then $\sigma$ that takes 1 2 3 4 5 and rearranges it to 1 4 5 2 3
do you understand?
$\sigma$ is just a function $\sigma : \{1,2,3,4,5\}\to \{1,2,3,4,5\}$. When we write $$\sigma=\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 1 & 4 & 5 & 2 & 3\end{pmatrix}$$ we mean $\sigma(1)=1$, $\sigma(2)=4$, $\sigma(3)=5$ and so on
For instance, in $S_5$, when you see something like $\sigma=(243)$ that means that $sigma$ do $$\begin{matrix}1\to 1\\ 2\to 4\\ 4\to 3\\ 3\to 2\\ 5\to 5\end{matrix}$$
it should be read as $2\to 4\to 3\to 2$ and leave the $1$ and $5$ unchanged
that's because things like $(1,2,4)$ or $(1,3)$ are called cycles
so does (1 2 4) means whatever's in the first spot needs to go to the 2nd spot, whatever's in 2nd spot goes to 4th spot, whatever's in 4th spot goes to 1st spot?
@JohnSmith to put things clear we write $(k_1,k_2,k_3,k_4)$ it means: whatever in the first spot go to the second spot, the second spot go to the third spot, the third spot go the fourth spot,..., the last spot go to the first spot
@robjohn I follow his answer because I was trying to understand you (6) equation :-)
@JohnSmith Here is how can you decompose a given permutation on a product of cycles: consider for instance $\sigma=\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 4 & 6 & 5 & 2 & 7 & 3 \end{pmatrix}$ in $S_7$
@JohnSmith, then calculate. $\sigma(1)=1$ stop. $\sigma(2)=4$, $\sigma^2(2)=5$, $\sigma^3(2)=2$, stop. $\sigma(3)=6$, $\sigma^2(3)=7$, $\sigma^3(3)=3$, stop. Now we have exhausted all the numbers in 1 2 3 4 5 6 7, so you product of cycles is: $\sigma=(2,4,5)(3,6,7)(1)$, so the associated partition is $3+3+1$
@JohnSmith by definition a partition of $n$ is $k-$tuple $(n_1,n_2,\ldots,n_k)$ such that $n_1\geq n_2\geq\ldots \geq n_1$, $n_j\geq 1$ and $n=n_1+n_2+\ldots+n_k$
I just became familiar with a few of the theorems and tools of complex analysis and got a grasp of it conceptually, plus maybe some actual study of the basics (like cauchy-riemann). You can get by just with some familiarity.
@Peter: I bet you don't have a bookmarks bar up do you?
I'm fairly certain I have a sign error in one of my answers, but I ended up with a correct result so either I'm imagining things or there's another sign error that cancels it out in the end.
I just look at the formulas on Wikipedia and go with the flow, most of the time. (Most of the formulas are easy to prove once you see them, although a few have given me trouble).
@MarianoSuárezAlvarez I honestly thought my multiset question had a good chance of being closed as "not a real question". I an pleased but surprised that it was so well-received.
