How do we know quickly that $S_3$ does not have a normal subgroup of order $2$? I was doing the following problem discussed in this Math.SE question:


True or false: There is a non-trivial group homomorphism from $S_3$ to $\mathbb{Z}/3\mathbb{Z}$.


The trick to it is seeing that that the kernel of the homomorphism must have order $2$, and $S_3$ does not have a normal subgroup of order $2$. So, my question is: 

How do we know quickly that $S_3$ does not have a normal subgroup of order $2$?

I did see this in more a computational method (suppose $S_3$ does and that then the subgroup must be of $\langle y\rangle $, $\langle xy\rangle $, and $\langle x^2y\rangle $ where $y = (12)$ and $x = (123)$). However, this method can be rather tedious.
 A: The question, as it currently is, is somehow difficult to answer. Since $S_3$ is such a small group, people are very familiar with it and something like this will come naturally and you don't even realize how you know it...
There are of course numerous different ways to do that. Let me mention this method.
We know that there are three different elements of order $2$ in $S_3$. These then generate three different subgroups of order $2$. Therefore none of them can be normal, since they must all be conjugate to each other, by Sylow's theorems.

If we have in mind some more general picture, then it might be useful to know what are all normal subgroups of $S_n$ (for a general $n$), or what happens to $D_n$ (dihedral groups, another possible generalization of $S_3$), etc.
A: A subgroup of order $2$ must be generated by an element of order $2$, which in $S_3$ means a transposition, wlog $(12)$; conjugate by another transposition, wlog $(23)$, and you make $(13)$ and have escaped the subgroup. So it can't have been normal.

No computation was required for that, if you remember the fact that "conjugation is the same as changing the world in which the group is acting": to conjugate by $(23)$, simply replace all instances of $2$ by $3$ and of $3$ by $2$, turning $(12)$ into $(13)$.
A: If it had a normal subgroup of order $2$, then $S_3$ would have to be the direct product $\Bbb Z_3\times\Bbb Z_2$, hence abelian. (Note that we do know that the subgroup of order $3$, being of index $2$, must be normal.)
A: Consider any group $G$ of order $6$ with a normal subgroup $P=\{e,a\}$ of order $2$.
Then $g^{-1}ag=a$ for all $g\in G$ and so $a\in Z(G)$.
Suppose  $b\in G-Z(G)$. Then $<Z(G),b>$ is an abelian subgroup of $G$ properly containing $Z(G)$. Therefore $G$ itself is abelian and cannot be $S_3$.
