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 $n \geq 2$.


3 Answers 3


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 $H_k=\langle Cl_{S_n}((1 \ 2 \ 3 \ 4 \cdots \ k-1 \ k)) \rangle$.


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,

  • 1
    $\begingroup$ Excellent! Looks good to me. Thanks for that. $\endgroup$ Apr 1, 2012 at 10:08

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.

  • $\begingroup$ This is the easiest way to see it. $\endgroup$
    – user641
    Apr 1, 2012 at 16:05

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