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 $a^2\neq e$?

In $S_3$, for every element $a$ with $a^2\ne e$, we have $a^3=e$. If we map $a$ to $1$ instead of $0$, this creates a problem, because $1+1+1\ne 0$ in $\Bbb Z / 2\Bbb Z$.
