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On the nilpotency class of certain wreath products
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It is said in a paper of P. Hall that if you have a wreath product $S=Awr B$ where $A$ is a cyclic group of order $p^r$ and $B$ is a cyclic group of order $p^s$, then the nilpotency class of $S$ is as follows: $$rp^s-(r-1)p^s-1.$$ He says that this can be directly established, however he does not say how a part from the easy case of $s=1$.
Any ideas how to deduce it directly?
abstract-algebra group-theory
edited Jul 18 at 21:18
Bernard
110k635103
asked Jul 18 at 21:08
W4cc0
1,81811025
add a comment |Â
up vote
5
down vote
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It is said in a paper of P. Hall that if you have a wreath product $S=Awr B$ where $A$ is a cyclic group of order $p^r$ and $B$ is a cyclic group of order $p^s$, then the nilpotency class of $S$ is as follows: $$rp^s-(r-1)p^s-1.$$ He says that this can be directly established, however he does not say how a part from the easy case of $s=1$.
Any ideas how to deduce it directly?
abstract-algebra group-theory
edited Jul 18 at 21:18
Bernard
110k635103
asked Jul 18 at 21:08
W4cc0
1,81811025
Have you tried generalizing Halls argument to the case $s>1$? Is there a specific place you get stuck?
– Steve D
Jul 19 at 0:52
1
I think is not possible to generalize that argument: you can sure find that $G_i=langle G_i+1,, alpha_irangle$ but then you cannot deduce the exact nilpotency class.
– W4cc0
Jul 19 at 4:44
OK, I think I have a proof of this, but I won't be able to write the whole thing up until this weekend. The main idea is this: let $x$ be the generator of the cyclic group of order $p^r$, and $y$ the generator of the cyclic group of order $p^s$. Let $n$ be the number Hall gives. Then the commutator $[x,y,ldots,y]$ -- where there are $n-1$ $y$'s -- is non-trivial. However, if there are $n$ $y$'s, then it is trivial, and that is the jumping-off point to showing all commutators of length $n+1$ are trivial.
– Steve D
Jul 19 at 20:05
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
It is said in a paper of P. Hall that if you have a wreath product $S=Awr B$ where $A$ is a cyclic group of order $p^r$ and $B$ is a cyclic group of order $p^s$, then the nilpotency class of $S$ is as follows: $$rp^s-(r-1)p^s-1.$$ He says that this can be directly established, however he does not say how a part from the easy case of $s=1$.
Any ideas how to deduce it directly?
abstract-algebra group-theory
edited Jul 18 at 21:18
Bernard
110k635103
asked Jul 18 at 21:08
W4cc0
1,81811025
It is said in a paper of P. Hall that if you have a wreath product $S=Awr B$ where $A$ is a cyclic group of order $p^r$ and $B$ is a cyclic group of order $p^s$, then the nilpotency class of $S$ is as follows: $$rp^s-(r-1)p^s-1.$$ He says that this can be directly established, however he does not say how a part from the easy case of $s=1$.
Any ideas how to deduce it directly?
abstract-algebra group-theory
edited Jul 18 at 21:18
Bernard
110k635103
asked Jul 18 at 21:08
W4cc0
1,81811025
edited Jul 18 at 21:18
Bernard
110k635103
edited Jul 18 at 21:18
Bernard
110k635103
edited Jul 18 at 21:18
Bernard
110k635103
110k635103
asked Jul 18 at 21:08
W4cc0
1,81811025
asked Jul 18 at 21:08
W4cc0
1,81811025
asked Jul 18 at 21:08
W4cc0
1,81811025
1,81811025
Have you tried generalizing Halls argument to the case $s>1$? Is there a specific place you get stuck?
– Steve D
Jul 19 at 0:52
1
I think is not possible to generalize that argument: you can sure find that $G_i=langle G_i+1,, alpha_irangle$ but then you cannot deduce the exact nilpotency class.
– W4cc0
Jul 19 at 4:44
OK, I think I have a proof of this, but I won't be able to write the whole thing up until this weekend. The main idea is this: let $x$ be the generator of the cyclic group of order $p^r$, and $y$ the generator of the cyclic group of order $p^s$. Let $n$ be the number Hall gives. Then the commutator $[x,y,ldots,y]$ -- where there are $n-1$ $y$'s -- is non-trivial. However, if there are $n$ $y$'s, then it is trivial, and that is the jumping-off point to showing all commutators of length $n+1$ are trivial.
– Steve D
Jul 19 at 20:05
add a comment |Â
Have you tried generalizing Halls argument to the case $s>1$? Is there a specific place you get stuck?
– Steve D
Jul 19 at 0:52
1
I think is not possible to generalize that argument: you can sure find that $G_i=langle G_i+1,, alpha_irangle$ but then you cannot deduce the exact nilpotency class.
– W4cc0
Jul 19 at 4:44
OK, I think I have a proof of this, but I won't be able to write the whole thing up until this weekend. The main idea is this: let $x$ be the generator of the cyclic group of order $p^r$, and $y$ the generator of the cyclic group of order $p^s$. Let $n$ be the number Hall gives. Then the commutator $[x,y,ldots,y]$ -- where there are $n-1$ $y$'s -- is non-trivial. However, if there are $n$ $y$'s, then it is trivial, and that is the jumping-off point to showing all commutators of length $n+1$ are trivial.
– Steve D
Jul 19 at 20:05
Have you tried generalizing Halls argument to the case $s>1$? Is there a specific place you get stuck?
– Steve D
Jul 19 at 0:52
Have you tried generalizing Halls argument to the case $s>1$? Is there a specific place you get stuck?
– Steve D
Jul 19 at 0:52
1
1
I think is not possible to generalize that argument: you can sure find that $G_i=langle G_i+1,, alpha_irangle$ but then you cannot deduce the exact nilpotency class.
– W4cc0
Jul 19 at 4:44
I think is not possible to generalize that argument: you can sure find that $G_i=langle G_i+1,, alpha_irangle$ but then you cannot deduce the exact nilpotency class.
– W4cc0
Jul 19 at 4:44
OK, I think I have a proof of this, but I won't be able to write the whole thing up until this weekend. The main idea is this: let $x$ be the generator of the cyclic group of order $p^r$, and $y$ the generator of the cyclic group of order $p^s$. Let $n$ be the number Hall gives. Then the commutator $[x,y,ldots,y]$ -- where there are $n-1$ $y$'s -- is non-trivial. However, if there are $n$ $y$'s, then it is trivial, and that is the jumping-off point to showing all commutators of length $n+1$ are trivial.
– Steve D
Jul 19 at 20:05
OK, I think I have a proof of this, but I won't be able to write the whole thing up until this weekend. The main idea is this: let $x$ be the generator of the cyclic group of order $p^r$, and $y$ the generator of the cyclic group of order $p^s$. Let $n$ be the number Hall gives. Then the commutator $[x,y,ldots,y]$ -- where there are $n-1$ $y$'s -- is non-trivial. However, if there are $n$ $y$'s, then it is trivial, and that is the jumping-off point to showing all commutators of length $n+1$ are trivial.
– Steve D
Jul 19 at 20:05
add a comment |Â
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Let's write $G=Kltimes H$, where $K$ is a cyclic group of order $p^s$ generated by $y$, and $H$ is the direct product of $p^s$ copies of (cyclic group of order $p^r$). We'll call the generators of $H$ $x_0, ldots, x_p^s-1$, with $x_i^y=x_i+1$ and all subscripts are treated $modp^s$. Because $H$ is abelian, we'll use additive notation for group elements. Finally, we write
$$ z_k = [x_0, y, ldots, y] $$
where there are $k$ $y$'s in that commutator.
Expression for $z_k$. It's easy to prove by induction that
$$ z_k = (-1)^ksum_i=0^k(-1)^ibinomkix_i $$
where again the subscripts are treated $modp^s$.
Why we care about $z_k$. For $k>0$, we have $z_k=0iff G_k=1$. One direction is obvious. Because $G_k$ is generated by commutators of length $k+1$, it's enough to show these are always trivial if we want to show $G_k$ is trivial. We will now repeatedly narrow the commutators of length $k+1$ we actually care about.
- Because $K$ is cyclic and $H$ is abelian, we have $[h_1y^m, h_2y^t]=[h_1, y^t]^y^m$. Induction then shows this works for longer lengths, and thus we really care about showing commutators of the form $[h_1, y^m_1, ldots, y^m_k]$ are trivial.
- Because $G'le H$ is abelian, we have $[h_1h_2,y^t]=[h_1, y^t][h_2, y^t]$. Again, induction shows this works for all lengths. Thus we only care about showing commutators of the form $[x_i, y^m_1, ldots, y^m_k]$ are trivial.
- Since $x_i=y^-ix_0y^i$, we have
$$ [x_i, y^m_1, ldots, y^m_k]=[x_0, y^m_1, ldots, y^m_k]^y^i$$
So we can actually focus on showing commutators of the form $[x_0, y^m_1, ldots, y^m_k]$ are trivial. - We can write
$$ [x_0, y^m]=[x_0, y](y^-1[x_0, y]y)cdots(y^-(m-1)[x_0, y]y^(m-1)) $$
And more generally, the same is true with $x_0$ replaced with any element of $H$. Since $G'le H$ and $z_k=[z_k-1,y]$, induction again shows that $[x_0, y^m_1, ldots, y^m_k]$ can be written as a long product of $z_k$ and its conjugates. Thus to show $G_k$ is trivial, it is enough for $z_k$ to be trivial.
So to find the nilpotency class of $G$, we're looking for the smallest $k$ with $z_k=0$. Originally, I had thought I had a combinatorial proof that the expression above for $z_k$ was trivial precisely when $k$ was the expression given by Hall. My proof was seriously flawed, however. Nevertheless, I found a nice article by Liebeck, "Concerning nilpotent wreath products", which not only establishes Hall's claim, but proves a generalization. I'll try now to distill Liebeck's argument for your specific case. Perhaps one day an MSE user more versed in combinatorics than I will come along and provide a nice elementary proof of when $z_k$ is trivial.
Let $R$ be the ring $mathbbZ[x]/langle x^p^s-1,p^rrangle$, which is a polynomial ring with coefficients$modp^r$ and exponents$modp^s$. Then the coefficient of $x_i$ in $z_k$ is the coefficient of $x^i$ in $(x-1)^k$. So we need to find the smallest $k$ with $(x-1)^kequiv0$ in $R$.
We have $kle rp^s-(r-1)p^s-1$. For let $n=p^s-1$ and $q=(p-1)n$. Note Hall's number is $rq+n$. Working in the field $mathbbZ/pmathbbZ$, we have
beginalign*
(x-1)^n &= x^n-1\
(x-1)^q &= sum_j=0^p-1x^jn
endalign*
In $R$, these mean
beginalign*
(x-1)^n &= pf(x) + x^n-1\
(x-1)^q &= pg(x) + sum_j=0^p-1x^jn
endalign*
Let's call that sum $phi(x)$. Then $x^nphi(x)=phi(x)$ in $R$, and so $phi(x)^2=pphi(x)$. Induction shows $phi(x)^r=p^r-1phi(x)$.
Putting this all together, we have
beginalign*
(x-1)^rq+n &= left(pg(x)+phi(x)right)^rcdotleft(pf(x)+x^n-1right)\
&= p^r-1h(x)phi(x)(pf(x)+x^n-1)\
&= p^r-1h(x)phi(x)(x^n-1)\
&= 0
endalign*
where we're repeatedly using $p^r=0$ in $R$, and in the last line that $phi(x)(x^n-1)=0$.
We have $kge rp^s-(r-1)p^s-1$. This will complete the proof. We have to show that $(x-1)^rq+n-1$ is not zero in $R$. Liebeck's argument is kind of messy here; I'll come back later and try and clean it up.
answered Jul 22 at 3:38
Steve D
2,042419
add a comment |Â
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let's write $G=Kltimes H$, where $K$ is a cyclic group of order $p^s$ generated by $y$, and $H$ is the direct product of $p^s$ copies of (cyclic group of order $p^r$). We'll call the generators of $H$ $x_0, ldots, x_p^s-1$, with $x_i^y=x_i+1$ and all subscripts are treated $modp^s$. Because $H$ is abelian, we'll use additive notation for group elements. Finally, we write
$$ z_k = [x_0, y, ldots, y] $$
where there are $k$ $y$'s in that commutator.
Expression for $z_k$. It's easy to prove by induction that
$$ z_k = (-1)^ksum_i=0^k(-1)^ibinomkix_i $$
where again the subscripts are treated $modp^s$.
Why we care about $z_k$. For $k>0$, we have $z_k=0iff G_k=1$. One direction is obvious. Because $G_k$ is generated by commutators of length $k+1$, it's enough to show these are always trivial if we want to show $G_k$ is trivial. We will now repeatedly narrow the commutators of length $k+1$ we actually care about.
- Because $K$ is cyclic and $H$ is abelian, we have $[h_1y^m, h_2y^t]=[h_1, y^t]^y^m$. Induction then shows this works for longer lengths, and thus we really care about showing commutators of the form $[h_1, y^m_1, ldots, y^m_k]$ are trivial.
- Because $G'le H$ is abelian, we have $[h_1h_2,y^t]=[h_1, y^t][h_2, y^t]$. Again, induction shows this works for all lengths. Thus we only care about showing commutators of the form $[x_i, y^m_1, ldots, y^m_k]$ are trivial.
- Since $x_i=y^-ix_0y^i$, we have
$$ [x_i, y^m_1, ldots, y^m_k]=[x_0, y^m_1, ldots, y^m_k]^y^i$$
So we can actually focus on showing commutators of the form $[x_0, y^m_1, ldots, y^m_k]$ are trivial. - We can write
$$ [x_0, y^m]=[x_0, y](y^-1[x_0, y]y)cdots(y^-(m-1)[x_0, y]y^(m-1)) $$
And more generally, the same is true with $x_0$ replaced with any element of $H$. Since $G'le H$ and $z_k=[z_k-1,y]$, induction again shows that $[x_0, y^m_1, ldots, y^m_k]$ can be written as a long product of $z_k$ and its conjugates. Thus to show $G_k$ is trivial, it is enough for $z_k$ to be trivial.
So to find the nilpotency class of $G$, we're looking for the smallest $k$ with $z_k=0$. Originally, I had thought I had a combinatorial proof that the expression above for $z_k$ was trivial precisely when $k$ was the expression given by Hall. My proof was seriously flawed, however. Nevertheless, I found a nice article by Liebeck, "Concerning nilpotent wreath products", which not only establishes Hall's claim, but proves a generalization. I'll try now to distill Liebeck's argument for your specific case. Perhaps one day an MSE user more versed in combinatorics than I will come along and provide a nice elementary proof of when $z_k$ is trivial.
Let $R$ be the ring $mathbbZ[x]/langle x^p^s-1,p^rrangle$, which is a polynomial ring with coefficients$modp^r$ and exponents$modp^s$. Then the coefficient of $x_i$ in $z_k$ is the coefficient of $x^i$ in $(x-1)^k$. So we need to find the smallest $k$ with $(x-1)^kequiv0$ in $R$.
We have $kle rp^s-(r-1)p^s-1$. For let $n=p^s-1$ and $q=(p-1)n$. Note Hall's number is $rq+n$. Working in the field $mathbbZ/pmathbbZ$, we have
beginalign*
(x-1)^n &= x^n-1\
(x-1)^q &= sum_j=0^p-1x^jn
endalign*
In $R$, these mean
beginalign*
(x-1)^n &= pf(x) + x^n-1\
(x-1)^q &= pg(x) + sum_j=0^p-1x^jn
endalign*
Let's call that sum $phi(x)$. Then $x^nphi(x)=phi(x)$ in $R$, and so $phi(x)^2=pphi(x)$. Induction shows $phi(x)^r=p^r-1phi(x)$.
Putting this all together, we have
beginalign*
(x-1)^rq+n &= left(pg(x)+phi(x)right)^rcdotleft(pf(x)+x^n-1right)\
&= p^r-1h(x)phi(x)(pf(x)+x^n-1)\
&= p^r-1h(x)phi(x)(x^n-1)\
&= 0
endalign*
where we're repeatedly using $p^r=0$ in $R$, and in the last line that $phi(x)(x^n-1)=0$.
We have $kge rp^s-(r-1)p^s-1$. This will complete the proof. We have to show that $(x-1)^rq+n-1$ is not zero in $R$. Liebeck's argument is kind of messy here; I'll come back later and try and clean it up.
answered Jul 22 at 3:38
Steve D
2,042419
add a comment |Â
up vote
3
down vote
accepted
Let's write $G=Kltimes H$, where $K$ is a cyclic group of order $p^s$ generated by $y$, and $H$ is the direct product of $p^s$ copies of (cyclic group of order $p^r$). We'll call the generators of $H$ $x_0, ldots, x_p^s-1$, with $x_i^y=x_i+1$ and all subscripts are treated $modp^s$. Because $H$ is abelian, we'll use additive notation for group elements. Finally, we write
$$ z_k = [x_0, y, ldots, y] $$
where there are $k$ $y$'s in that commutator.
Expression for $z_k$. It's easy to prove by induction that
$$ z_k = (-1)^ksum_i=0^k(-1)^ibinomkix_i $$
where again the subscripts are treated $modp^s$.
Why we care about $z_k$. For $k>0$, we have $z_k=0iff G_k=1$. One direction is obvious. Because $G_k$ is generated by commutators of length $k+1$, it's enough to show these are always trivial if we want to show $G_k$ is trivial. We will now repeatedly narrow the commutators of length $k+1$ we actually care about.
- Because $K$ is cyclic and $H$ is abelian, we have $[h_1y^m, h_2y^t]=[h_1, y^t]^y^m$. Induction then shows this works for longer lengths, and thus we really care about showing commutators of the form $[h_1, y^m_1, ldots, y^m_k]$ are trivial.
- Because $G'le H$ is abelian, we have $[h_1h_2,y^t]=[h_1, y^t][h_2, y^t]$. Again, induction shows this works for all lengths. Thus we only care about showing commutators of the form $[x_i, y^m_1, ldots, y^m_k]$ are trivial.
- Since $x_i=y^-ix_0y^i$, we have
$$ [x_i, y^m_1, ldots, y^m_k]=[x_0, y^m_1, ldots, y^m_k]^y^i$$
So we can actually focus on showing commutators of the form $[x_0, y^m_1, ldots, y^m_k]$ are trivial. - We can write
$$ [x_0, y^m]=[x_0, y](y^-1[x_0, y]y)cdots(y^-(m-1)[x_0, y]y^(m-1)) $$
And more generally, the same is true with $x_0$ replaced with any element of $H$. Since $G'le H$ and $z_k=[z_k-1,y]$, induction again shows that $[x_0, y^m_1, ldots, y^m_k]$ can be written as a long product of $z_k$ and its conjugates. Thus to show $G_k$ is trivial, it is enough for $z_k$ to be trivial.
So to find the nilpotency class of $G$, we're looking for the smallest $k$ with $z_k=0$. Originally, I had thought I had a combinatorial proof that the expression above for $z_k$ was trivial precisely when $k$ was the expression given by Hall. My proof was seriously flawed, however. Nevertheless, I found a nice article by Liebeck, "Concerning nilpotent wreath products", which not only establishes Hall's claim, but proves a generalization. I'll try now to distill Liebeck's argument for your specific case. Perhaps one day an MSE user more versed in combinatorics than I will come along and provide a nice elementary proof of when $z_k$ is trivial.
Let $R$ be the ring $mathbbZ[x]/langle x^p^s-1,p^rrangle$, which is a polynomial ring with coefficients$modp^r$ and exponents$modp^s$. Then the coefficient of $x_i$ in $z_k$ is the coefficient of $x^i$ in $(x-1)^k$. So we need to find the smallest $k$ with $(x-1)^kequiv0$ in $R$.
We have $kle rp^s-(r-1)p^s-1$. For let $n=p^s-1$ and $q=(p-1)n$. Note Hall's number is $rq+n$. Working in the field $mathbbZ/pmathbbZ$, we have
beginalign*
(x-1)^n &= x^n-1\
(x-1)^q &= sum_j=0^p-1x^jn
endalign*
In $R$, these mean
beginalign*
(x-1)^n &= pf(x) + x^n-1\
(x-1)^q &= pg(x) + sum_j=0^p-1x^jn
endalign*
Let's call that sum $phi(x)$. Then $x^nphi(x)=phi(x)$ in $R$, and so $phi(x)^2=pphi(x)$. Induction shows $phi(x)^r=p^r-1phi(x)$.
Putting this all together, we have
beginalign*
(x-1)^rq+n &= left(pg(x)+phi(x)right)^rcdotleft(pf(x)+x^n-1right)\
&= p^r-1h(x)phi(x)(pf(x)+x^n-1)\
&= p^r-1h(x)phi(x)(x^n-1)\
&= 0
endalign*
where we're repeatedly using $p^r=0$ in $R$, and in the last line that $phi(x)(x^n-1)=0$.
We have $kge rp^s-(r-1)p^s-1$. This will complete the proof. We have to show that $(x-1)^rq+n-1$ is not zero in $R$. Liebeck's argument is kind of messy here; I'll come back later and try and clean it up.
answered Jul 22 at 3:38
Steve D
2,042419
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let's write $G=Kltimes H$, where $K$ is a cyclic group of order $p^s$ generated by $y$, and $H$ is the direct product of $p^s$ copies of (cyclic group of order $p^r$). We'll call the generators of $H$ $x_0, ldots, x_p^s-1$, with $x_i^y=x_i+1$ and all subscripts are treated $modp^s$. Because $H$ is abelian, we'll use additive notation for group elements. Finally, we write
$$ z_k = [x_0, y, ldots, y] $$
where there are $k$ $y$'s in that commutator.
Expression for $z_k$. It's easy to prove by induction that
$$ z_k = (-1)^ksum_i=0^k(-1)^ibinomkix_i $$
where again the subscripts are treated $modp^s$.
Why we care about $z_k$. For $k>0$, we have $z_k=0iff G_k=1$. One direction is obvious. Because $G_k$ is generated by commutators of length $k+1$, it's enough to show these are always trivial if we want to show $G_k$ is trivial. We will now repeatedly narrow the commutators of length $k+1$ we actually care about.
- Because $K$ is cyclic and $H$ is abelian, we have $[h_1y^m, h_2y^t]=[h_1, y^t]^y^m$. Induction then shows this works for longer lengths, and thus we really care about showing commutators of the form $[h_1, y^m_1, ldots, y^m_k]$ are trivial.
- Because $G'le H$ is abelian, we have $[h_1h_2,y^t]=[h_1, y^t][h_2, y^t]$. Again, induction shows this works for all lengths. Thus we only care about showing commutators of the form $[x_i, y^m_1, ldots, y^m_k]$ are trivial.
- Since $x_i=y^-ix_0y^i$, we have
$$ [x_i, y^m_1, ldots, y^m_k]=[x_0, y^m_1, ldots, y^m_k]^y^i$$
So we can actually focus on showing commutators of the form $[x_0, y^m_1, ldots, y^m_k]$ are trivial. - We can write
$$ [x_0, y^m]=[x_0, y](y^-1[x_0, y]y)cdots(y^-(m-1)[x_0, y]y^(m-1)) $$
And more generally, the same is true with $x_0$ replaced with any element of $H$. Since $G'le H$ and $z_k=[z_k-1,y]$, induction again shows that $[x_0, y^m_1, ldots, y^m_k]$ can be written as a long product of $z_k$ and its conjugates. Thus to show $G_k$ is trivial, it is enough for $z_k$ to be trivial.
So to find the nilpotency class of $G$, we're looking for the smallest $k$ with $z_k=0$. Originally, I had thought I had a combinatorial proof that the expression above for $z_k$ was trivial precisely when $k$ was the expression given by Hall. My proof was seriously flawed, however. Nevertheless, I found a nice article by Liebeck, "Concerning nilpotent wreath products", which not only establishes Hall's claim, but proves a generalization. I'll try now to distill Liebeck's argument for your specific case. Perhaps one day an MSE user more versed in combinatorics than I will come along and provide a nice elementary proof of when $z_k$ is trivial.
Let $R$ be the ring $mathbbZ[x]/langle x^p^s-1,p^rrangle$, which is a polynomial ring with coefficients$modp^r$ and exponents$modp^s$. Then the coefficient of $x_i$ in $z_k$ is the coefficient of $x^i$ in $(x-1)^k$. So we need to find the smallest $k$ with $(x-1)^kequiv0$ in $R$.
We have $kle rp^s-(r-1)p^s-1$. For let $n=p^s-1$ and $q=(p-1)n$. Note Hall's number is $rq+n$. Working in the field $mathbbZ/pmathbbZ$, we have
beginalign*
(x-1)^n &= x^n-1\
(x-1)^q &= sum_j=0^p-1x^jn
endalign*
In $R$, these mean
beginalign*
(x-1)^n &= pf(x) + x^n-1\
(x-1)^q &= pg(x) + sum_j=0^p-1x^jn
endalign*
Let's call that sum $phi(x)$. Then $x^nphi(x)=phi(x)$ in $R$, and so $phi(x)^2=pphi(x)$. Induction shows $phi(x)^r=p^r-1phi(x)$.
Putting this all together, we have
beginalign*
(x-1)^rq+n &= left(pg(x)+phi(x)right)^rcdotleft(pf(x)+x^n-1right)\
&= p^r-1h(x)phi(x)(pf(x)+x^n-1)\
&= p^r-1h(x)phi(x)(x^n-1)\
&= 0
endalign*
where we're repeatedly using $p^r=0$ in $R$, and in the last line that $phi(x)(x^n-1)=0$.
We have $kge rp^s-(r-1)p^s-1$. This will complete the proof. We have to show that $(x-1)^rq+n-1$ is not zero in $R$. Liebeck's argument is kind of messy here; I'll come back later and try and clean it up.
answered Jul 22 at 3:38
Steve D
2,042419
Let's write $G=Kltimes H$, where $K$ is a cyclic group of order $p^s$ generated by $y$, and $H$ is the direct product of $p^s$ copies of (cyclic group of order $p^r$). We'll call the generators of $H$ $x_0, ldots, x_p^s-1$, with $x_i^y=x_i+1$ and all subscripts are treated $modp^s$. Because $H$ is abelian, we'll use additive notation for group elements. Finally, we write
$$ z_k = [x_0, y, ldots, y] $$
where there are $k$ $y$'s in that commutator.
Expression for $z_k$. It's easy to prove by induction that
$$ z_k = (-1)^ksum_i=0^k(-1)^ibinomkix_i $$
where again the subscripts are treated $modp^s$.
Why we care about $z_k$. For $k>0$, we have $z_k=0iff G_k=1$. One direction is obvious. Because $G_k$ is generated by commutators of length $k+1$, it's enough to show these are always trivial if we want to show $G_k$ is trivial. We will now repeatedly narrow the commutators of length $k+1$ we actually care about.
- Because $K$ is cyclic and $H$ is abelian, we have $[h_1y^m, h_2y^t]=[h_1, y^t]^y^m$. Induction then shows this works for longer lengths, and thus we really care about showing commutators of the form $[h_1, y^m_1, ldots, y^m_k]$ are trivial.
- Because $G'le H$ is abelian, we have $[h_1h_2,y^t]=[h_1, y^t][h_2, y^t]$. Again, induction shows this works for all lengths. Thus we only care about showing commutators of the form $[x_i, y^m_1, ldots, y^m_k]$ are trivial.
- Since $x_i=y^-ix_0y^i$, we have
$$ [x_i, y^m_1, ldots, y^m_k]=[x_0, y^m_1, ldots, y^m_k]^y^i$$
So we can actually focus on showing commutators of the form $[x_0, y^m_1, ldots, y^m_k]$ are trivial. - We can write
$$ [x_0, y^m]=[x_0, y](y^-1[x_0, y]y)cdots(y^-(m-1)[x_0, y]y^(m-1)) $$
And more generally, the same is true with $x_0$ replaced with any element of $H$. Since $G'le H$ and $z_k=[z_k-1,y]$, induction again shows that $[x_0, y^m_1, ldots, y^m_k]$ can be written as a long product of $z_k$ and its conjugates. Thus to show $G_k$ is trivial, it is enough for $z_k$ to be trivial.
So to find the nilpotency class of $G$, we're looking for the smallest $k$ with $z_k=0$. Originally, I had thought I had a combinatorial proof that the expression above for $z_k$ was trivial precisely when $k$ was the expression given by Hall. My proof was seriously flawed, however. Nevertheless, I found a nice article by Liebeck, "Concerning nilpotent wreath products", which not only establishes Hall's claim, but proves a generalization. I'll try now to distill Liebeck's argument for your specific case. Perhaps one day an MSE user more versed in combinatorics than I will come along and provide a nice elementary proof of when $z_k$ is trivial.
Let $R$ be the ring $mathbbZ[x]/langle x^p^s-1,p^rrangle$, which is a polynomial ring with coefficients$modp^r$ and exponents$modp^s$. Then the coefficient of $x_i$ in $z_k$ is the coefficient of $x^i$ in $(x-1)^k$. So we need to find the smallest $k$ with $(x-1)^kequiv0$ in $R$.
We have $kle rp^s-(r-1)p^s-1$. For let $n=p^s-1$ and $q=(p-1)n$. Note Hall's number is $rq+n$. Working in the field $mathbbZ/pmathbbZ$, we have
beginalign*
(x-1)^n &= x^n-1\
(x-1)^q &= sum_j=0^p-1x^jn
endalign*
In $R$, these mean
beginalign*
(x-1)^n &= pf(x) + x^n-1\
(x-1)^q &= pg(x) + sum_j=0^p-1x^jn
endalign*
Let's call that sum $phi(x)$. Then $x^nphi(x)=phi(x)$ in $R$, and so $phi(x)^2=pphi(x)$. Induction shows $phi(x)^r=p^r-1phi(x)$.
Putting this all together, we have
beginalign*
(x-1)^rq+n &= left(pg(x)+phi(x)right)^rcdotleft(pf(x)+x^n-1right)\
&= p^r-1h(x)phi(x)(pf(x)+x^n-1)\
&= p^r-1h(x)phi(x)(x^n-1)\
&= 0
endalign*
where we're repeatedly using $p^r=0$ in $R$, and in the last line that $phi(x)(x^n-1)=0$.
We have $kge rp^s-(r-1)p^s-1$. This will complete the proof. We have to show that $(x-1)^rq+n-1$ is not zero in $R$. Liebeck's argument is kind of messy here; I'll come back later and try and clean it up.
answered Jul 22 at 3:38
Steve D
2,042419
answered Jul 22 at 3:38
Steve D
2,042419
answered Jul 22 at 3:38
Steve D
2,042419
answered Jul 22 at 3:38
Steve D
2,042419
2,042419
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Have you tried generalizing Halls argument to the case $s>1$? Is there a specific place you get stuck?
– Steve D
Jul 19 at 0:52
1
I think is not possible to generalize that argument: you can sure find that $G_i=langle G_i+1,, alpha_irangle$ but then you cannot deduce the exact nilpotency class.
– W4cc0
Jul 19 at 4:44
OK, I think I have a proof of this, but I won't be able to write the whole thing up until this weekend. The main idea is this: let $x$ be the generator of the cyclic group of order $p^r$, and $y$ the generator of the cyclic group of order $p^s$. Let $n$ be the number Hall gives. Then the commutator $[x,y,ldots,y]$ -- where there are $n-1$ $y$'s -- is non-trivial. However, if there are $n$ $y$'s, then it is trivial, and that is the jumping-off point to showing all commutators of length $n+1$ are trivial.
– Steve D
Jul 19 at 20:05