So you have to show that there is a subgroup of some order that does not divide 33?
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 is not a subgroup of G. E.g. there are no subgroups of G so that . Thus there must be some element of order 3.
Why? Are you saying that if G contains an element of order 33 then it is obvious that it contains an element of order 3? I think you should show that as well.
Excuse me but what do you mean by , , etc.?Then every non-identity element of G has order 3 or 11. If every non-identity element has order 11, then the set is not a subgroup of G. E.g. there are no subgroups of G so that . Thus there must be some element of order 3.