Need help understanding product of cycles I need to understand how product of cycles work. The textbook i am referring gives explanation only for simple products and just answer for bigger ones. i would be very thankful if someone can explain me how to work around products like this : 
$(1532)(14)(35)$ 
some relevant link where i can read and understand will also work.
Aman
 A: The idea here is to repeatedly evaluate until we get back to where we started. We can start with $1$ and see where that takes us. $(3\:5)$ leaves $1$ alone, then $(1\:4)$ sends $1$ to $4,$ and then $(1\:5\:3\:2)$ leaves $4$ alone. Thus, $(1\:5\:3\:2)(1\:4)(3\:5)$ sends $1$ to $4$.
Next, we evaluate at $4$. $(3\:5)$ leaves $4$ alone, then $(1\:4)$ sends $4$ to $1,$ and then $(1\:5\:3\:2)$ sends $1$ to $5.$ Thus, $(1\:5\:3\:2)(1\:4)(3\:5)$ sends $4$ to $5$.
Next, we evaluate at $5$. $(3\:5)$ sends $5$ to $3$, then $(1\:4)$ leaves $3$ alone, and then $(1\:5\:3\:2)$ sends $3$ to $2.$ Thus, $(1\:5\:3\:2)(1\:4)(3\:5)$ sends $5$ to $2$.
Next, we evaluate at $2$. $(3\:5)$ leaves $2$ alone, as does $(1\:4),$ and then $(1\:5\:3\:2)$ sends $2$ to $1.$ Thus, $(1\:5\:3\:2)(1\:4)(3\:5)$ sends $2$ to $1$, and so we're back where we started.
The only thing left to check is what happens to $3$, and you should be able to see that $3$ is sent to $5,$ then $5$ is left alone, then $5$ is sent to $3,$ and so $(1\:5\:3\:2)(1\:4)(3\:5)$ effectively leaves $3$ alone.
Thus, $$(1\:5\:3\:2)(1\:4)(3\:5)=(1\:4\:5\:2)(3)=(1\:4\:5\:2).$$
More generally, we'll do the same kind of thing until we have all the available numbers put into disjoint cycles.

Added: Babak's post reminds me that some textbooks do a left-to-right composition. My answer above assumes a right-to-left composition, so that $$(1\:3)(2\:3\:5)=(1\:3\:5\:2),$$ for example. If instead, you're using left-to-right composition, so that $$(1\:3)(2\:3\:5)=(1\:5\:2\:3),$$ then we'll take a similar approach to the above, but addressing the cycles in the opposite order. In particular, for example, $(1\:5\:3\:2)$ sends $1$ to $5,$ then $(1\:4)$ leaves $5$ alone, then $(3\:5)$ sends $5$ to $3$. Thus, $(1\:5\:3\:2)(1\:4)(3\:5)$ sends $1$ to $3.$ (And so on.)
You'll need to determine which convention your textbook is using.
A: The notation varies sometimes, but the way I find easiest to think of it is (using your example):
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\ 
5 & 1 & 2 & 4 & 3
\end{pmatrix}\circ
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
4 & 2 & 3 & 1 & 5 \end{pmatrix}
\circ
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
1 & 2 & 5 & 4 & 3 
\end{pmatrix}.$$
The way you read the above is as a composition of functions hitting your permutation, right to left. So, the first permutation moves three to five and five to three, which is abbreviated $(35)$ or $(53)$ (either one). Then you move onto the next permutation. I find it easiest to think in my head (e.g.) "four goes to one and one goes to five, so four goes to five." For further reading and a better explanation, check http://people.math.sfu.ca/~jtmulhol/math302/notes/4-Permutations-CycleForm.pdf . 
A: $$\begin{array}{}
I:(1\,5\,3\,2)&II:(1\,4)&III:(3\,5)\\ \hline
1\mapsto 5&1\mapsto 4&1\mapsto 1\\
2\mapsto 1&2\mapsto 2&2\mapsto 2\\
3\mapsto 2&3\mapsto 3&3\mapsto 5\\
4\mapsto 4&4\mapsto 1&4\mapsto 4\\
5\mapsto 3&5\mapsto 5&5\mapsto 3
\end{array}$$
$I\times II$: $(1\,5\,3\,2)(1\,4)$
$$\begin{align*}
&(1\overset{I}\longmapsto 5)(5\overset{II}\longmapsto 5)\implies(1\overset{I\times II}\longmapsto  5)\\
&(2\overset{I}\longmapsto 1)(1\overset{II}\longmapsto 4)\implies(2\overset{I\times II}\longmapsto  4)\\
&(3\overset{I}\longmapsto 2)(2\overset{II}\longmapsto 2)\implies(3\overset{I\times II}\longmapsto  2)\\
&(4\overset{I}\longmapsto 4)(4\overset{II}\longmapsto 1)\implies(4\overset{I\times II}\longmapsto  1)\\
&(5\overset{I}\longmapsto 3)(3\overset{II}\longmapsto 3)\implies(5\overset{I\times II}\longmapsto  3)\\
\end{align*}$$
$(I\times II)\times III$:
$$\begin{align*}
&(1\overset{I\times II}\longmapsto  5)(5\overset{III}\longmapsto  3)\implies(1\longmapsto  3)\\
&(2\overset{I\times II}\longmapsto  4)(4\overset{III}\longmapsto  4)\implies(2\longmapsto  4)\\
&(3\overset{I\times II}\longmapsto  2)(2\overset{III}\longmapsto  2)\implies(3\longmapsto  2)\\
&(4\overset{I\times II}\longmapsto  1)(1\overset{III}\longmapsto  1)\implies(4\longmapsto  1)\\
&(5\overset{I\times II}\longmapsto  3)(3\overset{III}\longmapsto  5)\implies(5\longmapsto  5)\\
\end{align*}$$
So it is: $$1\longrightarrow 3\\2\longrightarrow 4\\3\longrightarrow 2\\4\longrightarrow 1\\5\longrightarrow 5\\.......\\(1,3,2,4)$$
A: Note that different authors define multiplication of permutations differently; it's either left-to-right or right-to-left.  The distinction is whether we define $f \circ g$ as $x \mapsto f(g(x))$ or $x \mapsto g(f(x))$.  Check with your notes/books as to which you should use.
Left-to-right
We can draw the points and where they map to as follows:

Then we just follow the paths to find
\begin{align*}
1 & \mapsto 3 \\
2 & \mapsto 4 \\
3 & \mapsto 2 \\
4 & \mapsto 1 \\
5 & \mapsto 5. \\
\end{align*}
We can convert this to cycle notation $(1324)(5)$, or $(1324)$ for short.
Right-to-left
The drawing now looks like:

Again, we just follow the paths to find
\begin{align*}
1 & \mapsto 4 \\
2 & \mapsto 1 \\
3 & \mapsto 3 \\
4 & \mapsto 5 \\
5 & \mapsto 2. \\
\end{align*}
We can convert this to cycle notation $(1452)(5)$, or $(1452)$ for short.
