Multiplication in Permutation Groups Written in Cyclic Notation I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if
$$
  a=(1\,3\,5\,2),\quad b=(2\,5\,6),\quad c=(1\,6\,3\,4),
$$ 
then why does $ab=(1\,3\,5\,6)$ and $ac=(1\,6\,5\,2)(3\,4)$?

I didn't find this method written as an answer here, so I am just adding it.
You might convert these cyclic notations to Cauchy's two-line notations. Carefully observing your $a$ and $b$, the permutation groups can be written in the following form:
$$a=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 1 & 5 & 4 & 2 & 6
\end{pmatrix} \\
b=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
1 & 5 & 3 & 4 & 6 & 2
\end{pmatrix}
$$
Now, under the multiplication operation, we will have:
$$a\cdot b=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 1 & 5 & 4 & 2 & 6
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
1 & 5 & 3 & 4 & 6 & 2
\end{pmatrix}
$$
As @Arturo explained in his answer, "you perform the permutation in $b$ first and then in $a$".
So, in $b$, $1\to1$ and then in $a$, $1\to3$. Thus, in $a\cdot b$, $1\to3$.
Similarly, in $b$, $2\to5$ and then in $a$, $5\to2$. Thus, in $a\cdot b$, $2\to2$ itself. Continuing in a similar process, you'll find that
$$a\cdot b=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6\\
3 & 2 & 5 & 4 & 6 & 1
\end{pmatrix}$$
Now convert this final form to the cyclic form. Note that we have, in $a\cdot b$, $1\to3$. Thus $$(1,3$$
Again, $3\to5$. Thus $$(1,3,5$$
Note again that $5\to6$ and $6\to1$, making our cyclic notation as $$\left(1,3,5,6\right)$$
Also, note that the rest of the elements in the group map to themselves. Thus we can ignore their presence in our cyclic notation. This makes
$$a\cdot b=\left(1,3,5,6\right)$$
Now can you similarly proceed for $a\cdot c$ also?
