1. ## Problem: Abelian/non-abelian subgroups

Hi,
Can anyone help with the following please?

A non-Abelian group can have both Abelian and non-Abelian subgroups. The symmetric groups Sn are non-Abelian for n ≥ 3.
a. Prove that the alternating group, An, which is a subgroup of Sn, is non-Abelian for n ≥ 3.
b. Construct an Abelian and a non-Abelian subgroup of S4 (excluding the trivial cases G = {e} and G = S4).

For a) i was thinking i need to find two subgroups say (1 2 3...) and (1 3 2...) (in cycle notation) and show that they are not commutative but (1 2 3)(1 3 2) = (1)(2)(3)... and (1 3 2) (1 2 3) = (1)(2)(3)...so this example is commutative but yields a result which is not an element of A4 (the set of even permutations). Am I on the right lines with this?

b) I'm unsure of but solving a) would help I assume.

2. Originally Posted by jackiemoon
Hi,
a. Prove that the alternating group, An, which is a subgroup of Sn, is non-Abelian for n ≥ 3.
Try $\displaystyle (123)(132)$ and $\displaystyle (132)(123)$.

b. Construct an Abelian and a non-Abelian subgroup of S4 (excluding the trivial cases G = {e} and G = S4).
The subgroup $\displaystyle \left< (1234) \right>$ is cyclic.
While, $\displaystyle S(1)$ the permutations that fix $\displaystyle 1$ is non-abelian.