Are there any Symmetric Groups that are cyclic?

When $n \geq 3$ the symmetric group $S_n$ is not Abelian (exercise), so in particular isn't cyclic.

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.

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$^\dagger$Not only is the symmetric group not abelian—worse, it has trivial center. See here.

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 $S_n$.