17
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ what is $Z(G)$ ? $\endgroup$ Jul 11, 2019 at 1:57
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 11, 2019 at 2:08
  • 2
    $\begingroup$ @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 $g = a^ib^jc^k$ . $\endgroup$
    – Mike
    Jul 11, 2019 at 2:13
  • 1
    $\begingroup$ @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 $\langle \overline{a},\overline{b} \rangle = \{ \overline{a}^j,\overline{b}^j:i,j \in \mathbb{Z} \}$ $\endgroup$
    – Mike
    Jul 11, 2019 at 3:05
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 11, 2019 at 7:42

2 Answers 2

7
$\begingroup$

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 $$a^n b^n=(ab)^n [b,a]^{n\choose 2}$$.

(See https://en.wikipedia.org/wiki/Commutator).

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 $p^2, p$.

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Mike
    Jul 11, 2019 at 15:18
  • 1
    $\begingroup$ @Mike: It means $G'\subseteq Z(G)$ in general. In the particular case of nonabelian groups of order $p^3$ you also get equality. $\endgroup$ Jul 11, 2019 at 16:28
  • $\begingroup$ @ArturoMagidin Thank you $\endgroup$
    – Mike
    Jul 11, 2019 at 16:29
  • $\begingroup$ @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. $\endgroup$
    – Mike
    Jul 11, 2019 at 16:37
  • $\begingroup$ 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. $\endgroup$ Jul 11, 2019 at 17:18
3
$\begingroup$

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: $$ \sum_{s=0}^{p-1} (p+1)^s = \frac{(p+1)^p - 1}{p} \equiv p \pmod{p^2} $$

$\endgroup$
5
  • $\begingroup$ What is the set $A$? $\endgroup$
    – Mike
    Jul 11, 2019 at 13:38
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 11, 2019 at 16:28
  • 1
    $\begingroup$ Also, \pmod{p} will produce the correct spacing and parentheses. $\endgroup$ Jul 11, 2019 at 16:31
  • $\begingroup$ thanks for that mathjax info Arturo $\endgroup$ Jul 11, 2019 at 23:30
  • $\begingroup$ @Mike apologies, $A = \langle a \rangle$. will correct the text $\endgroup$ Jul 12, 2019 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.