= Failure of the converse to Lagrange's theorem
{title2=$6\mid|A_4|,\quad A_4\text{ has no subgroup of order }6$}
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.
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