Suppose G has order 33. Show that it contains an element of order 3.

So G can contains elements of order 3,11, and 33. If G contains an element of order 33, then this will generate the whole group. So suppose G contains no element of order 33. Then every non-identity element of G has order 3 or 11. If every non-identity element has order 11, then the set $\displaystyle H = \{e, a_{1}, \dots, \underbrace{a_{11}, a_{1}a_{2}, \dots, a_{1}a_{11} \dots}_{11^{11} \ \text{terms}} \} $ is not a subgroup of G. E.g. there are no subgroups of G so that $\displaystyle |H| > |G| $. Thus there must be some element of order 3.