Intuition behind symmetric groups

Question

I am having a hard time understanding the intuition behind symmetric groups, and in particular, their elements. Consider the group $S_3$, with elements $id, (1, 2), (2, 3), (1, 3), (1, 2, 3), (1, 3, 2).$ Now, I fully understand that the group needs 6 elements since there are $3!$ ways to rearrange six elements, what I do not understand is the norm where these elements are used, or how I am to remember it without memorizing it, which is not convenient (nor good habit) at all. The most direct way of answering my question would be to answer this thoroughly: if $(2, 3)$ is an element of $S_3$, why isn't $(2, 3, 1)$?

Answer 1

You ask about the element $(2,3,1)$. What this element means is that the second position is sent to the third position, the third to the first, and the first to the second. But notice this is the same as $(1,2,3)$ and $(3,1,2)$.

So $(2,3,1)$ is in $S_3$, and you'd already mentioned it - but you called it $(1,2,3)$ above.

Answer 2

The intuition behind say $S_3$ is just how many bijective map you can have from a set $\{1,2,3\}\to \{1,2,3\}$

I leave it to you that there can be $6$ such function, and you must write them explicitly to get a clear picture.

$f_1=Id,f_2,f_3,f_4,f_5,f_6$ say and then we use usual composition of function as group operation.

$f_2=(123)$ say so $f_2$ sends $1\to2\to3\to1$

$(1,2)$ means $1\to2\to1$ and $3\to 3$