hmm.. but if n_3 divides 7=21/3, and n_3=1mod(7), then n_3 is either 1 or 7. Why do you claim it has three different groups of order 3? How comes it doesn't have only one?

My bad: i misread the question. You said there're three different elements of order 3 and I read "three different subgroups of order 3". It never minds: if there are more than 2 elements of order 3 then there are more than 1 subgroup of order 3 and thus the argument stays the same.

And you're right: there are either 1 or 7 subgroups of order 3 in a group of order 21...

Tonio
And, how do you use the fact that there are at least 3 elements of order 3 in the group, in order to find out there exist three different groups of order 3?

Thanks