Micromass' big July Challenge

[fresh_42]

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 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.

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.

 

[micromass]

You can't define a group only by giving the presentation. You have then no guarantee that ##a\neq b## as generators of the group. There are many similar issues. So what you've proven is that there are at most ##2## nonabelian groups. But a presentation alone doesn't suffice in defining a group.

 

[fresh_42]

You can't define a group only by giving the presentation.

Why not? A group ##G## defined as ##<a,b \;|\; \mathcal{R}_i (a,b)>## is defined as the quotient of the free group ##\mathcal{F}(a,b)## and its normal subgroup generated by all relations ##\mathcal{R}_i (a,b)## (and their conjugates).

I have shown that ##G## of order ##p^2## is isomorphic to a group in ##\{C_p^2,C_{p^2} \}##, the abelian groups ##G## of order ##p^3## are isomorphic to a group in ##\{C_p^3,C_{p^2} \times C_p, C_{p^3} \}## and the non-abelian groups ##G## of order ##p^3## are isomorphic to ##G \cong (G\diagup Z) \ltimes Z \cong (C_p \times C_p) \ltimes C_p ## with the center ##Z=G' = [G,G]## of ##G## and cyclic groups ##C_p##. The representations only show the two basic non-abelian possibilities of the structure of the semi-direct product.

Do you mean that I have to show that ##G \diagup Z## is isomorphic to a subgroup of ##G##?

 

[micromass]

Why not? A group ##G## defined as ##<a,b \;|\; \mathcal{R}_i (a,b)>## is defined as the quotient of the free group ##\mathcal{F}(a,b)## and its normal subgroup generated by all relations ##\mathcal{R}_i (a,b)## (and their conjugates).

You have no guarantee about the order of this group, neither about whether the generators are distinct or not.

I have shown that ##G## of order ##p^2## is isomorphic to a group in ##\{C_p^2,C_{p^2} \}##, the abelian groups ##G## of order ##p^3## are isomorphic to a group in ##\{C_p^3,C_{p^2} \times C_p, C_{p^3} \}## and the non-abelian groups ##G## of order ##p^3## are isomorphic to ##G \cong (G\diagup Z) \ltimes Z \cong (C_p \times C_p) \ltimes C_p ## with the center ##Z=G' = [G,G]## of ##G## and cyclic groups ##C_p##. The representations only show the two basic non-abelian possibilities of the structure of the semi-direct product.

Right, if you say the group is ##(C_p\times C_p)\ltimes C_p##, then you know that this is a group of ##p^3##. So THAT is a good answer. Just giving the presentation gives you no guarantees about the orders.[/QUOTE]

 

[fresh_42]

Ok, you're right. E.g. if ##a## is of order ##4## in ##D_4## then ##a^2## generates the center which isn't obvious by the representation itself. I thought I could bypass this by showing ##Z = [G,G]##.