What is the group generated by the conjugacy class containing $(12\ldots n)$ in $S_n$? This question is motivated by this answer to a question about groups generated by conjugacy classes.
Let $n \geq 1$ and $S_n$ be the symmetric group on $\{1,2,\ldots,n\}$.  Define the n-cycle $\alpha=\pmatrix{1&2&\cdots &n}$, and let $\operatorname{Cl}(\alpha)=\{\beta \alpha \beta^{-1}:\beta \in S_n\}$ denote the conjugacy class of $\alpha$.

Question: What is the group $\langle \operatorname{Cl}(\alpha) \rangle$?

Observation 1: If $n$ is odd and $n \geq 3$, then $\alpha$ and its conjugates are even permutations (since they all have the same cycle structure), so $\langle \operatorname{Cl}(\alpha) \rangle$ contains only even permutations, and thus is a subgroup of the alternating group $A_n < S_n$.
Observation 2: Judging from some computations in GAP, it looks like $\langle \operatorname{Cl}(\alpha) \rangle=A_n$ for odd $n \geq 3$ and $\langle \operatorname{Cl}(\alpha) \rangle=S_n$ for even $n \geq 2$.

The following is even true: let $1 < k \leq n$, and $H_k$ the subgroup of $S_n$ generated by the $k$-cycles. If $k$ is odd then $H_k = A_n$ and if $k$ is even then $H_k = S_n$.
Note (added 6 years later, ahum, apologies!!) I never supplied a proof, so here it is. It is well-known and to be found in every standard introductory book on group theory, that $S_n$ is generated by $2$-cycles and $A_n$ is generated by $3$-cycles. So let's assume $3 \lt k \leq n$. Observe the following. Put $\sigma=(k \ 2 \ 3 \ 4  \cdots \ k-1 \ 1)$ and $\tau= (k \ k-1 \ k-2 \ \cdots \ 3 \ 2 \ 1)$. Then $\sigma, \tau \in H_k$ and $\sigma \tau=(1 \ 2 \ k)$. By conjugating $(1 \ 2 \ k)$ with the appropriate element, say $\lambda$ of $S_n$, you can "reach" any $3$-cycle, while conjugating $\sigma$ and $\tau$ with such a $\lambda$ does not change their cycle structure. In other words, $A_n \subseteq H_k$ for any $k$. Since $|S_n:A_n|=2$, it follows that either $H_k=A_n$ or $H_k=S_n$. If $k$ is odd, then all elements of $H_k$ are even cycles, so then the first case applies. If $k$ is even then $H_k$ contains odd cycles and the latter case holds true.
Finally, observe that of course $H_k=\langle Cl_{S_n}((1 \ 2 \ 3 \ 4  \cdots \ k-1 \ k)) \rangle$.
