I am trying to show that a non-abelian group $G$ of order $p^3$ is isomorphic to one of two groups constructed on page 48 of these group theory notes (see examples 3.14 and 3.15 on that page).

Here is some of my work so far:

Since $G$ is a $p$-group, it must have non-trivial center; but, as $G$ is non-abelian, the order of $Z := Z(G)$ must be $p$. This means that $G/Z \cong \Bbb{Z}_p \times \Bbb{Z}_p$, so $G/Z$ is generated by two elements. Let $x,y$ be the elements whose images (under the canonical projection map) in $G/Z$ generate $G/Z$. Since $G/Z$ is abelian (all $p^2$ groups are abelian), $[x,y] \in Z$. If $[x,y]$ were the identity, then $G$ would actually be abelian, so $[x,y] \neq 1$, in which case the only possible order of it is $p$. Hence, $Z = \langle [x,y] \rangle$.

At this point, my idea was to break up the problem into two cases: both $x$ and $y$ have order $p$, or at least one has order For the first case, I thought I was basically finished. I showed that $x$, $y$ and $[x,y]$ satisfied the three defining relations in example 3.15:

$$a^p = b^p = c^p = 1 \text{ , } ab = cac^{-1} \text{ , } [b,a] = 1 = [b,c],$$

where, in my situation, $a = x$, $b = [x,y]$, and $c =y$. But then it quickly occured to me that $[x,y] \in \langle x,y \rangle$, so the commutator cannot be the third generator.

As for the case, I wasn't able to come up with anything. I have been working on this problem for a rather long time; I could use some guidance. Initially I was vainly following the hint in the back of the book; but then someone in the chatroom kindly pointed out that the author made a false claim in the hint.

  • $\begingroup$ There is no issue; generating sets/relations need not be as small as possible. Note that even in the original description, you do not actually need $b$, since $b$ can be replaced with $b=a^{-1}cac^{-1}$ (from the equality $ab=cac^{-1}$), and so you can describe the group using just two generators. That is, $b\in\langle a,c\rangle$ in the original description, so there is no problem with you having your third “generator” be in the subgroup generated by the first two. $\endgroup$ – Arturo Magidin Mar 23 at 21:59
  • $\begingroup$ Your final paragraph seem to be incomplete. “As for the case”.... what case? Where one of $x$ and $y$ has order $p^2$? Say $x$ has order $p^2$ and $y$ has order $p$. Since $x$ and $y$ project to different things, you must have $\langle x\rangle\cap \langle y\rangle = \{e\}$, and so $\langle x\rangle\langle y\rangle = G$ (by a simple counting argument). So every element is of the form $x^ay^b$, $0\leq a\lt p^2$, $0\leq b\lt p$. Show that $[x,y]\in\langle x^p\rangle$. $\endgroup$ – Arturo Magidin Mar 23 at 22:02
  • $\begingroup$ $[x,y]$ lies in $\langle x,y\rangle$; but, it can be taken as third generator! unlike in abelian groups, the third generator is not necessarily outside the subgroup $\langle x,y\rangle$. [In fact, $\langle x,y\rangle$ is whole group $G$ in your case, whether $o(x)=p^2,o(y)=p$ or both have order $p$, or both have order $p^2$! Thus, the third generator has to be in $G=\langle x,y\rangle$; where is problem? $G$ is non-abelian! $\endgroup$ – Beginner Mar 24 at 5:39
  • $\begingroup$ @Beginner "unlike in abelian groups"? Generating sets never have to be minimal. For example, the following is a perfectly fine presentation of an abelian group (the Klein 4-group, which can be generated by two elements as well): $\langle x,y,z\mid x^2=y^2=z^2=1, xy=z, xy=yx, xz=zx, yz=zy\rangle$. $\endgroup$ – Arturo Magidin Mar 24 at 5:49
  • $\begingroup$ yes; that is fine. My rough comment was, in finite abelian groups, the usual generating sets are minimal ones (of course all elements of $G$ form generating set.) The user193319 asked this problem of picking third generator outside $\langle x,y\rangle$; I thought, he might be considering situations of abelian groups. In (one of) the proof of structure of finite abelian groups, generators are chosen in such fashion. $\endgroup$ – Beginner Mar 24 at 5:51

(1) $G/Z(G)$ is generated by $xZ(G)$ and $yZ(G)$; $Z(G)$ is generated by $[x,y]$ (also note $Z(G)=G'$).

(2) Suppose that not both $x,y$ have order $p$. This implies two cases: $o(x)=p^2$, $o(y)=p$ or $o(x)=o(y)=p^2$. [The case $o(x)=p, o(y)=p^2$ has considered, can you see?]

(2.1) Suppose $o(x)=p^2$ and $o(y)=p$. Then $\langle y\rangle$ is subgroup of order $p$, and $\langle y\rangle\cap Z(G)=1$.

Since $xZ(G)$ is of order $p$, so $x^p\in Z(G)$, and $x^p\neq 1$ (since $o(x)=p^2)$, so without loss, we can take $x^p=[x,y]$. So $\langle x\rangle$ is group of order $p^2$, it has unique subgroup of order $p$ namely $\langle x^p\rangle$ which is $\langle [x,y]\rangle$, which is $Z(G)$, and $\langle y\rangle$ intersects with this trivialy; i.e. $\langle x\rangle\cap\langle y\rangle=1$. So $G=\langle x\rangle.\langle y\rangle=\langle x,y\rangle$, where $x,y$ satisfy relations (which perhaps you want) $$x^{p^2}=1, y^p=1, x^p=[x,y]=x^{-1}y^{-1}xy \mbox{ i.e. } x^{1+p}=y^{-1}xy.$$

(2.2) Suppose $o(x)=o(y)=p^2$. Then point is we can modify one of these two generators to get a new generator of order $p$. How to do this? This requires commutator calculus: since $[x,y]\in Z(G)$, so $[x,y]$ commutes with both $x$ and $y$. You can prove that $(xy)^n=x^ny^n[y,x]^{\binom{n}{2}}$. (Prove this, use induction on $n$.) Now we use this. Since $o(X)=o(y)=p^2$, and $x^p,y^p\in Z(G)=\langle [x,y]\rangle$, without loss, we can assume that $$x^p=y^p=[x,y].$$ Then $$(yx^{-1})^p=y^px^{-p}[x^{-1},y]^{\binom{p}{2}}.$$ Since $[x^{-1},y]$ is a member of $G'=Z(G)$, it is of order $p$ and for odd $p$, the integer $\binom{p}{2}=p(p-1)/2$ is multiple of $p$, so $[x^{-1},y]^{\binom{p}{2}}=1$. Also $y^px^{-p}=1$ (why?) Thus we get $$(yx^{-1})^p=1.$$ So, instead of $y$ you take new generator $y_1=yx^{-1}$; this is of order $p$, and we move in case (2.1), which is done!

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    $\begingroup$ Regarding case 2.1, why can we without loss of generality assume $x^p = [x,y]$? You also make this assumption about $x^p$ and $y^p$ in case 2.2., so hopefully you could clarify that, too. $\endgroup$ – user193319 Mar 24 at 13:17
  • $\begingroup$ Also, why does $Z = G'$? $\endgroup$ – user193319 Mar 24 at 17:58
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    $\begingroup$ @user193319 Write $[x,y]=x^{bp}$; this is possible since $[x,y]\in\langle x\rangle$ and $|x|=p^2$. Also, $b$ is invertible mod. $p$, because $[x,y]\ne 1$. Thus, there exists a multiplicative inverse $c$ of $b$ mod. $p$. Now, because $x$ and $y$ commute with $[x,y]$, we have $[x,y^c]=[x,y]^c$. Furthermore, $[x,y]^c=x^p$. Replacing $y$ with $y^c$, we get the desired equality. $\endgroup$ – Karl Kronenfeld Mar 24 at 23:46
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    $\begingroup$ @user193319: $G’$ is nontrivial because $G$ is not abelian; $G/G’$ is abelian, and any normal subgroup $N$ of $G$ such that $G/N$ is abelian must contain $G’$. Therefore, $G’\subseteq Z(G)$, and since $Z(G)$ is order $p$ and $G’$ is of order greater than $1$, the two must be equal. $\endgroup$ – Arturo Magidin Mar 24 at 23:49

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