# Constructing a group of order $2187=3^7$

I am trying to look for the $$2$$-generated groups of order $$3^7$$ and class $$4$$ all whose upper central series quotients are elementary abelian of order 9 except the center which has order $$3$$.

A small check through GAP reveals there is a unique one which is a semi-direct product of $$C_{81}$$ and $$C_{27}$$, namely SmallGroup($$2187,194$$). I am trying to get a structural argument for this.

The second center is $$2$$-generated abelian of order $$27$$. Is it possible to construct the third center using the cohomology argument from this without knowing how the generators of the group behave?

Apologies, if the question is too easy.

$${\mathrm{\bf{Revised~notes}}}$$: The group SmallGroup($$2187,194$$) turns out to be powerful (Thanks to Derek!).

• I take it that the cyclic group of order $81$ is the normal one. Do you mean that you want a structural explanation of why this is the ONLY example? Or a structural explanation of why this example fulfils your conditions? – Geoff Robinson Sep 7 '20 at 8:59
• I want a structural explanation of why this is the ONLY example. – Siddhartha Sep 7 '20 at 10:07
• Is your assertion about the structure of the second centre something that you know how to prove already, or are you just assuming that to be true? (It has to be abelian of order $27$, but is it clear that it is not elementary abelian?) – Derek Holt Sep 7 '20 at 11:03
• @Derek: the GAP code showed it and hence I wasn't suspicious about it. But you are right, I have a different and long reasoning that implies $Z_2(G)$ is $2$-generated abelian of order $27$. – Siddhartha Sep 7 '20 at 11:59
• I just tried some calculations with $\mathtt{SmallGroup}(2187,194)$, and I found that $G' < G^p$, which seems to contradict the first of your additional notes. (Or did you mean that it is not true that $G' \le G^p$, in which case why be so negative?) – Derek Holt Sep 7 '20 at 13:47

I think I can see how to prove this now under the assumption that $$G$$ is powerful. I think the same approach would work without that assumption, but would involve eliminating more cases.

We are given that $$G$$ is a $$2$$-generated group, and that the upper central series of $$G$$ is $$1=Z_0 < Z_1 < Z_2 < Z_3 < Z_4 = G$$ with $$|Z_1|=3$$, $$|Z_2|=27$$, $$|Z_3| = 243$$, and $$|Z_4|=|G|=2187$$, with $$Z_2/Z_1$$, $$Z_3/Z_2$$ and $$Z_4/Z_3$$ elementary abelian

My approach is to identify the quotients $$G/Z_i$$ for $$i=3,2,1,0$$. I will just sketch the proof for now, and I can fill in details later if necessary.

We know that $$G/Z_3 = C_p^2$$ is elementary abelian. and it is not hard to see that $$G/Z_2 \cong \langle a,b \mid a^9=1, [[b,a],a] = [[b,a],b] = 1, b^3 = 1\ {\rm or}\ [b,a] \rangle.$$ Since the group with $$b^3=1$$ is not powerful, we can assume that $$b^3 = [b,a]$$ and in fact $$G/Z_2 \cong \langle a,b \mid a^9=b^9=1, a^{-1}ba=b^4 \rangle.$$

The hardest part is to identify $$G/Z_1$$, but using the facts that it is a powerful 2-generated group with centre is $$Z_2/Z_1 \cong C_p^2$$, it can be shown that $$G/Z_1 \cong \langle a,b \mid a^{27}=b^{27}=1, a^{-1}ba=b^4 \rangle.$$

The final step, showing that $$G \cong \langle a,b \mid a^{27}=b^{81}=1, a^{-1}ba=b^4 \rangle$$ is similar but easier.

$${\bf Edit\!:}$$ Since we would prefer to prove this without assuming that $$G$$ is powerful, we need to eliminate the possibility that $$b^3=1$$ in $$G/Z_2$$. So assume that $$b^3=1$$ in $$G/Z_2$$ and now replace $$a,b$$ by inverse images in $$G/Z_1$$.

Suppose that $$a^{-1}ba = bt$$ in $$G/Z_1$$. Then $$a^{-1}b^3a = (bt)^3 = b^3t^3[b,t]^3 = b^3t^3$$, because $$[b,t] \in Z_2/Z_1$$. So, since $$b^3 \in Z_2/Z_1$$, we have $$t^3=1$$.

But then $$b^{-1}a^{-1}b = ta^{-1}$$, so $$b^{-1}a^{-3}b = t^3 a^{-3} = a^{-3}$$, and so $$a^3 \in Z_2/Z_1$$, contrary to assumption, because $$a$$ has order 9 in $$G/Z_2$$.

So we have determined $$G/Z_2$$ up to isomorphism without assuming that $$G$$ is powerful.

A similar, but more complicated calculation, reveals that there is one other non-powerful (powerless?) possibility for $$G/Z_1$$ other than the one above, which is $$\langle a,b,t \mid a^{9} = b^{27} = 1, a^{-1}ba=b^4t, t^3=[a,t]=[b,t]=1 \rangle,$$ where the generator $$t$$ is redundant. (This is $$\mathtt{SmallGroup}(739,32)$$, and the other (correct) option above is $$\mathtt{SmallGroup}(739,22)$$.

A similar argument to the one above shows that that this other option does not extend further to a group $$G$$ of order $$2187$$ with the required properties.

Although it is helpful to check things by computer, this can all be done by hand, and I can add further details if necessary. Of course whether you really gain much additional insight by doing it by hand rather than computer is an interesting question!

• I think I just mislead you again. We are trying to prove that the assumptions implies $G$ must be powerful. – Siddhartha Sep 7 '20 at 16:02
• Some of those $Z_2$'s should be $Z_1$'s, right? – Noah Snyder Sep 7 '20 at 16:11
• @Derek: Just a quick cross check. Writing ${\overline{G}} := G/Z$ we have a group of class $3$ order $729$ with upper central series quotients all elementary abelian of order $9$. Then ${\overline{G}}/{Z({\overline{G}})}$ is capable of order $81$. It seems that there are only two choices, as you wrote: SmallGroup($81,3$) and ($81,4$) depending on $[b,a] = 1$ or $[b,a] = b^3$. The Epicentre of ($81,3$) seems to contain a central element. – Siddhartha Sep 7 '20 at 17:08
• @NoahSnyder Yes I have attempted to correct the indexes – Derek Holt Sep 7 '20 at 17:53
• @Siddhartha Yes, there are just two choices for $G/Z_2$ (of order 81) and I rejected $\mathtt{SmallGroup}(81,3)$ because it is not powerful. But since you don't me to assume that $G$ is powerful, I will have to reconsider that case. – Derek Holt Sep 7 '20 at 17:57