Let $f: G \to H$ be a group homomorphism such that the order of $\ker(f)$ is $2$. Show that any element of $H$ has either no pre-image or exactly two pre-images under $f$.
I have been trying to prove this - since $|\ker f|= 2$, $\ker f$ contains two elements say $k_1, k_2$.
Now we need to show each element of $H$ contains exactly $2$ or no pre-images.
Suppose $y \in H$ such that $y$ has $3$ pre-images, say $x_1,x_2,x_3$. Then $f(x_1)=f(x_2)=f(x_3)$.
Since $f$ is a homomorphism, $x₁\cdot x_2^{−1},x_2\cdot x_3^{-1},x_3\cdot x_1^{-1}$ belong to $\ker f$.
Without loss of generality suppose $x_2\cdot x_3^{-1} = x_1\cdot x_2^{−1} = k_1$, and this implies $x_1=x_2$.
Is this ok?