# Can $S_n$ be a cyclic group?

Some notes before the question:

1- there are many questions in MSE asking about elements generating $S_n$ but they all involve more than one transposition to generate $S_n$ for example "$S_n$ is generated by elements of the form $(1k)$ [more than one element because $k$ varies] or "... generated by $\{(1,2), (1,2,3,...,n) \}$".

2- By cyclic or generated by, considering the same meaning of generating a group by a single element and so called the cyclic group.

My question is : for which $i$ and $i$, $(i \ \ j)$ generated $S_n$, i.e. $S_n = \langle (i \ \ j) \rangle$ i.e. $S_n = {\{ (i \ \ j)^a | \text{fixed} \ i, j \ \text{and varying}\ a \ \in {\{1, \dots n}\} \ }\}$? And if the answer is no single transposition can generated $S_n$, is it possible with two transpositions and if so for what numbers i, j, k, m $S_n = \langle (i \ \ j), (k \ \ m) \rangle$ ?

I would appreciate any simple clear detailed explanation.

• Any cyclic group is abelian, and very few $S_n$ are abelian. – Santiago Canez Mar 30 '16 at 16:33
• A cyclic group is necessarily abelian, but $S_n$ is not except for $n<3$. – Captain Lama Mar 30 '16 at 16:33
• Yes, two is possible : take $(12)$ and $(12\cdots n)$. – Captain Lama Mar 30 '16 at 16:35
• I read "two generators", I forgot you wanted them to be transpositions. If you want that, then you need $n-1$ of them, for instance $(1i)$ for $1<i\leqslant n$. – Captain Lama Mar 30 '16 at 16:38
• Clearly $2$ is not possible because all numbers need to appear in your transpositions if you want them to generate the whole group. – Captain Lama Mar 30 '16 at 16:38

As it was said in the comments (by Captain Lama and Santiago Canez), this is not possible if $n ≥ 3$ because the transpositions $(1 \; 2)$ and $(2 \; 3)$ do not commute. So $S_n$ is not abelian, and therefore it is not cyclic (i.e. it can't be generated by $1$ element). You only have $S_2 = \langle (1 \; 2) \rangle$.
For $n ≥ 5$, $S_n$ is not generated by $2$ transpositions: let $(a \; b)$ and $(c \; d)$ two of them, and let $x \in \{1,...,n\}$ with $x \not \in \{a,b,c,d\}$. Then $(a \; x)$ is not in the subgroup $G$ generated by $(a \; b)$ and $(c \; d)$, because every element of $G$ fixes $x$.
However, for $n=3$, you have $S_3 = \langle (1 \; 2), (1 \; 3) \rangle$.
• Is writing $S_n = {\{ (i \ \ j)^{a_1} (k \ \ m)^{a_2} (i \ \ j)^{a_3} \dots | \text{fixed} \ i, j, k, m \ \text{and varying}\ a_i \ \in {\{1, \dots n}\} \ }\}$ possible? It has nothing to do with abelian since powers of same transpositions don't sum-up. – Liebe Mar 30 '16 at 16:39
• @Liebe : you are right for the description of the subgroup generated by $(i \; j)$ and $(k \; m)$ (you can have $0≤a_i≤1$ here). The first part of my answer is about one generator, and this has something to do with abelian. But for two generators, this is indeed different. – Watson Mar 30 '16 at 16:44
• Sorry one q. What is the least number of transpositions to generate $S_n$? It seems that in your example for n=5 if we bring a third transposition (i.e. incl. 1) so they may generate $S_n$ so the least number is 3? Thank you – Liebe Mar 30 '16 at 16:49
• @Liebe The argument shows that no fewer that $n-1$ transpositions can generate $S_n$ and indeed, it is generated by either the $n-1$ transpositions containing some fixed element, or the $n-1$ transpositions exchanging some $i$ and $i+1$. – Tobias Kildetoft Mar 30 '16 at 17:19