2\rightarrow 1$
     • Disjoint cycle is $(1 2)$
   • $3$ is fixed
   • $4 \rightarrow 2 \rightarrow 1, 1 \rightarrow 4$
     • Disjoint cycle is $(41)$
   • $5 \rightarrow 7 \rightarrow 8, 8\rightarrow 5$
     • Disjoint cycle is $(58)$
   • $7 \rightarrow 88 \rightarrow 5, 5 \rightarrow 7$
     • Disjoint cycle is $(75)$
   • Thus our answer is $p^2 = (12)(41)(58)(75)$

Answer 1

Your first solution is correct, but your second approach does not make sense. Setting $s = (1 4 2)$ and $t = (5 7 8)$, you want to compute $stst$. Thus, starting from $1$, it should be $$ (stst)(1) = sts(1) = st(4) = s(4) = 2 $$ Instead, when you start by getting $1 \to 4 \to 2$ (and then $2 \to 1$) you rather compute $ss(1) = 2$ (and then $s(2) = 1$), which is wrong.