How to Construct a Non-abelian Group with p^3 Elements

Recently, while working through the exercises in Chapter 6 of Silverman’s Abstract Algebra: An Integrated Approach, I came across the following challenge problem.

6.16 (d): Challenge Problem. Construct a non-abelian group of order $p^{3}$ for every prime $p$ .

First attempts

We know that when $p = 2$ , two examples immediately come to mind: $D_{4}$ and $Q_{8}$ . But for other primes $p$ , this does not seem obvious at all—at least not for someone encountering this problem for the first time. The difficulty of the problem lies in the “non-abelian” condition: it is a rather tricky requirement, and at this point I did not yet have any handy tools for dealing with it.

So I recalled an exercise from our university textbook:

Let $G = (a, b) ∣ a, b \in R \land a \neq 0$ , and define $(a, b) \circ (c, d) = (a c, a d + b)$ . Prove that $G$ forms a group under the operation $\circ$ .

A quick check shows that this operation is indeed noncommutative. So this gives a non-abelian group on $R^{2}$ , with identity element $(1, 0)$ . But here the underlying set is $R^{2}$ ; how can we “adapt” it to a group of order $p^{3}$ ?

One direct idea is to let $a \in C_{p^{2}}, b \in C_{p}$ , and simply copy the textbook operation. However, this does not work, because some elements do not have inverses. For example, you cannot find an inverse for the element $(0, 0)$ , since $(0, 0) \circ (c, d) = (0, b) \neq (1, 0)$ , so this construction does not form a group.

So my exploration got stuck. It seemed that we might need to impose some additional restrictions on the domains of $a$ and $b$ , but how exactly to do that was far from clear. At this point, this line of thought appeared to hit a dead end.

The semidirect product approach

Then one day, while randomly flipping through later parts of Silverman, I noticed a new symbol: $⋊$ , called the semidirect product. It looks like an ordinary direct product that has somehow been “twisted,” so the resulting group ought to be noncommutative.

So I tried to construct $C_{p^{2}} ⋊ C_{p}$ . But how exactly is such a construction carried out? The textbook explanation was long and cumbersome, and I was too lazy to read it carefully, so I asked GPT for the construction of a semidirect product, and got the following answer:

$φ : H \to Aut (N)$ $N ⋊_{φ} H : (n_{1}, h_{1}) (n_{2}, h_{2}) = (n_{1} \cdot φ (h_{1}) (n_{2}), h_{1} \cdot h_{2})$

I had only just encountered the notation $Aut (N)$ for the first time in exercise 6.26 on Sylow theory. It seemed to mean “automorphisms,” and then one also has to prove that $Aut (C_{n})$ is isomorphic to $(Z / n Z)^{\times}$ . So I gradually understood that $Aut$ is actually a set of maps, whose objects are “maps” rather than “elements,” and these maps are required to be isomorphisms.

Then I thought about it heuristically. For a cyclic group $C_{n}$ , suppose its generator is $g$ . If $g$ is sent to $g^{d}$ (where $d ∣ n$ ), then every original element $g^{k}$ gets sent to $g^{k d}$ , and there are only $n / d$ distinct elements of the form $g^{k d}$ . So such a map certainly cannot be bijective, and hence cannot be an isomorphism. Therefore a natural conclusion is that $gcd (d, n) = 1$ , which corresponds exactly to the elements of $(Z / n Z)^{\times}$ .

Now let us return to $Aut (N)$ , namely $Aut (C_{p^{2}})$ . By the same reasoning, we immediately get $| Aut (C_{p^{2}}) | = (p - 1) p$ , and its elements are $ϕ_{k} : g \to g^{k} ∣ gcd (k, p^{2}) = 1$ .

The next question is this: we need a homomorphism $H \to Aut (N)$ , and the image $Image (H)$ is a subgroup of $Aut (N)$ ; meanwhile, by the First Isomorphism Theorem, $| Image (H) | ∣ | H |$ . Hence $| Image (H) | ∣ gcd (| H |, | Aut (N) |) = p$ , so $Image (H)$ can only be $e$ or $C_{p}$ . The former is obviously the trivial map $id$ , corresponding to the direct product, which does not satisfy the non-abelian requirement. Therefore, we are forced to consider how to construct a subgroup of $Aut (N)$ isomorphic to $C_{p}$ .

Let us restate the goal: within the set $ϕ_{k} : g \to g^{k} ∣ gcd (k, p^{2}) = 1$ , we want to choose $p$ elements forming a cyclic group of order $p$ .

At first glance, this does not seem easy. To gain some intuition, I tried the case $p = 3$ . Then $Aut (C_{3^{2}}) = g \to g, g \to g^{2}, g \to g^{4}, g \to g^{5}, g \to g^{7}, g \to g^{8}$ . First of all, $g \to g$ must be in the subgroup. Then the remaining two elements are $g \to g^{k}$ and $g \to (g^{k})^{k} = g^{k^{2}}$ . Let us check whether closure holds:

$k = 2$ , in which case the group would be $g \to g, g \to g^{2}, g \to g^{4}$ . Clearly $(g^{2})^{4} = g^{8}$ is not in the set, so this is ruled out.
$k = 4$ , in which case the group would be $g \to g, g \to g^{4}, g \to g^{16} = g^{7}$ . And $(g^{4})^{16} = g^{64} = g, (g^{16})^{16} = g^{256} = g^{4}$ , so closure holds, and this is a valid choice.

Thus for $p = 3$ , the valid exponents are $1, 4, 7$ . At this point, a clever reader may boldly guess that for arbitrary $p$ , the corresponding values should be $1, p + 1, 2 p + 1, \dots, p (p - 1) + 1$ . Let us check whether this really works—which is equivalent to the following classical number-theoretic congruence:

(1 + p)^{k} \equiv 1 + k p (\mod p^{2})

The proof is easy by the binomial theorem:

(1 + p)^{k} = 1 + k p + (\binom{k}{2}) p^{2} + \dots \equiv 1 + k p (\mod p^{2})

Therefore, we have found a valid homomorphism $φ : H \to Aut (N)$ satisfying:

φ (h_{1}) (n_{2}) = n_{2}^{1 + h_{1} p}

Substituting this into the original semidirect product formula gives:

(n_{1}, h_{1}) (n_{2}, h_{2}) = (n_{1} \cdot φ (h_{1}) (n_{2}), h_{1} \cdot h_{2}) = (n_{1} \cdot n_{2}^{1 + h_{1} p}, h_{1} \cdot h_{2})

But there is actually a slight problem with this notation, because $h_{1} p$ is an operation between a group element and an integer, and such an operation is not defined. So we need to specify explicitly that $n \in Z / p^{2} Z, h \in Z / p Z$ , and then replace the original multiplication by addition, and exponentiation by multiplication. This yields:

(a, b) (c, d) = (a + c (1 + b p), b + d)

If we check associativity (you can use a computer algebra system to reduce the pain of hand computation), we find that whether we associate left or right, the result is $(x, y) (a, b) (c, d) = (x + a (1 + y p) + c (1 + y p + b p), y + b + d)$ .

As for commutativity, $(0, 1) (1, 0) = (1 + p, 1), (1, 0) (0, 1) = (1, 1)$ , which are not equal, so the group is non-abelian.

The identity element is easily seen to be $(0, 0)$ . As for inverses, we seek $(x, y)$ such that $(a, b) (x, y) = (0, 0)$ . Clearly $y = - b$ , and then we have $a + x (1 + b p) = 0 (\mod p^{2})$ . Since $(1 + b p)$ is not divisible by $p$ , it has an inverse modulo $p^{2}$ , hence $x = - a (1 + b p)^{- 1} (\mod p^{2})$ .

In summary, we have successfully constructed a non-abelian group of order $p^{3}$ .

Relation to the textbook construction

Although I had solved the problem, I was still curious whether the textbook route was really a dead end. In fact, if one looks carefully, the structure of the textbook operation is somewhat similar to my answer, so let us try to transform the textbook operation a bit:

(a, b) \circ (c, d) = (a c, a d + b)

Set $x = b, y = a, u = d, v = c$ , and then the equation becomes:

(x, y) * (u, v) = (y u + x, y v)

This at least confirms that it bears some resemblance to the formula we derived earlier, $(a + c (1 + b p), b + d)$ . But there are still some differences in the details: for instance, the element $y$ probably needs to be modified, and the term $y v$ on the right somehow has to be turned into addition.

Note that our operation is taking place in $C_{p^{2}} ⋊ C_{p}$ , so $x, u \in Z / p^{2} Z$ . Then what about $y, v$ ? We can try to reinterpret the earlier homomorphism $φ (h_{1}) (n_{2}) = n_{2}^{1 + h_{1} p}$ in additive form, namely by setting

φ (h_{1}) = 1 + h_{1} p (\mod p^{2})

As shown earlier, all $1 + h_{1} p$ form a subgroup of $(Z / p^{2} Z)^{\times}$ of size $p$ . Therefore we may let $U = 1 + k p (\mod p^{2}) : k \in Z / p Z ≅ C_{p}$ , and take $y, v \in U$ . In this way we indeed obtain a valid non-abelian group of order $p^{3}$ .

To summarize, the situation now is:

$U = 1 + k p (\mod p^{2}) : k \in Z / p Z$
$(x, y) * (u, v) = (y u + x, y v)$
$x, u \in Z / p^{2} Z, y, v \in U$ .

Of course this is not the end yet. In order to simplify it into exactly the same form as our previous answer, we need to reparametrize $U$ as $C_{p}$ . So we make another change of variables, setting:

y = 1 + b p, v = 1 + d p, b, d \in Z / p Z

Thus:

y u + x = (1 + b p) u + x

y v = (1 + b p) (1 + d p) \equiv 1 + (b + d) p (\mod p^{2})

And the latter corresponds to $b + d (\mod p)$ , so in fact the original formula has already transformed into:

(x, b) * (u, d) = (x + (1 + b p) u, b + d)

Finally, substituting back $x = a, u = c$ gives:

(a, b) * (c, d) = (a + (1 + b p) c, b + d)

which is exactly the result obtained in the previous section. GPT’s comment on this was:

Your group of order $p^{3}$ is a specialization of the textbook’s “affine-type” group law to a particular $p$ -subgroup of the acting group; the two formulas differ only by coordinate order and reparametrization.

Are there any others?

In fact, for odd primes $p$ , there are only $5$ groups of order $p^{3}$ , of which three are abelian:

C_{p^{3}}, C_{p^{2}} \times C_{p}, C_{p} \times C_{p} \times C_{p}

and two are non-abelian. Namely the Heisenberg group $H$ :

H = {(\begin{matrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{matrix}) : a, b, c \in F_{p}}

and the one we constructed:

C_{p^{2}} ⋊ C_{p} : (a, b) (c, d) = (a + c (1 + b p),, b + d)

We can prove that for odd primes $p$ , $H$ is not isomorphic to $C_{p^{2}} ⋊ C_{p}$ , because $C_{p^{2}} ⋊ C_{p}$ has an element of order $p^{2}$ , namely $(1, 0)$ . Now take $X \in H$ and consider the order of $X$ . Let

N = (\begin{matrix} 0 & a & c \\ 0 & 0 & b \\ 0 & 0 & 0 \end{matrix})

Then:

X^{p} = (I + N)^{p} = I + (\binom{p}{1}) N + (\binom{p}{2}) N^{2} + (\binom{p}{3}) N^{3} + \dots

Observe that for $k \geq 3$ , $N^{k} = 0$ , so $(\binom{p}{3}) N^{3}$ and all subsequent terms vanish. Also, $(\binom{p}{1}) = p \equiv 0 (\mod p)$ , so we obtain:

X^{p} = I + (\binom{p}{2}) N^{2}

Since $(\binom{p}{2}) = \frac{p (p - 1)}{2}$ , we have $\frac{p (p - 1)}{2} \equiv 0 (\mod p)$ if and only if $p > 2$ . In that case $X^{p} = I$ , which shows that $H$ is not isomorphic to $C_{p^{2}} ⋊ C_{p}$ .

Finally, what happens when $p = 2$ ? One can prove that both $H$ and $C_{2^{2}} ⋊ C_{2}$ are isomorphic to the $D_{4}$ mentioned at the beginning of this post. Together with $Q_{8}$ , there are still five types in total. This shows that $Q_{8}$ is really an outlier: it cannot be seen at a glance from the most naive semidirect product intuition.