Permutation group products (23)(12)(34)=(1243)? I just wanted to ask why it makes sense that $(23)(12)(34)=(1243)$. Note I'm going from right to left.

I'm trying to understand the concepts trying to find a true "method" to follow in all cases. My reasoning at the moment is: $1 \mapsto 2$ but since it doesn't close we must then leave the cycle open right (I'm assuming this is the case for when a cycle doesn't close unless when it returns to itself). Then we have $(12$. To proceed we consider where $2$ goes to. Then we see that $2 \mapsto 3, 3 \mapsto 4$. I'm getting the idea that because we see that $3$ much like a "path" to $4$ then the mapping is then $2 \mapsto 4$, giving us $(124$. Then since we know $4\mapsto 3$ we just close with $(1243)$. To me however it feels more accurate to see that I have $1 \mapsto 2, 2\mapsto 3$. Then we see that we have $(13$. Then since we went through all cycles once we then start at 2 since it hasn't closed then we go from $2 \mapsto 3, 3 \mapsto 4$, then we would have $(1324)$. What am I not seeing? I'm trying to imagine functions where maybe each cycle is its own function and sends to another cycle but I never seem to find anything consistent.

Thanks in advance for any of your clarifications.
I would also get $(1342)$ if I were to calculate that product, following basically the same argument you did.

If you got $(1243)$ from WolframAlpha, note what it says in the input interpretation field:


  the product $\sigma\tau$ represents the permutation $\tau(\sigma(\cdot))$


meaning they apply permutations from left to right (first $\sigma$, then $\tau$). In that light, their calculations are entirely correct, but they do not get the same answer as you do because they interpret the product of permutations differently.

At least they are nice enough to tell you what convention they use.
Just plug in numbers and see where they end up when you finish going through (23)(12)(34) a single time.

Plug in 1: It first gets sent to 2 by (12), then that 2 gets sent to 3 by (23), so we end at 3. Hence, 1 -> 3, so we start with (13 . [sic]

Plug in 3: It goes to 4 via (34), but the 4 doesn’t get sent anywhere after that (34). Hence, 3 -> 4, and we now have (134 . [sic]

Plug in 4: It goes to 3 via (34), and that 3 goes to 2 by (23). Hence, 4 -> 2, and we have (1342 . [sic]

Plug in 2: It goes to 1 via (12), and that 1 doesn’t go anywhere else. Hence, 2 -> 1. Since 1 is the first entry of (1342 , this means we have closed the loop—literally, we close the loop and get (1342) (by adding a parenthesis :P).