Are there any Symmetric Groups that are cyclic? Are there any Symmetric Groups that are cyclic?
Because I have been doing some problems and I tend to notice that the problems I do that involve the symmetric group are not cyclic meaning they do not have a generator which generates the set. 
So are there any cases in which any of the symmetric group is cyclic? If not, then why are none of them cyclic?
 A: Here is another way to show this (NOT necessarily the best way, from a pedagogical point of view, not by a long shot):
$S_n$ is cyclic if and only if it has an element of order $n!$.
Now, decomposing a permutation $\sigma \in S_n$ into disjoint cycles is the same as defining a partition of $n$ (A partition of $n$ is a (finite) sequence of integers $1 \leq a_1 \leq a_2 \leq \cdots \leq a_k \leq n$ such that $a_1 + a_2 +\cdots + a_k = n$), if we assign fixed elements of $\{1,2,\dots,n\}$ under $\sigma$ to $1$-cycles, and write the cycles in order of increasing length.
The order of any permutation $\sigma$ then depends ONLY on the induced partition (call it $\pi_{\sigma}(n)$). If we define $\text{lcm}(\pi_{\sigma}(n)) = \text{lcm}(a_1,a_2,\dots,a_k)$, when $k > 1$ and $\text{lcm}(\pi_{\sigma}) = a_1$ when $k = 1$, then:
$|\sigma| = \text{lcm}(\pi_{\sigma}(n))$
For example, in $S_6$, if $\sigma = (2\ 5)(3\ 4\ 6) = (1)(2\ 5)(3\ 4\ 6)$, the induced partition is $(1,2,3)$: note that $1+2+3 = 6$, and the order of $\sigma$ is indeed $6 = \text{lcm}(1,2,3)$.
Theorem: if $n > 2$ then $\text{lcm}(\pi(n)) < n!$ for any partition $\pi$ of $n$.
The proof is by (complete) induction on $n$. Our base case is $n = 3$. There are $3$ possible partitions of $3$:
$\pi_1 = (1,1,1), \pi_2 = (1,2),\pi_3 =  (3)$. We compute:
$\text{lcm}(\pi_1) = 1$,
$\text{lcm}(\pi_2) = 2$,
$\text{lcm}(\pi_3) = 3$. In all cases, $\text{lcm}(\pi_j) < 6 = 3!$, for $j = 1,2,3$.
So we assume that $\text{lcm}(\pi) < m!$ for all partitions $\pi$ of $2 < m < n$.
Note that if $\pi$ is a partition of $n$ with $k > 1$, say $\pi = (a_1,a_2,\dots,a_k)$, that $\pi' = (a_2,\dots,a_k)$ is a partition of $n - a_1 < n$. Hence:
$\text{lcm}(\pi) = \text{lcm}(a_1,a_2,\dots,a_k) = \text{lcm}(a_1,\text{lcm}(a_2,\dots,a_k)) = \text{lcm}(a_1,\text{lcm}(\pi')) \leq a_1\cdot\text{lcm}(\pi')$
$< a_1\cdot (n-a_1)!$ (by our induction hypothesis)
$\leq a_1\cdot (n - 1)! \leq n\cdot (n-1)! = n!$
Otherwise $\pi = (n)$ and $\text{lcm}(\pi) = n < n!$, since $n > 2$. This concludes the induction proof.
Corollary: for $n > 2,\ S_n$ is not cyclic.
Let $\sigma \in S_n$. Then $|\sigma| = \text{lcm}(\pi_{\sigma})$, and $\pi_{\sigma}$ is a partition of $n$, so $|\sigma| < n!$, and $\sigma$ does not generate $S_n$.
A: Just to provide a complete answer (in reply to the comments under the question) the following. The question does not ask for "up to isomorphism", so let me go for describing all cyclic symmetric groups.
The symmetric group of the empty set, and any symmetric group of a singleton set are all trivial groups, and therefore cyclic groups. The symmetric group $S(X)$ of any set $X$ with $\#X=2$ has $\#S(X)=2$, so $S(X)$ is cyclic, and generated by the transposition of the two elements of$~X$. This completes the list of cyclic symmetric groups. Whenever $X$ has at least $3$ distinct elements, choose $3$ of them and call them $a,b,c$ respectively, then $S(X)$ cannot be cyclic: it contains the transpositions $(a~b)$ and $(b~c)$ which do not commute, whereas cyclic groups are necessarily commutative.
A: When $n \geq 3$ the symmetric group $S_n$ is not Abelian (exercise), so in particular isn't cyclic.
A: There are a number of ways one could answer this, so here's my shot:
First, it is true that $S_1$ and $S_2$ are cyclic—these have only $1$ and $2$ elements respectively, so nothing surprising here.  So let's consider the symmetric group on three or more elements.

Claim: $S_n$ is not cyclic for $n \geq 3$.

To begin, it is a good exercise to show that subgroups of cyclic groups are necessarily cyclic.  Given this, prove that $S_3$ is not cyclic and note that $S_n \subset S_m$ for all $n \leq m$. 
As others are saying, another (probably easier) way to prove the claim is to show that cyclic groups are abelian$^\dagger$ and to show that subgroups of abelian groups are abelian.  Then it suffices to show that $S_3$ is not abelian. 

$^\dagger$Not only is the symmetric group not abelian—worse, it has trivial center.  See here.
A: This is my approach, the idea is very elementary, I enjoyed thinking it, it may be little intuitive (if there is something wrong, please warn me).
Lemma 1: Every permutation can be written by product of transpositions.
Proof:  Let $f=\left (f_n \; f_{n-1}\; ... f_3 \;f_2\;f_1\right)$ and for arbitrary $n\in\mathbb N^+$ $f$ is equal to $(f_{2}\;f_{1})(f_{3}\;f_{1})(f_{4}\;f_{1})....(f_{n-1}\;f_{1})(f_{n}\;f_{1})=\left (f_n \; f_{n-1}\; ... f_3 \;f_2\;f_1\right)$$\Box$
Lemma 2: Every permutation is written either odd or even number of transpositions
Proof: Omitted, it can be found, it is a little bit long.$\Box$
Lemma 3: For $n>2$ every $S_n$ has even and odd permutations.
Proof: $S_3$ has $(123)$ which is even and $(12)$ which is odd. Since $\forall n>3$,$S_n$ consists $S_3$, $S_n$ has even and odd permutations.$\Box$
Lemma 4: Every power of a permutation is the same polarity. That is, $f$ be permutation, and $P(x)$ be polarity function that shows permutations odd or even.
$P(f)=P(f^2)=....=P(f^d)$ where $f^d=id$
Proof: Using order and $lcm(n,m)$ lemma, I prove only for odd pemutation's powers.
Assume that $f$ is odd permutation, that is order of $f$ is even. $o(f^2)=lcm(o(f),o(f))=o(f)$ that is order of $f^2$ is even and so permutation $f^2$ is odd permutation. $\Box$
Theorem: $S_n$ is not cyclic $\forall n>3$
Proof: Assume that $S_n$ is cyclic for some $n>3$ so there must be an element that generates $S_n$, let call this element as $g$ and by assumption $<g>=S_n$ but it is wrong because $g$ can only generate like its polarity (that is, if $g$ is even permutation, $g$ can only generate even permutation but there is odds as well, and for $g$ is odd, logic is the same.)
A: Proof is wrong since for $S_5$ there exists $(123)(45)\in S_5$ but this is order $10$.(contradiction)
Theorem: $S_n$ is not cyclic $\forall n>3$
Proof: $|S_n|=n!$, pick an element and assume that  $g\in S_n$ is generator of $S_n$, $<g>=S_n$ and such that $o(<g>)=n!$ but it is impossible since $S_n$'s elements can have only at most order $n$, contradiction. $\Box$
