I was doing some computation with $A_3$, the alternating group on 3 letters. I know it has to be abelian, even cyclic, but when I carried out the actual computation... I couldn't make sense out of it. I think I misunderstood the entire concept of cyclic permutation, so please help me out...
So, in $A_3$, there are: $(1,2),(2,3),(1,3)$. All of these have to permute, but... (ACCORDING TO MY MOST LIKELY WRONG CALCULATION,)
$(1,2)(2,3) = (1,3,2)$, but $(2,3)(1,2)=(1,2,3)$. What??
Likewise, $(1,2)(1,3)=(1,2,3)$, but $(1,3)(1,2)=(1,3,2)$.
What am I doing wrong??? By the way, I use array notation for computing the permutation. So when I do $(1,2)(1,3)$, I write first $2,1,3$, and apply to it $(1,3)$ to get $2,3,1$, and so final permutation becomes $(1,2,3)$. Am I doing something wrong? If so, please correct me once and for all...
Thanks.
$S_3$ is non-abelian.
$A_3$ is the subgroup of $S_3$ that has an even number of transpositions. (1,2) and (2,3) are not in $A_3$ but (1,2)(2,3) is. (1,2)(2,3) = (1,3,2) and (2,3)(1,2) = (1,2,3) because $S_3$ is non-abelian.
(1,2,3)^2 = (1,3,2) and (1,2,3)^3 = e. $A_3$ is cyclic.
Hope this helps.