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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 $n \geq 2$.

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3 Answers

up vote 3 down vote accepted

Your conjecture is right. First, $\text{CL}(\alpha)$ is precisely all the $n$-cycles, from the simple fact that if $\beta \in S_n$, then $$ \beta(12\dots n)\beta^{-1} = (\beta(1) \beta(2) \dots \beta(n)). $$ Suppose $n \ge 5$ is odd. Then the $n$-cycle is an even permutation, so we can write it as a product of even permutations. The product $(abc d \dots n) (n \dots d bc a) = (bdc)$ gives us an arbitrary $3$-cycle, so that the $n$-cycles generate $A_n$. But since every $n$-cycle is even this is all it can generate. (For $n=3$, the $n$-cycles are $3$-cycles so they obviously generate $A_3$).

Suppose $n \ge 4$ is even. Then the $n$-cycle is an odd permutation. Taking the product $(abc \dots n)(n \dots cab) = (acb)$ gives us again $A_n$, but then $A_n$ is normal in $S_n$ (it has index $2$), so that any odd permutation will be generated by any $n$-cycle multiplied by some unique element of $A_n$. Therefore for $n$ even one gets $S_n$.

Hope that helps,

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Excellent! Looks good to me. Thanks for that. –  Douglas S. Stones Apr 1 '12 at 10:08
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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 $H_k = S_n$.

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If $S$ is a subset of a group $G$ satisfying

$\bullet$ $gSg^{-1}\subset S$ for all $g$ in $G$ and

$\bullet$ $s^{-1}\in S$ for all $s$ in $S$,

then the subgroup generated by $S$ is normal.

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This is the easiest way to see it. –  user641 Apr 1 '12 at 16:05
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