mjhjmtu

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 $4mapsto 3$ we just close with $(1243)$. To me however it feels more accurate to see that I have $1 mapsto 2, 2mapsto 3$. 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 $sigmatau$ 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.

For me, cycle notation is always confusing when multiplying, so I always try to write the whole permutation out. I will also go right to left, so we start with $(34)$:

$$left(beginmatrix 1 & 2 & 3 & 4 \ 1 & 2 & 4 & 3endmatrixright)$$

Then, we apply $(12)$:

$$left(beginmatrix 1 & 2 & 3 & 4 \ 2 & 1 & 4 & 3endmatrixright)$$

Finally, we apply $(23)$:

$$left(beginmatrix 1 & 2 & 3 & 4 \ 3 & 1 & 4 & 2endmatrixright)$$

Thus, $1 to 3 to 4 to 2 to 1$, so we get the cycle $(1342)$. This is the same answer you got, so I hope this alternative method gives you more confidence that your reasoning is correct!

You want to clarify $$(23)(12)(34)=(1243)$$

We need to move from left to right through these permutations.

Well, let us start with $1$

$1$ does not move under $(2,3)$ but it goes to $2$ under $(1,2)$ and does not move under $(3,4)$.

As the result, we have $$ 1to 2$$
Now let us see what happens to $2$

Well $2$ goes to $3$ under $(2,3)$ and $3$ does not move under $(1,2)$ and maps to $4$ under $(3,4)$
As the result $$ 2to 4$$

Let us see what happens to $4$

Well , $4$ does not move under $(2,3)$ and $(1,2)$ but maps to $3$ under $(3,4)$.

As the result $$ 4to 3$$

Let us see what happens to $3$

Well , $3$ moves to $2$ under $(2,3)$ and $2$ moves to
$1$ under $(1,2)$ and $1$ does not move under $(3,4)$.

As the result $$ 3to 1$$

The final result is $$ (23)(12)(34)=(1243)$$

Which is what we want to prove.

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).