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 $ac=(1\,6\,5\,2)(3\,4)$?
 A: There is a small example on this page . Basically multiplication of permutation groups is applying permutations from right to left on an unaltered sequence.
A: The method I use for multiplying permutations like this is to think of each cycle as a set of mappings.  a (in your example) maps 1 to 3, 3 to 5, 5 to 2, and 2 to 1.  Also, remember that ab means "apply b, then apply a."  So, here, we want to see where ab maps each number 1-6.
Start with 1: b fixes 1 (maps it to itself) and a maps 1 to 3. 
So we can begin writing ab = (13...
Now do 3: b fixes 3, and a maps 3 to 5.
Put a 5 in: ab = (135...
Now 5: b maps 5 to 6 and a fixes 6, so ab = (1356...
Now 6: b maps 6 to 2 and a maps 2 to 1, so ab = (13561... = (1356).  
Notice that ab fixes 4 since both a and b fix 4, but ab actually also fixes 2.  This is because b maps 2 to 5, and a maps 5 right back to 2.
Hopefully you can use this method to check the other products. 
A: You are thinking of the permutations as functions, so when you write "$ab$", you mean that you perform the permutation $b$ first, and the permutation $a$ second.
Here's one way to do it: write the disjoint cycle expressions for both $a$ and $b$, in the given order:
$$(1,3,5,2)(2,5,6)$$
Now, moving from right to left, see what happens to each number in each cycle.
For instance, start with $1$, so we write $1$ down: 
$$(1,$$
The first cycle, $(2,5,6)$, does nothing to $1$, so it stays $1$. Then the next cycle, $(1,3,5,2)$, sends $1$ to $3$. So, in total, $1$ is sent to $3$. We write
$$(1,3,$$
Now consider $3$. The first cycle, $(2,5,6)$, does nothing to $3$. The second maps $3$ to $5$. So the product maps $3$ to $5$. So now we have
$$(1,3,5,$$
Now $5$. The first cycle, $(2,5,6)$, sends $5$ to $6$; the second cycle does nothing to $6$, so in total, $5$ is sent to $6$. So for the product we now have
$$(1,3,5,6,$$
Next, what happens to $6$? The first cycle sends $6$ to $2$; and then the next cycle sends $2$ to $1$. So $6$ is sent to $1$, and because we started out with $1$, this now closes the cycle we have; thus, we also close the bracket. So the product so far is
$$(1,3,5,6)$$
Now we consider the "next" number that hasn't been described yet, $2$. The first cycle, $(2,5,6)$, sends $2$ to $5$; then we check what the next cycle does to $5$, which is that it sends it back to $2$. So $2$ maps to $2$, and since we started out with $2$, it again closes the cycle. So now we have
$$(1,3,5,6)(2)$$
Finally we check what happens $4$, as it's the remaining number: $(2,5,6)$ fixes $4$ (it doesn't do anything to it – it remains as it is), as does $(1,3,5,2)$, so $4$ is overall fixed. So now finally we have:
$$ab = (1,3,5,2)(2,5,6)=(1,3,5,6)(2)(4) = (1,3,5,6)$$
$$\therefore (1,3,5,2)(2,5,6)=(1,3,5,6)$$
It's similar with $ac$. Here we have:
$$(1,3,5,2)(1,6,3,4).$$
First consider $1$: the first cycle maps it to $6$, the second cycle fixes $6$. So $1\mapsto 6$. Then $6$ is sent to $3$ by the first cycle, and $3$ to $5$ by the second cycle (reading right to left, remember), so $6\mapsto 5$. Then $5$ is fixed by the first cycle and sent to $2$ by the second cycle, so $5\mapsto 2$. Then $2$ is fixed by the first cycle and sent to $1$ by the second, which means $2\mapsto 1$, closing the cycle: we have $(1,6,5,2)$. The next number not already covered is $3$; $3$ is mapped to $4$ by the first cycle (by $b$), and $4$ is fixed by $a$, so $3\mapsto 4$. Then $4$ is sent to $1$ by the first cycle, and $1$ is sent to $3$ by the second cycle, so this closes this second cycle as $(3,4)$. Putting the two together we get
$$(1,3,5,2)(1,6,3,4) = (1,6,5,2)(3,4)$$
as given.
A: 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?
