Multiplication of transpositions? I can't seem to understand how the multiplication of two transpositions yield the results below:
$(x b)(x a) = (x a)(a b) \\
(c a)(x a) = (x c)(c a)$
I can't figure it out for the life of me. I'm trying to learn permutation multiplication on my own, and I feel confident in my abilities, but here I feel as though I'm missing something essential. Could someone please enlighten me as to the process (in depth) of multiplying these transpositions and getting the desired results?
 A: Without loss of generality, suppose the only elements in the set are in fact $a,b,x$ for the first case.  Try writing each as a single permutation.
$(x~b)(x~a)$ read from right to left says:
$\begin{array}{c}a\mapsto x\mapsto b\\ x\mapsto a\mapsto a\\ b\mapsto b\mapsto x\end{array}$
I.e. $(x~b)(x~a) = \begin{pmatrix}a&b&x\\b&x&a\end{pmatrix}$
On the other hand, $(x~a)(a~b)$ read from right to left says:
$\begin{array}{c}a\mapsto b\mapsto b\\ b\mapsto a\mapsto x\\ x\mapsto x\mapsto a\end{array}$
So we have $(x~a)(a~b)=\begin{pmatrix} a&b&x\\b&x&a\end{pmatrix}$
These are indeed equal.  We have then $(x~b)(x~a)= (x~a)(a~b)$
Try reading through the second example to see if it is true.
A: I'll assume $a$, $b$ and $x$ to be distinct.
The trick for computing the composition of permutations is to start from an arbitrary element, say $x$, and follow its image under the permutations to compose; then write the final element next to $x$ and go on. If an element doesn't appear in a cycle, it is fixed by it. The image of an element under a cycle is the element next to it (on the right) or the initial one if at the right there's a ).
For instance, the cycle $(xa)$ represents the permutation that sends $x$ to $a$, $a$ to $x$ and leaves all other elements fixed.
Depending on conventions you start from the left or from the right; when composition of functions is the usual one, one starts from the right; so with $(xb)(xa)$
$x\mapsto a\mapsto a$
$a\mapsto x\mapsto b$
$b\mapsto b\mapsto x$
When we get back to the element we started with, we stop the cycle and start again if there's still an uncovered element. In this case we obtain the cycle $(xab)$.
If we consider $(xa)(ab)$ we get
$x\mapsto x\mapsto a$
$a\mapsto b\mapsto b$
$b\mapsto a\mapsto x$
So the composition is again the cycle $(xab)$ and we've proved equality.
