How to multiply in Sym(X) Could someone show me how to multiply in say $S_4$. I know how to multiply say $(4321)(2341)$ but when it comes to ones that do not contain $4$ terms, like $(34)(231)$, I have no idea how to handle this. We do from left to right. 
 A: Don't go from left to right.  Go from right to left.
For $(34)(231)$, start with the right pair of parentheses.  Pick a number from between that pair and write it in a new pair of parentheses.  I will first choose $2$.  So I write:
$(2)$
Now, where does $2$ go?  It goes to $3$ in the right pair, but then if you work your way to the left pair of parentheses we see in that pair that $3$ goes to $4$.  So, ultimately, $2$ goes to $4$.  So in your new pair of parentheses, I would write $4$ after $2$.
$(24)$
Now, we need to see where $4$ goes.  So start again with the right most pair of parentheses.  Since $4$ is not present in there, it means $4$ goes to itself under that pair.  Then move to the left, and see where $4$ goes.  $4$ goes to $3$, so ultimately, $4$ goes to $3$.  So I would write $3$ after $4$.
$(243)$
Now, where does $3$ go?  Well, in the right most pair of parentheses, $3$ goes to $1$, and in the left pair, $1$ isn't present, so $1$ goes to $1$ in that pair. So, ultimately, $3$ goes to $1$, so I will write $1$ next to the $3$.
$(2431)$
Now where does $1$ go?  In the right pair, $1$ goes to $2$, and in the left pair $2$ is not present, so $2$ goes to itself there.  So ultimately, $1$ goes to $2$.  So I would need to write $2$ after $1$, but notice that in what I've written, $1$ already goes to $2$, so I don't need to change anything. 
So we have $(2431)$.  Since every distinct integer in $(34)(231)$ is used, we are done.  If any where not used, I would start a new pair to the right of the old pair, like $(2431)()$ and pick one of the unused integers, and go from there.
A: Let me explain this in your given example. First, the product $(3,4)(2,3,1)$ is composition $(3,4) \circ (2,3,1)$. From the composition perspective, let's write $f = (3,4)$ and $g = (2,3,1)$. In this case, we have
$$
\begin{array}{c|c|c|c}
x & g(x) & f\circ g(x) & \text{conclusion}\\ \hline
1 & g(1) = 2 & f(2) = 2 & 1 \to 2\\
2 & g(2) = 3 & f(3) = 4 & 2 \to 4\\
3 & g(3) = 1 & f(1) = 1 & 3 \to 1\\
4 & g(4) = 4 & f(4) = 3 & 4 \to 3
\end{array}
$$
Using the "conclusion" column, we see that $(3,4)(2,3,1) = (1,2,4,3)$.
A: $(34)(231)$ means:
$(1,2,3,4)\to (2,3,1,4) \to (2,4,1,3)$
A: The usual function notation means  fog = f(g).. f acts on g.
But (older) group theorists say the "multiplication" (f)(g) of two permutations means g(f).
 Of course fog is not equal to (f)(g) = gof.
The "cycle" notation (34) means 3 -> 4; and 4-> 3 ..all other entries remain unchanged.
(231) means 2-> 3; 3 -> 1 ; 1 -> 2. That is (1234) -> (2314).
In usual function notation
(34)(231) = (1243) o (2314) =   (2341)
In group theorist notation
(34)(231) = (2314)(1243) = (2413).
