multiplication of permutation

Question

I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation.

For example, $a=(1\ 3\ 5\ 2)$, $b=(2\ 5\ 6)$, $c=(1\ 6\ 3\ 4)$, $ab=(1\ 3\ 5\ 6)$, $ac=(1\ 6\ 5\ 2)(3\ 4)$.

Why?

Answer 1

Just look at where each letter ends up after applying the permutation. For example, $2 \mapsto 5$ by $b$, and $5\mapsto 2$ via $a$, so $ab$ fixes $2$. Likewise, $1$ is fixed by $b$ and sent to $3$ by $a$, so we have (overall) that $1 \mapsto 3$. Continuing in this manner, we find that $3 \mapsto 5$, $4 \mapsto 4$, $5 \mapsto 6$ and $6 \mapsto 1$, giving the desired cycle in the product.

Answer 2

$a=(1352), b=(256), c=(1634), ab=(1356), ac=(1652)(34)$

$a=(1352)$ means swap position 1 with 3, 3 with 5, then 5 with 2.

$b=(256)$ means swap position 2 with 5, then 5 with 6.

So $ab=(1352)(256)$ means swap position 1 with 3, 3 with 5, 5 with 2, 2 with 5, then 5 with 6. Which can be simplified since the emphasized step reverses the one prior.

Thus $ab=(1352)(256)=(135256)=(1356)$

The same cannot be done with $ac$ because the end point of a is not the start of c.

$ac=(1652)(34)$ means: swap 1 with 6, 6 with 5, 5 with 2, then 3 with 4. (Note: you don't swap 2 with 3.

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$a=(1352)\equiv a=\begin{pmatrix}A&B&C&D&E&F\\C&A&E&D&B&F\end{pmatrix}$

$b=(256)\equiv b=\begin{pmatrix}A&B&C&D&E&F\\A&E&C&D&F&B\end{pmatrix}\equiv b=\begin{pmatrix}C&A&E&D&B&F\\C&B&E&D&F&A\end{pmatrix}$

$\therefore ab=(1352)(256)\equiv ab=\begin{pmatrix}A&B&C&D&E&F\\C&B&E&D&F&A\end{pmatrix}\equiv ab =(1356)$