Proof for every permutation will have a cyclic notation to it or not? Assuming you do not know about Cyclic Notation Let's first see what it is Suppose we have some Permutation of Length 4 which is $[4,1,2,3]$ So We can map it by saying Number at Position $1$ is 4 , Number at Position $2$ is $1$ , and so on 
So It can be written that 
$1\xrightarrow{}4$
$2\xrightarrow{}1$
$3\xrightarrow{}2$
$4\xrightarrow{}3$
So if we try to make a Graph out of the given connections we Get a Cycle out of it which is as follows 
$1\xrightarrow{}4\xrightarrow{}3\xrightarrow{}2\xrightarrow{}1$ can be translated in Cyclic Notation as  
$(1,4,3,2) $
I was watching this video about Cycle Notations of a permutations. I am a beginner in Abstract Algebra , It has not been easy for me to visualize it that why is it always True that A Permutation will have a cyclic notation , Isn't It possible that We do not have a cycle being made ?
 A: Not every permutation is a cycle, but every permutation is by necessity a product of disjoint cycles. For example, consider the permutation $$\begin{bmatrix}1&2&3&4&5&6&7&8&9\\3&4&2&6&1&5&8&7&9\end{bmatrix}$$ To break into cycles, just pick a number a follow its path:

*

*$1 \to 3 \to 2 \to 4 \to 6 \to 5 \to 1$. This gives the cycle $(132465)$. This includes everything from $1$ to $6$, so the next unaccounted for value is $7$:

*$7 \to 8 \to 7$. This gives the cycle $(78)$. This accounts for $7$ and $8$, leaving $9$.

*$9 \to 9$, which is the cycle $(9)$.

So in all, $[342615879] = (132465)(78)(9)$, a product of three disjoint cycles.
And it always has to work like this. When you start following the trail for $1$, there only a finite number of values for it to pass through. It has to repeat a value at some time. Further that value has to be $1$, because if the first repeat was some later value $n$, then the predecessor of $n$ on its first and second visits would be different, and that is not allowed in a permutation. So the trail from $1$ eventually closes in a cycle. Now pick any number not in that cycle, and follow it instead. Eventually it must lead back to the original number for the same reason as before. So you get another cycle. This continues until you run out of numbers to trace.
So every finite permutation has to be a product of disjoint cycles.
