The compound shown is 1-chloroprop-1-ene (no geometry specified). Which statements about the isomers of its molecular formula $\mathrm{C}{}_3\mathrm{H}{}_5\mathrm{Cl}$ are correct?

$\mathrm{C}{}_3\mathrm{H}{}_5\mathrm{Cl}$ has one degree of unsaturation. Constitutional isomers: 3-chloroprop-1-ene (allyl chloride), 1-chloroprop-1-ene, 2-chloroprop-1-ene and chlorocyclopropane: 4. In 1-chloroprop-1-ene each alkene carbon carries two different groups (H/Cl and H/$\mathrm{CH}{}_3$), so cis and trans forms exist. In 2-chloroprop-1-ene the terminal carbon carries two hydrogens, so there is no geometrical isomerism. Chlorocyclopropane has no stereocentre. Total = 4 + 1 = 5.