Converse of Lagrange's theorem fails for the alternating group A4

Counterexample

The converse of Lagrange's theorem would say this. If $G$ is a finite group and $d$ is a positive divisor of $|G|$, then $G$ has a subgroup of order $d$. This is false. Let $A_4$ be the group of the even permutations of $\{1, 2, 3, 4\}$, under composition. It has order $12$, and $6$ divides $12$, but $A_4$ has no subgroup of order $6$.

Remarks

The elements of $A_4$ are the identity, the three products of two disjoint transpositions, and the eight $3$-cycles. A subgroup $H$ of order $6$ would have index $2$, and thus it would contain the square of each element of $A_4$. The square of each $3$-cycle is a $3$-cycle, and each $3$-cycle is such a square. So $H$ would contain all eight $3$-cycles, which is too many. The group $A_4$ is the smallest group for which the converse fails.

Explanations

L1 human

Lagrange's theorem says which sizes a subgroup can have, but it does not promise that each allowed size occurs. The group $A_4$ has $12$ elements, and $6$ is an allowed size, but no subgroup has $6$ elements. For cyclic groups the converse is true, so a counterexample must be a group that is not cyclic.