Solution
= Solution
\b[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 \b[such a kernel is impossible]. This argument does not assume that an arbitrary order-two <subgroup> is normal.