I'm trying to proof that if a group $G$ has order $6$, then it is either $\mathbb{Z}_{6}$ or $S_{3}$. I know that there are a lot of solutions to this on the internet, but I want to know why I found such errors as below.
My attempt: $|G| = 6 = 2.3 $
If $G$ has element of order $6$, then it is cyclic, then it is abelian, then, by Fundamental Theorem of Finite Abelian Groups, $G = \mathbb{Z}_2 \times \mathbb{Z}_3 $, since there are only those $2$ groups of orders $2$ and $3$.
If $G$ doesn't have element of order $6$, then, since by Sylow theorem the number of 3-Sylow subgroups is $1$, I can write $ G = \mathbb{Z}_{3} \rtimes H $, where H is a subgroup of order $2$ of $Aut{\mathbb{Z}_{3}} = S_{2}$. So $ G = \mathbb{Z}_{3} \rtimes S_{2} $.
But I tried examinating the orders of this semidirect product and concluded that $G$ can't be $S_{3}$:
$o(0_{3},(,))=1$
$o(0_{3},(1,2))=2$
$o(1_{3},(,))=3$
$o(1_{3},(1,2))=6$
$o(2_{3},(,))=3$
$o(2_{3},(1,2))=6$
Obviously I've done something wrong, because those orders aren't for the elements of $S_{3}$ (And weirdly are for $\mathbb{Z}_{6}$; I guess my mistake was on the semidirect product).
Can someone explain this mistake to me? How can I find $S_{3}$ by what I tried to do?
Thanks.
I think you're computing the group operation in the semidirect product wrong. Writing elements of $S_2$ as $\mathbb{Z}_2$, we have $(1_3,1_2)\cdot(1_3,1_2) = (1_3 + 1_2\cdot 1_3,1_2 + 1_2) = (1_3 + 2_3,0_2) = (0_3,0_2)$. Perhaps you accidentally did multiplication in $\mathbb{Z}_3$?