Nonabelian group of order p^3

1. Apr 6, 2012

joeblow

For any prime p how do I show that there is a nonabelian group of order p^3?

Since we are dealing with a p-group (call it G), its center is nontrivial (i.e., of order p,p^2, or p^3). Obviously, the center cannot have order p^3 (otherwise it's abelian). Also, if its center has order p^2, then $|G/Z(G)|=p$ implying that G is cyclic and thus abelian. Another no-go.

Thus, $|Z(G)|=p$ and $Z(G) \cong \mathbb{Z}_p$.

Now, I want G to be nonabelian, so I want a G containing Z_p such that $G/\mathbb{Z}_p \cong \mathbb{Z}_p \times \mathbb{Z}_p$ (which would be the only noncyclic group of order p^2).

From here, I have no idea how to recover the original G. Also, semidirect products are off-limits for this problem. (I'm going through an old algebra book and the semidirect product is not introduced till after the section from which this problem comes.)

Any ideas?

2. Apr 6, 2012

morphism

First off, you seem to be misquoting the result that states "G/Z(G) cyclic $\implies$ G abelian". I guess you're using this to conclude that $G/Z(G) = Z_p \times Z_p$ in your fourth paragraph.

Anyway, if all you need to know is that a nonabelian group of order p^3 exists, then you can do this: choose x and y in G that generate G/Z(G). Show that they in fact generate G. Since x and y necessarily don't commute (why?), G must be nonabelian.

(In fact, if you go one step further and study the product xy, then you can probably write down G as a semidirect product.)

Last edited: Apr 6, 2012
3. Apr 6, 2012

DonAntonio

First of all, I have no idea how to help you without using semidirect products. Even writing down the specific Cayley Table would be, at

least for me, equivalent to use the semidirect product without mentioning it, or writing the group as generators and relations...all the same.

Here you have some intent at this, but eventually it comes down to semidirect products: http://www.math.chalmers.se/Math/Grundutb/GU/MMA200/A09/group3.pdf

The next paper also gives it a shot, though it begins by stating and proving that there are 5 non-isomorphic groups of order p^3, which is hardly "first principles". He also uses the Heisenberg groups:

Anyway, what old book are you using and where in it is this problem?

DonAntonio

4. Apr 6, 2012

Hurkyl

Staff Emeritus
I can understand not invoking that machinery in your proof, but I see no reason why you wouldn't use the basic idea.

5. Apr 6, 2012

joeblow

@morphism- In the fourth paragraph, since $|G/Z(G)|=p^2$, it must be abelian. That leaves two possibilities: either $G/Z(G) \cong \mathbb{Z}_{p^2}$ or $G/Z(G) \cong \mathbb{Z}_p \times \mathbb{Z}_p$. Since $\mathbb{Z}_{p^2}$ is cyclic, G/Z can only be isomorphic to it if G is abelian, right?

If this is so, and G/Z is abelian, how can I ever find generators of G/Z that don't commute?

@DonAntonio- It's from _Abstract Algebra_ by John Beachy and William Blair. Upon further inspection, semidirect products are not covered in the book at all. (!) I took the class from Prof. Beachy who did cover them during the course, but he must have given us supplemental notes.

@Hurkyl- This problem came from a section on conjugacy and the class equation. I figure there must be something easier since developing direct products would not have much to do with the topic in the section.

6. Apr 6, 2012

morphism

Yep, that's right. I made that comment because you said "|G/Z(G)|=p implying that G is cyclic and thus abelian" in the second paragraph (what you should have said is "implying that G/Z is cyclic and thus G must be abelian"), so I wanted to make sure you understood the argument.

True - the images of x and y in G/Z will commute. But that doesn't mean x and y (which are elements of G) must commute - and in fact, I'm claiming that they cannot.

7. Apr 12, 2012

joeblow

By the correspondence theorem, I can show that the normal subgroups of G/Z(G) ($\mathbb{Z}_p \times 0$ and $0 \times \mathbb{Z}_p$) correspond to the normal subgroups of G containing Z(G). Since there are two normal subgroups of G/Z(G), there are two distinct normal subgroups of G containing Z(G). The mere existence of these two normal subgroups implies that there are elements in G that are not also in Z(G). Thus, G is nonabelian.

Does this sound like a valid argument? Somehow I doubt it.

Edit: Nope. Totally begged the question. Maybe I can show that the normal subgroups mentioned are such that the intersection is 0 and the sum of the two groups has p^3 elements.

Last edited: Apr 12, 2012
8. Apr 12, 2012

mathwonk

have you played with the unit group of integral quaternions? {±1, ±i, ±j, ±k}.

9. Apr 12, 2012

morphism

If you want to continue along the lines of my hint, then try to mimic the proof of the fact that G/Z cyclic => G abelian.

10. Apr 14, 2012

joeblow

We can assume that the center is nontrivial and thus if G has any hope of being nonabelian at all, then $G/Z(G)\cong \mathbb{Z}_p\times \mathbb{Z}_p$. However, taking inverse images of the generators of $\mathbb{Z}_p\times\mathbb{Z}_p$ assumes that G is nonabelian to begin with, doesn't it? So, I must not quite understand what you're telling me.