Corollary 6.12 suggests that groups of prime order $p$ must somehow look like ${\mathbb Z}_p$.
The group $A_4$ has order 12; however, it can be shown that it does not possess a subgroup of order 6. According to Lagrange's Theorem, subgroups of a group of order 12 can have orders of either 1, 2, 3, 4, or 6. However, we are not guaranteed that subgroups of every possible order exist. To prove that $A_4$ has no subgroup of order 6, we will assume that it does have such a subgroup $H$ and show that a contradiction must occur. Since $A_4$ contains eight 3-cycles, we know that $H$ must contain a 3-cycle. We will show that if $H$ contains one 3-cycle, then it must contain more than 6 elements.