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I don't know other group names. The case ##p=2## would suggest a generalized quaternion group for the ##A's##, but those are dicyclic, and higher dihedral groups for the ##B's##, but those contain reflexions. However, I admit I did everything to avoid group characters and the unit circle. I thought that might get too complicated, resp. too far from standard knowledge. (To be honest: it has been so long ago that I studied them that I've forgotten most of it.) Maybe there is another way to show the result by more emphasis on conjugation classes and orbits.micromass said:If you write groups as a presentation, then you risk the groups to be trivial, or not having the order you want. So how are you sure that ##A## and ##B## have order ##p^3##?
I know that ##G \diagup Z## is not cyclic of order ##p^2##, i.e. ##G \diagup Z ≅ C_p^2##. (The other cases have been ruled out or treated before.) So it is generated by two elements ##aZ## and ##bZ## of order ##p \, (mod \, Z)## and therefore I have ##p^2## elements modulo ##Z##.
And ##G' = [G,G] = Z## is cyclic of order ##p##. So I get ##p \cdot p^2 = p^3## elements in ##G##.
Since ##G## is not abelian and ##a \notin Z## has been chosen in the w.l.o.g statement, I can choose an element ##b \notin Z## of order ##p## that doesn't commute with ##a##, i.e. ##1 \neq [a,b] ## generates ##Z = G' ≅ C_p##. For this purpose I can take a representative ##b## of the set ##G \diagup{<a>}##. If such a ##b## would commute with ##a## it would be a central element of ##G##. If all those elements were of order ##p^2## then ##G \diagup Z \ncong C_p^2.##
Finally, the groups ##A## and ##B## cannot be isomorphic since ##B## contains an element of order ##p^2## and ##A## doesn't.