Multiplication of transpositions?

Question

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?

score 3 · Answer 1 · answered Feb 10 '16 at 21:11

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.

egreg · Accepted Answer · 2016-02-10T21:47:29.550

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.

@cbutler16 No, I started from the right; under $(ab)$ the image of $x$ is $x$. — egreg, Feb 10 '16 at 21:46
But for $(x a)(a b)$, $a$ goes to $b$, $b$ goes to $a$, and finally $a$ goes to $x$. So wouldn't you have the permutation $( a b x)$ if you began multiplying from the right? — , Feb 10 '16 at 21:53

Multiplication of transpositions?

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