Hi guys (i gotta stop saying that, since we now have JaneBennet ),

ahem, Hi Mathematicians,

Is there anyone familiar with permutations in Abstract Algebra (if you are, skip to the "Problem:" section, the problem is in bold font, if not, you may read through this very brief introduction to permutations). permutations are one-to-one mapping diagrams with the natural numbers as the domain, kind of. for instance,

$\displaystyle \left( \begin{array}{ccc} 1 & 2 & 3 \\ 2 & 1 & 3 \end{array} \right)$

is an $\displaystyle S_3$* permutation in which $\displaystyle 1 \mapsto 2$, $\displaystyle 2 \mapsto 1$ and $\displaystyle 3 \mapsto 3$

we may also write these as cycles. that is, the above can be written as $\displaystyle (1~2)(3) = (1~2)$ we omit (3) because it maps to itself and this is understood.

how did we get that form? we did the following.

start with the lowest number in the domain, obviously 1, here

so we write (1

but 1 maps to 2, so we write (1 2

and 2 maps back to 1, this is a cycle, so we close brackets here, because we go back to the beginning. so we have (1 2)

now we start at the next lowest number, 3.

but 3 maps to itself, so we have, (1 2)(3), which we can write as (1 2), because it is understood if any number in the domain is omitted, it maps to itself.

hope that crash course helped someone to understand the problem...

Problem:

I need to show that every element of $\displaystyle S_n$ is a 2-cycle or can be written as a product of 2-cycles. I am given a "suggestion,"$\displaystyle \bold{(a_1a_2 \cdots a_k) = (a_1a_k) \cdots (a_1a_3)(a_1a_2)}$

*) we say a permutation is of group $\displaystyle S_n$ if it is of the form

$\displaystyle \left( \begin{array}{cccc} 1 & 2 & \cdots & n \\ \_\_ & \_\_ & \cdots & \_\_ \end{array} \right)$