# Groups of order $p^3$

The following is exercise 8 (section 2.6) in Algebra by Hungerford:

Let $$p$$ be an odd prime. Prove that there are at most two nonabelian groups of order $$p^3$$. (One has generators $$a,b$$ satisying $$|a| = p^2; |b|=p;b^{-1}ab = a^{1+p}$$ and the other has generators $$a,b,c$$ satisfying $$|a|=|b| = |c|= p;c = a^{-1}b^{-1}ab;ca=ac;cb=bc$$)

I realize that there are many links online classifying groups of order $$p^3$$. However, most of the solutions I have seen have required several pages of work. Since this is just an exercise, I suspect there is a shorter solution. I have come up with the following:

We know that $$Z(G) = p$$ because $$G$$ is a $$p$$-group and if $$|Z(G)| = p^2$$ or $$p^3$$, then $$G/Z(G)$$ would be cyclic. Thus, $$G$$ would be abelian, contrary to hypothesis. So, let $$c$$ be a generator for $$Z(G)$$. Consider the quotient group $$G/Z(G)$$ which has order $$p^2$$. Hence, $$G/Z(G) \cong \mathbb{Z}_p \times \mathbb{Z}_p$$. Let $$\overline{a},\overline{b}$$ be generators for $$G/Z(G)$$ (the overlines denote passage to the quotient). Now, this means that every element in $$G$$ can be written as $$a^ib^jc^k$$ for $$i,j,k \in \mathbb{Z}$$. Since $$c \in Z(G)$$ we cannot have that $$a,b$$ commute (this is important later).

Notice that $$\overline{a},\overline{b}$$ both have order $$p$$ in $$G/Z(G)$$. Thus, $$a^p,b^p \in Z(G)$$. We consider cases of $$a^p$$ and $$b^p$$.

Case 1: both $$a^p = b^p = 1$$

Then, since $$G/Z(G)$$ is abelian, the commutator subgroup, $$G'$$, is contained in $$Z(G)$$ so that $$|G'| = 1$$ or $$p$$. But, $$|G'| = 1$$ implies $$G$$ is abelian, contrary to hypothesis. So, $$|G'| = p$$ and consequently $$G' = Z(G)$$. Thus, since $$a,b$$ do not commute, $$a^{-1}b^{-1}ab$$ is nontrivial and $$Z(G) = \langle a^{-1}b^{-1}ab\rangle$$ (this is by the previous exercise where we showed that the commutator subgroup is generated by elements of this form). Thus, by possibly switching generators for $$Z(G)$$ this group satisfies the second pair of generators and relations in the question.

Case 2: $$a^p \neq 1$$ and $$b^p = 1$$ (or vice versa)

In this case $$a^p$$ is a nontrivial element in $$Z(G)$$ and since $$Z(G)$$ is cyclic of order $$p$$, $$c = a^p$$ and so $$|a| = p^2$$. This gives rise to the first pair of generators and relations.

It is here where I am stuck. These should be the only two, but I have not found a reason why we cannot have $$a^p \neq 1$$ and $$b^p \neq 1$$. How can I guarantee that this case does not occur (or gives one of the two cases above)? I assume this has something to do with the oddness of $$p$$, as I have not used this anywhere in the argument. Moreoever, any other suggestions or corrections on the current work would be greatly appreciated.

Thank You

• what is $Z(G)$ ? Jul 11, 2019 at 1:57
• I do not think it is true that for every element in $G$ can be written as $a^ib^jc^k$ for $i,j,k\in \mathbb{Z}$ if $G$ is non-Abelian. Jul 11, 2019 at 2:08
• @zongxiangyi Why so? Since $G/Z(G) = \langle \overline{a},\overline{b}\rangle$ any coset represenative $gZ(G) = a^iZ(G)b^jZ(G) = a^ib^jZ(G)$. Now, since $Z(G) = \langle c \rangle$ this implies $g = a^ib^jc^k$ .
– Mike
Jul 11, 2019 at 2:13
• @zongxiangyi I see your point, but we also have $c$. Looking at the line above, if you agree that $gZ(G) = a^ib^iZ(G)$, then you should believe that $g = a^ib^jc^k$. The reason why $gZ(G) = a^ib^iZ(G)$ is true is because the quotient group is abelian and thus $\langle \overline{a},\overline{b} \rangle = \{ \overline{a}^j,\overline{b}^j:i,j \in \mathbb{Z} \}$
– Mike
Jul 11, 2019 at 3:05
• You cannot guarantee that you do not have $a^p\neq 1$ and $b^p\neq 1$. However, if that is the case, you should be able to replace one of them, say $b$, with a different element of order $p$, whose image together with that of $a$ generates the quotient. That is, you should be able to tweak your generators to get back to the second case. Jul 11, 2019 at 7:42

As @Arturo Magidin suggests in a comment, you only need to tweak your generators.

For suppose $$a$$ and $$b$$ have order $$p^2$$; then by taking suitable powers we can assume $$a^p=c$$, $$b^p=c^{-1}$$ where $$c$$ is a generator of the centre.

Now use the magic fact: as the derived group is central we have $$a^n b^n=(ab)^n [b,a]^{n\choose 2}$$.

Now take $$n=p$$, an odd prime, so that $$p$$ divides $$p\choose 2$$: we will get $$c c^{-1}=(ab)^p$$ and so can replace the generators $$a,b$$ by new generators $$a,ab$$ of orders $$p^2, p$$.

• Thank you for your response. I am looking at the wikipedia page and am wondering what it means for the commutator subgroup to be central? Does this mean the $G' = Z(G)$ or something along these lines?
– Mike
Jul 11, 2019 at 15:18
• @Mike: It means $G'\subseteq Z(G)$ in general. In the particular case of nonabelian groups of order $p^3$ you also get equality. Jul 11, 2019 at 16:28
• @ArturoMagidin Thank you
– Mike
Jul 11, 2019 at 16:29
• @ancientmatematician Once again thank you for your response, it definitely solves my problem. I do not think I would have thought to even look at the "magic fact" that you showed.
– Mike
Jul 11, 2019 at 16:37
• Not my fact of course! But I cut my mathematical teeth reading Higman's 1959 enumeration of $p$-groups, and the groups of order $p^3$ are but a very special case of $p$-groups of $\Phi$-class 2, which Higman enumerates. Jul 11, 2019 at 17:18

Case 2

Suppose $$G$$ has an element $$a$$ of order $$p^2$$. This case seems to require a certain minimum of cabalistic manipulation, I have not encountered a way of deriving the required result without a bit of slog. The following account still leaves a little arithmetic to be verified by the reader.

let $$A = \langle a \rangle$$. then $$Z \subset A$$ since $$x \not \in \langle a\rangle \implies \langle a,x\rangle = G$$ and G is non-abelian. so $$Z = \langle a^p\rangle$$.

continue with $$x$$ representing an element in $$G \setminus A$$, and use an overbar to indicate images in the factor group $$G/Z$$.

since $$\bar G$$ is abelian the inner automorphism of $$\bar A$$ corresponding to $$\bar x$$ is trivial. i.e. $$\bar x^{-1} \bar a \bar x = \bar a$$.

the pre-image of this relationship in $$G$$ is $$(xa^{mp})^{-1}a^{np+1}xa^{mp} = a^{n'p+1}$$ for some $$m,n,n' \in [1,p) \cap \mathbb{N}$$. As $$\langle a^p\rangle = Z$$ this simplifies to: $$x^{-1}ax = a^{jp + 1} \tag{1}$$

where $$j=n'-n$$.

now (1), together with $$a^{p^2}=1$$ implies that: $$x^{-k}ax^k= a^{(jp+1)^k} = a^{kjp+1}$$

for some $$k$$ we have $$kj \equiv 1 \pmod{p}$$, so set $$b_1 = x^k$$

now we have $$b_1^{-1}ab_1 = a^{p+1}$$

it follows that: $$b_1^{-p}ab_1^p =a \tag{2}$$

construction of the other generator, forcing |b|= $$p$$

from (2) $$b_1^p$$ commutes with $$a$$ hence $$b_1^p = a^k$$ for some $$k$$. but now raising this to the power $$p$$ gives $$1=a^{kp}$$ so that $$k$$ is a multiple of $$p$$, say $$k=hp$$.

suppose $$b_1$$ has order $$p^2$$ with $$b_1^p = a^{hp}$$ define $$b = b_1a^{-h}$$. Since $$a^jb_1 = b_1a^{j(p+1)}$$ we may write $$a^{-h}b_1 = b_1a^{-h(p+1)}$$ hence: $$b^p = (b_1a^{-h})^p = b_1^pa^{-h\sum_{s=0}^{p-1} (p+1)^s}=b_1^pa^{-hp} = 1 \tag{2}$$

so $$G = \langle a,b\rangle$$, $$|a|=p^2$$, $$|b|=p$$, $$b^{-1}ab = a^{p+1}$$.

the last equality in (2) follows from: $$\sum_{s=0}^{p-1} (p+1)^s = \frac{(p+1)^p - 1}{p} \equiv p \pmod{p^2}$$

• What is the set $A$?
– Mike
Jul 11, 2019 at 13:38
• Please use \langle and \rangle, not < and >. It also means you don't need to add that weird extra thing space you are adding manually. Jul 11, 2019 at 16:28
• Also, \pmod{p} will produce the correct spacing and parentheses. Jul 11, 2019 at 16:31
• thanks for that mathjax info Arturo Jul 11, 2019 at 23:30
• @Mike apologies, $A = \langle a \rangle$. will correct the text Jul 12, 2019 at 13:24