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!).



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


