Group homomorphisms $S_3\to\mathbb{Z}/2\mathbb{Z}$ Knowing that $S_3=\{\text{id},\sigma,\sigma^2,\tau,\tau\circ\sigma,\tau\circ\sigma^2\}$ ($\tau=(1 2), \sigma=(123)$), why does a group homomorphism $f:S_3\to\mathbb{Z}/2\mathbb{Z}$ satisfy $f(a)=0$ for all $a\in S_3$ such that $a^2\neq e$?

If $a^2\neq e,$ then the order of $a$ must be 3. The order of $f(a)$ must then divide 3, but it must also divide 2 (as element of $\mathbb{Z}/2\mathbb{Z}$). Thus, $f(a)$ must have order 1.
