 Failure of the converse to Lagrange's theorem

Divisibility of a proposed subgroup order is necessary by Lagrange's theorem but is not sufficient in arbitrary finite groups. The alternating group $A_4$ has order $12$ and no subgroup of order $6$. Such a subgroup would have index two, hence be a normal subgroup. The quotient would have order two, so every element of order three would map to its identity. But $A_4$ has eight distinct three-cycles, which together with the identity cannot fit in a subgroup of order six. This obstruction contrasts with cyclic groups, which have a subgroup of every divisor order.