Algebra 5.14 has \begin{eqnarray*}
&&2p^{5}+7p^{4}+19p^{3}+49p^{2}+128p+256+(p^{2}+7p+29)\gcd (p-1,3) \\
&&\;\;\;+(p^{2}+7p+24)\gcd (p-1,4)+(p+3)\gcd (p-1,5) \end{eqnarray*}%
immediate descendants of order $p^{7}$ and $p$-class $3$.
Algebra 5.14 has presentation \[ \langle a,b,c\,|\,cb,\,pa,\,pb,\,pc,\,\text{class }2\rangle , \]
and so if $L$ is an immediate descendant of 5.14 of order $p^7$ then $L_2$
is generated by $ba$, $ca$ modulo $L_3$, and $L_3$ has order $p^2$ and is
generated by $baa$, $bab$, $bac$, $caa$, $cab$. And $cb$, $pa$, $pb$, $pc\in
L_3$. The commutator structure is the same as one of 7.65 -- 7.88 from the
list of nilpotent Lie algebras over $\mathbb{Z}_p$. So we may assume that
one of the following holds:
Since the total number of descendants of 5.14 of order $p^{7}$ is of order $%
2p^{5}$, we need presentations with at least 5 parameters in some of these
cases. In each case the commutator structure is determined, and so to give a
presentation for the Lie rings we only need to specify $pa,pb,pc$. These
powers take values in $L_{3}$, which has order $p^{2}$, so we need 2
coefficients for each of $pa,pb,pc$. For the sake of simplicity I give a
single presentation with 6 parameters for each of the 24 different
commutator structures defined above, and I give the conditions for two sets
of parameters to define isomorphic Lie rings. In most of the cases I was
able to \textquotedblleft solve\textquotedblright\ the isomorphism problem
in the sense that I was able to produce a number of presentations with fewer
parameters, and with simple conditions on the parameters. But I was not able
to do this in every case.
The file notes5.14.m gives \textsc{Magma} programs to compute each
isomorphism class. The programs have complexity at most $p^{5}$, in the
sense that they have nested loops and the statements in the innermost loops
are executed a maximum of $O(p^{5})$ times. The programs run reasonably fast
for $p<20$, but you need to take a deep breath before running them for $p>20$%
. Apart from anything else the shear number of groups becomes overwhelming
pretty quickly. My classification of the nilpotent Lie rings of order $p^{7}$
has been criticized on the grounds that the Lie rings for any given prime
have to be computed \textquotedblleft on the fly\textquotedblright .
However, as I observed above, you need some presentations with at least 5
parameters, and even if you had five parameters independently taking all
values between $0$ and $p-1$ you would still need a program of complexity $%
p^{5}$ to print them out.
Here $L_{3}$ is generated by $baa$ and $bab$, and if we let \[ \left( \begin{array}{l}
pa \\
pb \\
pc% \end{array}% \right) =A\left( \begin{array}{l}
baa \\
bab% \end{array}% \right) \]%
then the isomorphism classes of algebras satisfying these commutator
relations correspond to the orbits of $3\times 2$ matrices $A$ under the
action \[
A\rightarrow\left( \begin{array}{lll} \alpha & \beta & \gamma\\
0 & \lambda & \mu\\
0 & 0 & \xi% \end{array}% \right) A\left( \begin{array}{ll} \alpha ^{2}\lambda & \alpha\beta\lambda\\
0 & \alpha\lambda ^{2}% \end{array}% \right) ^{-1}. \]
This case is identical to Case 11, so again there are \[ \allowbreak (p^4+p^3+4p^2+p-1+\allowbreak (p^2+2p+3)\gcd (p-1,3)+(p+2)\gcd
(p-1,4))/2 \]
algebras here.
The total number of algebras in this case is \[ \allowbreak p^{3}+\frac{7}{2}p^{2}+\frac{17}{2}p+\frac{59}{2}+\frac{5}{2}% \gcd (p-1,3)+\frac{p+1}{2}\gcd (p-1,4) \]
\subsection{Case 16}
\[ \langle
a,b,c\,|% \,cb,bac,caa,cac-baa,pa-x_{1}baa-x_{2}bab,pb-x_{3}baa-x_{4}bab,pc-x_{5}baa-x_{6}bab\rangle . \]%
$L_{3}$ is generated by $baa$ and $bab$ and if we let \[ \left( \begin{array}{l}
pa \\
pb \\
pc% \end{array}% \right) =A\left( \begin{array}{l}
baa \\
bab% \end{array}% \right) \]%
then the isomorphism classes of algebras satisfying these commutator
relations correspond to the orbits of $3\times 2$ matrices $A$ under the
action \[
A\rightarrow\left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \alpha ^{-1}\gamma ^{2} & 0 \\
0 & 0 & \gamma% \end{array}% \right) A\left( \begin{array}{ll} \alpha\gamma ^{2} & 0 \\
0 & \alpha ^{-1}\gamma ^{4}% \end{array}% \right) ^{-1}. \]
So in general the number of orbits is \[ \allowbreak (p^4+2p^3+3p^2+4p+2)\frac{p-1}{\gcd (p-1,3)}+3p+4+(p^2+p+1)\gcd
(p-1,4)/2 \]
\subsection{Case 18}
\[ \langle a,b,c\,|\,cb,bac,caa-\omega
bab,cac-baa,pa-x_{1}baa-x_{2}bab,pb-x_{3}baa-x_{4}bab,pc-x_{5}baa-x_{6}bab% \rangle\;(p=1\func{mod}3). \]%
This case is very similar to Case 17, though we do not have as many
automorphisms. $L_{3}$ is generated by $baa$ and $bab$ and if we let \[ \left( \begin{array}{l}
pa \\
pb \\
pc% \end{array}% \right) =A\left( \begin{array}{l}
baa \\
bab% \end{array}% \right) \]%
then the isomorphism classes of algebras satisfying these commutator
relations correspond to the orbits of $3\times 2$ matrices $A$ under the
action \[
A\rightarrow\left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \alpha ^{-1}\gamma ^{2} & 0 \\
0 & 0 & \gamma% \end{array}% \right) A\left( \begin{array}{ll} \alpha\gamma ^{2} & 0 \\
0 & \alpha ^{2}\gamma% \end{array}% \right) ^{-1} \]%
with $\alpha ^{3}=\gamma ^{3}$. \[ \left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \alpha ^{-1}\gamma ^{2} & 0 \\
0 & 0 & \gamma% \end{array}% \right) \left( \begin{array}{ll}
x & y \\
z & t \\
u & v% \end{array}% \right) \left( \begin{array}{ll} \alpha\gamma ^{2} & 0 \\
0 & \alpha ^{2}\gamma% \end{array}% \right) ^{-1}=\allowbreak\left( \begin{array}{cc} \frac{x}{\gamma ^{2}} & \frac{1}{\alpha }\frac{y}{\gamma } \\ \frac{1}{\alpha ^{2}}z & \frac{1}{\alpha ^{3}}\gamma t \\ \frac{1}{\gamma }\frac{u}{\alpha } & \frac{v}{\alpha ^{2}}% \end{array}% \right) \]
The number of algebras is \[
(2p^5+2p^4+2p^3+2p^2+14p+17)/3 \]
Combining Case 17 and Case 18, the total number of algebras in the two cases
is \[ \allowbreak p^5+p^4+p^3+p^2-2p-\frac 32+(3p+\frac 72)\gcd
(p-1,3)+(p^2+p+1)\gcd (p-1,4)/2 \]
\subsection{Case 19}
\[ \langle
a,b,c\,|% \,cb,baa,caa,cac,pa-x_{1}bab-x_{2}bac,pb-x_{3}bab-x_{4}bac,pc-x_{5}bab-x_{6}bac\rangle . \]%
$L_{3}$ is generated by $bab$ and $bac$ and if we let \[ \left( \begin{array}{l}
pa \\
pb \\
pc% \end{array}% \right) =A\left( \begin{array}{l}
bab \\
bac% \end{array}% \right) \]%
then the isomorphism classes of algebras satisfying these commutator
relations correspond to the orbits of $3\times 2$ matrices $A$ under the
action \[
A\rightarrow\left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \beta & \gamma\\
0 & 0 & \delta% \end{array}% \right) A\left( \begin{array}{ll} \alpha\beta ^{2} & 2\alpha\beta\gamma\\
0 & \alpha\beta\delta% \end{array}% \right) ^{-1} \]
The total number of algebras in this case is \[
2p^{3}+6p^{2}+7p+7+(p+1)\gcd (p-1,4). \]
\subsection{Case 22}
\[ \langle a,b,c\,|\,cb,baa,caa,cac-\omega
bab,pa-x_{1}bab-x_{2}bac,pb-x_{3}bab-x_{4}bac,pc-x_{5}bab-x_{6}bac\rangle . \]%
$L_{3}$ is generated by $bab$ and $bac$ and if we let \[ \left( \begin{array}{l}
pa \\
pb \\
pc% \end{array}% \right) =A\left( \begin{array}{l}
bab \\
bac% \end{array}% \right) \]%
then the isomorphism classes of algebras satisfying these commutator
relations correspond to the orbits of $3\times 2$ matrices $A$ under the
action \[
A\rightarrow\left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \omega\beta & \pm\gamma\\
0 & \omega\gamma & \pm\omega\beta% \end{array}% \right) A\left( \begin{array}{ll} \omega\alpha (\omega\beta ^{2}+\gamma ^{2}) & \pm 2\omega\alpha\beta \gamma\\
2\omega ^{2}\alpha\beta\gamma & \pm\omega\alpha (\omega\beta
^{2}+\gamma ^{2})% \end{array}% \right) ^{-1}. \]
The total number of algebras in Case 22 is \[
(2p^{3}+3p^{2}+3p+13-\gcd (p-1,3)+(p+1)\gcd (p-1,4))/2. \]
\subsection{Case 23}
\[ \langle a,b,c\,|\,cb,baa,caa-bac,cac-\omega
bab,pa-x_{1}bab-x_{2}bac,pb-x_{3}bab-x_{4}bac,pc-x_{5}bab-x_{6}bac\rangle . \]%
$L_{3}$ is generated by $bab$ and $bac$ and if we let \[ \left( \begin{array}{l}
pa \\
pb \\
pc% \end{array}% \right) =A\left( \begin{array}{l}
bab \\
bac% \end{array}% \right) \]%
then the isomorphism classes of algebras satisfying these commutator
relations correspond to the orbits of $3\times 2$ matrices $A$ under the
action \[
A\rightarrow\left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \alpha & 0 \\
0 & 0 & \pm\alpha% \end{array}% \right) A\left( \begin{array}{ll} \alpha ^{3} & 0 \\
0 & \pm\alpha ^{3}% \end{array}% \right) ^{-1} \]%
or when $p=2\func{mod}3$ and $12\omega\beta ^{2}=-1$, \[
A\rightarrow\left( \begin{array}{lll}
4\omega\alpha\beta & -3\omega\alpha\beta & \frac{\alpha }{2} \\
0 & -2\omega\alpha\beta & \alpha\\
0 & \pm\omega\alpha & \mp 2\omega\alpha\beta% \end{array}% \right) A\left( \begin{array}{ll} \frac{8}{3}\omega ^{2}\alpha ^{3}\beta & \frac{4}{3}\omega\alpha ^{3} \\ \pm\frac{4}{3}\omega ^{2}\alpha ^{3} & \pm\frac{8}{3}\omega ^{2}\alpha
^{3}\beta% \end{array}% \right) ^{-1}. \]
Now \[ \left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \alpha & 0 \\
0 & 0 & \pm\alpha% \end{array} \right) \left( \begin{array}{ll}
x & y \\
z & t \\
u & v% \end{array} \right) \left( \begin{array}{ll} \alpha ^3 & 0 \\
0 & \pm\alpha ^3% \end{array} \right) ^{-1}=\allowbreak\left( \begin{array}{cc} \frac 1{\alpha ^2}x & \pm\frac 1{\alpha ^2}y \\ \frac 1{\alpha ^2}z & \pm\frac 1{\alpha ^2}t \\ \pm\frac 1{\alpha ^2}u & \frac 1{\alpha ^2}v% \end{array} \right) \]
and so if $p=1\func{mod}3$ there are $p^5+p^4+p^3+p^2+p+2+(p^2+p+1)\gcd
(p-1,4)/2$ algebras.
When $p=2\func{mod}3$ the number of algebras here is \[ \allowbreak\frac{1}{3}p^{5}+\frac{1}{3}p^{4}+\frac{1}{3}p^{3}+\frac{1}{3}%
p^{2}+p+2+(p^{2}+p+1)\gcd (p-1,4)/2. \]
\subsection{Case 24}
\[ \langle a,b,c\,|\,cb,baa,caa-kbab-bac,cac-\omega
bab,pa-x_{1}bab-x_{2}bac,pb-x_{3}bab-x_{4}bac,pc-x_{5}bab-x_{6}bac\rangle \;(p=2\func{mod}3). \]%
where $k$ is any (fixed) integer which is not a value of \[ \frac{\lambda (\lambda ^{2}+3\omega\mu ^{2})}{\mu (3\lambda ^{2}+\omega\mu
^{2})}\func{mod}p. \]%
$L_{3}$ is generated by $bab$ and $bac$ and if we let \[ \left( \begin{array}{l}
pa \\
pb \\
pc% \end{array}% \right) =A\left( \begin{array}{l}
bab \\
bac% \end{array}% \right) \]%
then the isomorphism classes of algebras satisfying these commutator
relations correspond to the orbits of $3\times 2$ matrices $A$ under the
action \[
A\rightarrow\left( \begin{array}{lll} \alpha & 0 & 0 \\
0 & \alpha & 0 \\
0 & 0 & \alpha% \end{array}% \right) A\left( \begin{array}{ll} \alpha ^{3} & 0 \\
0 & \alpha ^{3}% \end{array}% \right) ^{-1} \]%
and \[
A\rightarrow\left( \begin{array}{lll}
-4\alpha & k\alpha\beta +3\alpha & 3k\omega ^{-1}\alpha +\alpha\beta\\
0 & 2\alpha & 2\alpha\beta\\
0 & 2\omega\alpha\beta & 2\alpha% \end{array}% \right) A\left( \begin{array}{ll}
32\alpha ^{3} & -32\alpha ^{3}\beta\\
-32\omega\alpha ^{3}\beta & 32\alpha ^{3}% \end{array}% \right) ^{-1} \]%
where $\omega\beta ^{2}=-3$.
The number of orbits is \[ \allowbreak\frac{2}{3}p^{5}+\frac{2}{3}p^{4}+\frac{2}{3}p^{3}+\frac{2}{3}%
p^{2}+2p+3. \]
The total number of algebras from Case 23 and Case 24 is \[
p^{5}+p^{4}+p^{3}+p^{2}+4p+\frac{13}{2}-(p+\frac{3}{2})\gcd
(p-1,3)+(p^{2}+p+1)\gcd (p-1,4)/2. \]
The total number of algebras from cases 17, 18, 23 and 24 is \[
p^5+p^4+p^3+p^2+2p+5+(2p+2)\gcd (p-1,3)+(p^2+p+1)\gcd (p-1,4). \]
\end{document}
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