List all the elements of the alternating group $A_3$ written in cyclic notation.
I come up with
Identity $(1)$ Obviously $(123)$
List all the elements of the alternating group $A_3$ written in cyclic notation.
I come up with
Identity $(1)$ Obviously $(123)$
(1) (123) (132)
got it
Hint: If a group is cyclic, then find an element that is not the identity and find the group that is generated by that element. That is a subset of your group (and even more so a subgroup, albeit unrelated here). And if the group generated is the size of your desired group (you should know the size of $A_3$), then you have all the elements within your group. If you are unfamiliar with cyclic notation please reference the Wikipedia article.
The element of A3 = {(1), (1,2,3), (1,3,2)} The formula for finding how many elements A3 has is (3!/2)=6/2=3.