This is the question:
Decide whether the following sets generate $S_n$ or not:
The set of 2-cycles in $S_n$
The set of even permutations in $S_n$
The set of odd permutations in $S_n$
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.