This is the question:

Decide whether the following sets generate $S_n$ or not:

  1. The set of 2-cycles in $S_n$

  2. The set of even permutations in $S_n$

  3. The set of odd permutations in $S_n$

  4. The set of 3-cycles in $S_n$

Here's my justifications and attempts:

(1) is true since every permutation in $S_n$ can be written as a product of 2-cycles.

(2) is false since the set of even permutations is a subgroup of $S_n$ and it is closed. Thus, none of the odd permutation can be written as products of even permutations.

(3) The set of odd permutation contains the set of 2-cycles which generate $S_n$, thus, the set of odd permutation generates $S_n$.

I'm not sure what to conclude for (4). However, I can conclude that the set of 3-cycles in $S_n$ generates $A_n$ for every even permutation can be written as a product of 3-cycles using the fact that $(ab)(bc)=(abc)$ and $(ab)(cd)=(cba)(acd)$ but does that prevent it from generating $S_n$? I need some hints.


The 3-cycle set indeed only generates $A_n$, not $S_n$. 3-cycles and their compositions/inverses are even permutations, but $S_n$ contains odd permutations.


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