# Thread: Prove that for all p, there exists a non-abelian group of order p^3.

1. ## Prove that for all p, there exists a non-abelian group of order p^3.

Using the conjugacy class equation: $|G| = |Z(G)| + \sum [G: C(x)]$ i want to show that there exists a non-abelian group of order $p^3$. I know that $|G| = p^3$ and since i want a non-abelian group, $|Z(G)| = p$ or $p^2$. If |Z(G)| = p^2, then |G/Z(G)| = p but then that would mean G/Z(G) is cyclic with order p which cannot be since G/Z(G) is never a nontrivial cyclic group. therefore |Z(G)| = p. likewise each [G: C(x)] must be either p or p^2 since if it was $p^3$ then |C(x)| = 1 which contradicts the definition of the class equation, and if the index = 1, then |C(x)| = $p^3$ which means Z(G) = G. Then i say that each [G: C(x)] = p and if there are $p^2 - 1$ of them, then $(p^2 - 1)p + p = p^3 - p + p = p^3$ which is the correct order of the group. But also, if each [G: C(x)] = p^2, and if there are p-1 of them, then $(p-1)p^2 p = p^3 - p + p = p^3$ which is also the correct order.

i am having trouble with the last part, since it seems like since we have a sum, a bunch of the [G: C(x)] can be p while a bunch of other could be $p^2$ and when you add them all together and add them to $p$ we might somehow get $p^3$. in other words, there doesn't seem to be a reason why each conjugacy class must have the same number of elements. although i assumed so to simplify things and to check that it worked, i don't know how to deal with the case if not all of the conjugacy classes are the same order. Could someone help fix up the end of my proof? thanks.

2. ## Re: Prove that for all p, there exists a non-abelian group of order p^3.

This may not be what you are looking for, but a simpler way to prove the existence of a nonabelian group of order $p^3$ would be to give an explicit construction of such a group, for example the Heisenberg group over the field $\mathbb{F}_p$ of p elements.

3. ## Re: Prove that for all p, there exists a non-abelian group of order p^3.

note that every C(x) has to contain the center Z(G), since elements of Z(G) commute with ALL of G, so they certainly commute with x.

since x is not in the center by construction, |C(x)| > p.

this means we have to have |C(x)| = p^2, so [G:C(x)] = p.

EDIT: for a specific construction, consider this:

we know $\mathbb{Z}_p$ and $\mathbb{Z}_{p^2}$ exist.

consider the following maps $\sigma_k:\mathbb{Z}_{p^2} \to \mathbb{Z}_{p^2}$

defined by $\sigma_k(x) = (1+kp)x\ (\text{mod }p^2), k = 0,1,2,\dots,p-1$

note that each $\sigma_k \in \text{Aut}(\mathbb{Z}_{p^2})$.

we can define the following product on $\mathbb{Z}_{p^2} \times \mathbb{Z}_p$:

$(m_1,n_1)*(m_2,n_2) = (m_1+\sigma_{n_1}(m_2),n_1+n_2)$

note that $(1,0)*(1,1) = (1+1,1)$, while $(1,1)*(1,0) = (1+1+p,1)$

so this multiplication is not abelian, so we have a non-abelian group of order p^3.

4. ## Re: Prove that for all p, there exists a non-abelian group of order p^3.

Or, to say what is more general true if you can factor $n$ as $\ell k$ where $\ell\mid |\text{Aut}(\mathbb{Z}_k)|=\varphi(k)$ then you have a non-trivial homomorphism $f:\mathbb{Z}_\ell\to\text{Aut}(\mathbb{Z}_k)$ and thus have a non-trivial, and so nonabelian, semidirect product of order $n$. So, in your example, since $\varphi(p^2)=p(p-1)$ and $p\mid p(p-1)$ there exists nontrivial homomorphisms $f:\mathbb{Z}_p\to\text{Aut}(\mathbb{Z}_{p^2})$ and so nontrivial (thus nonabelian) semidirect products o order $p^3$--this is precisely what Deveno said.

This all said, I like Opalg's solution the best, since not only does it give a succinct answer but should get one acquainted with the ubiquitous Heisenberg group.

5. ## Re: Prove that for all p, there exists a non-abelian group of order p^3.

for the homomorphism f to be well defined, doesn't the order of the element that [1] maps to have to divide $\ell$? i'm confused on why $\ell | \varphi(k)$ and how that implies that f is a well defined homomorphism. is this a general result or does this only occur for the specific groups that you guys mentioned?

6. ## Re: Prove that for all p, there exists a non-abelian group of order p^3.

it happens for a wide class of groups. there is always a homomorphism N-->Aut(H) for any two groups N,H, but often this is only the trivial one:

$\phi(n) = \text{id}_H$ for all $n \in N$. suppose we have p = 2.

then p^3 = 8, so we're looking for a nontrivial homomorphism of Z2 into Aut(Z4). now φ(4) = 2, so we have just 2 automorphisms:

x-->x (this is x--->(1+0*2)x)
x-->3x (this is x--->(1+1*2)x).

note that x-->3x is really the same as x--> -x in Z4.

so we can re-write our multiplication in Z4 x Z2 as: (a,b)*(a',b') = (a+((-1)^b)(a'), b+b').

we have two types of elements: (c,0) and (c,1).

multiplying by (c,0) just "collects first coordinates": (c,0)*(a,b) = (c+a,b).

multiplying by (c,1) "twists the first coordinate, and flips the second": (c,1)*(a,b) = (c-a,b+1).

this is our old friend D4, with (1,0) being the rotation r, and (0,1) being the reflection s.

in drexel28's post, we want to have $\ell|\varphi(k)$ or else there may not be a non-trivial homomorphism,

since $\varphi(\mathbb{Z}_{\ell})$ is a subgroup of Aut(Zk) and so has to divide its order φ(k). if indeed $\ell|\varphi(k)$, then we can just send [1] to an element of an order that divides $\ell$.

7. ## Re: Prove that for all p, there exists a non-abelian group of order p^3.

By the way, is p assumed to be a prime here? You don't say so explicitely but the people who are responding seem to be assuming that p is a prime number.

8. ## Re: Prove that for all p, there exists a non-abelian group of order p^3.

Originally Posted by HallsofIvy
By the way, is p assumed to be a prime here? You don't say so explicitely but the people who are responding seem to be assuming that p is a prime number.
yes, that is the assumption, or else we cannot conclude that |Z(G)| = p. for example Q8 x Z27 has order 216 = 6^3, but it's center is <(-1,1)> which has order 54, not 6.

the reason we make this assumption even though the original poster failed to state this explicitly, is because this is typical of problems using the class equation.