 Solution

False. Any supposed kernel of a group homomorphism of order two would have a nonidentity central element by the introductory proof. Every element of order two in $S_3$ is a transposition, and none is central: conjugating $(ij)$ by a suitable three-cycle gives a different transposition. Thus the center of a group for $S_3$ is trivial, and such a kernel is impossible. This argument does not assume that an arbitrary order-two subgroup is normal.