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If I have a symmetric group $S_n$ with n distinct elements then is this group cyclic or is the group of permutations cyclic?

For example if n=3 and we have $S_3={1,2,3}$

Let $\sigma_i$ denote $i^{th}$ permutation i.e $\sigma_1={3,1,2} \\\sigma_2={2,3,1} \\\sigma_3={1,2,3}=\sigma_0$

So is permuation group $\sigma=(\sigma_0,\sigma_1,\sigma_2)$ the cyclic and symmetric group or is $S_3$ the symmetric and cyclic group.

From what my believe is that $\sigma=(\sigma_0,\sigma_1,\sigma_2)$ is a cyclic group of order 3 but why does my professor keep referring to $S_3$ as cyclic group?

and is there a notation for the mapping in cyclic and symmetric groups?

Orpheus
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  • Every group of order $3$ is cyclic, while $S_3$ is a noncyclic group of order $6$. In general, the symmetric group $S_n$ on $n$ objects has $n!$ elements, not $n$ elements. – Greg Martin Aug 09 '21 at 07:29
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    Your question looks unclear to me. How do you define $S_n$? The symmetric group $S_n$ on $n$ symbols is defined to be the set of all the $n!$ permutations of $n$ symbols with the operation of composition, while your definition only seems to talk about the cyclic subgroup of $S_3$ generated by the 3-cycle $(2~3~1)$, ie, $\langle (2~3~1)\rangle={(2~3~1),(3~1~2),(1~2~3)}$ – Prasun Biswas Aug 09 '21 at 07:37
  • $S_3={1,2,3}$ and I believe $\sigma$ is the set of mappings from the $S_3$ to the set of permutations, so if I multiply any element of $\sigma$ with another element of $\sigma$ , the product lies in $\sigma$ i.e. $\sigma_i \times \sigma_j=\sigma_k \in \sigma$ – Orpheus Aug 09 '21 at 07:47

3 Answers3

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For any $n\ge3$, $S_n$ is not Abelian.

Take $\sigma=(1~2)$, $\tau=(1~3)$ Then $\sigma\circ\tau=(1~3~2)$ and $\tau\circ \sigma= (1~2~3)$.

But all cyclic groups are abelian. Hence $S_n$ is not cyclic for all $n\ge 3$. But what can you say about $n=1,2$ ?

epsilon_delta
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Short answer is no. For example $S_3$ is not cyclic. (There is no $\sigma\in S_3$ whose order is $6 = |S_3|$, since $\sigma^3 = 1$ or $\sigma^2=1$.

Cayley's theorem: for every group $G$, there exist $H \leq \operatorname{Sym}(G)$ such that $G\cong H$. In this point of view, the symmetric groups are complicated enough to have all of group structure as its subgroup (if its order is big enough), thus it will not be cyclic. (Cyclic group is the most simple group structure.)

dust05
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Short Note: For a group G that is finite(let say order n) and cyclic there must exist at least one element(or exactly $\phi(n)$ elements https://en.wikipedia.org/wiki/Euler%27s_totient_function) of order n.

Now in the case of $S_n$ and we know that order of elements in $S_n$ is $lcm$ of disjoint cycles(every permutation is a product of disjoint cyclic) of $S_n$ and clearly it can't be equals to $n!$ so by the definition of a finite cyclic group we say $S_n$ is not a cyclic group for all $n\geq3$. for more What are the elements of order $n$ in symmetric group $S_n$?.

Mathematics learner
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