Let $G$ be a nilpotent group of class 3. Then for every $x ,y ,z$ in $G$, $[x,y,z][y,z,x][z,x,y]=1$. The 2019 Stack Overflow Developer Survey Results Are Innilpotent group of class 2Does every two-generated subgroups being nilpotent imply that the group itself is nilpotent?Show that this group is nilpotent.Can every torsion-free nilpotent group be ordered?When is every group of order $n$ nilpotent of class $leq c$?Nilpotency class of a nilpotent group.$G$ is nilpotent iff for any maximal subgroups $M$ and $N$ of $G$, $MN=NM$.Let $G$ be a finite group and $M,N lhd G$ such that $M leq Ncap Phi(G)$. Then $fracNM$ is nilpotent iff $N$ is nilpotent.Let $G$ be a polycyclic group and assume that every finite quotient of $G$ is nilpotent. Then $G$ is nilpotentLet $G$ be a $p$-group of nilpotency class at most 2, where $p$ is an odd prime. Then $ x in G$ is a subgroup of $G$.
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Let $G$ be a nilpotent group of class 3. Then for every $x ,y ,z$ in $G$, $[x,y,z][y,z,x][z,x,y]=1$.
The 2019 Stack Overflow Developer Survey Results Are Innilpotent group of class 2Does every two-generated subgroups being nilpotent imply that the group itself is nilpotent?Show that this group is nilpotent.Can every torsion-free nilpotent group be ordered?When is every group of order $n$ nilpotent of class $leq c$?Nilpotency class of a nilpotent group.$G$ is nilpotent iff for any maximal subgroups $M$ and $N$ of $G$, $MN=NM$.Let $G$ be a finite group and $M,N lhd G$ such that $M leq Ncap Phi(G)$. Then $fracNM$ is nilpotent iff $N$ is nilpotent.Let $G$ be a polycyclic group and assume that every finite quotient of $G$ is nilpotent. Then $G$ is nilpotentLet $G$ be a $p$-group of nilpotency class at most 2, where $p$ is an odd prime. Then $ x in G$ is a subgroup of $G$.
$begingroup$
Let $G$ be a nilpotent group of class 3. Then for every $x ,y ,z$ in $G$, $[x,y,z][y,z,x][z,x,y]=1$.
As $G$ is a nilpotent group of class 3, $[G,G,G,G]=1$ and $G^'$ is abelian.
I want to use Hall-Witt alliance but I don't know how to choose the elements.
Hall-Witt alliance : $[x,y^-1,z]^y[y,z^-1,x]^z[z,x^-1,y]^x=1$
We have $[x,y,z,y^-1] =1$. So $[x,y,z]=[x,y,z]^y^-1$. I don't know how to continue.
group-theory finite-groups nilpotent-groups
asked Oct 31 '18 at 7:50
YasminYasmin
303111
$endgroup$
|
show 10 more comments
$begingroup$
Let $G$ be a nilpotent group of class 3. Then for every $x ,y ,z$ in $G$, $[x,y,z][y,z,x][z,x,y]=1$.
As $G$ is a nilpotent group of class 3, $[G,G,G,G]=1$ and $G^'$ is abelian.
I want to use Hall-Witt alliance but I don't know how to choose the elements.
Hall-Witt alliance : $[x,y^-1,z]^y[y,z^-1,x]^z[z,x^-1,y]^x=1$
We have $[x,y,z,y^-1] =1$. So $[x,y,z]=[x,y,z]^y^-1$. I don't know how to continue.
group-theory finite-groups nilpotent-groups
asked Oct 31 '18 at 7:50
YasminYasmin
303111
$endgroup$
1
$begingroup$
You need to prove that $[x,y^-1,z] = [x,y,z]^-1$. More generally, in a group of class $3$, triple commutators $[x,y,z]$ are linear in all three variables i.e. $[wx,y,z] = [w,y,z][x,y,z]$, etc.
$endgroup$
– Derek Holt
Oct 31 '18 at 8:02
1
$begingroup$
This holds in any metabelian group, so in particular in a nilpotent group of class $3$, see Lemma 6.
$endgroup$
– Dietrich Burde
Oct 31 '18 at 9:59
1
$begingroup$
There are at least 4 conventions for the commutator ($[a,b]=aba^-1b^-1$, $bab^-1a^-1$, $a^-1b^-1ab$, $b^-1a^-1ba$), and for each of them there are two conventions for the double commutator: $[[a,b],c]$ and $[a,[b,c]]$. So it's always good to recall which definition you choose (even if if probably the given equality hold for each of these 8 conventions).
$endgroup$
– YCor
Nov 1 '18 at 22:26
1
$begingroup$
With the convention $[a,b]=a^-1b^-1ab$, there's a Hall identity $[a^b,[b,c]]cdot[b^c,[c,a]]cdot[c^a,[a,b]]$, which directly entails the desired equality in any metabelian group.
$endgroup$
– YCor
Nov 1 '18 at 23:19
1
$begingroup$
OK, so the Hall identity writes as $[a,b,c^a]cdot [c,a,b^c]cdot [b,c,a^b]=1$ (in any group).
$endgroup$
– YCor
Nov 2 '18 at 8:25
|
show 10 more comments
$begingroup$
Let $G$ be a nilpotent group of class 3. Then for every $x ,y ,z$ in $G$, $[x,y,z][y,z,x][z,x,y]=1$.
As $G$ is a nilpotent group of class 3, $[G,G,G,G]=1$ and $G^'$ is abelian.
I want to use Hall-Witt alliance but I don't know how to choose the elements.
Hall-Witt alliance : $[x,y^-1,z]^y[y,z^-1,x]^z[z,x^-1,y]^x=1$
We have $[x,y,z,y^-1] =1$. So $[x,y,z]=[x,y,z]^y^-1$. I don't know how to continue.
group-theory finite-groups nilpotent-groups
asked Oct 31 '18 at 7:50
YasminYasmin
303111
$endgroup$
Let $G$ be a nilpotent group of class 3. Then for every $x ,y ,z$ in $G$, $[x,y,z][y,z,x][z,x,y]=1$.
As $G$ is a nilpotent group of class 3, $[G,G,G,G]=1$ and $G^'$ is abelian.
I want to use Hall-Witt alliance but I don't know how to choose the elements.
Hall-Witt alliance : $[x,y^-1,z]^y[y,z^-1,x]^z[z,x^-1,y]^x=1$
We have $[x,y,z,y^-1] =1$. So $[x,y,z]=[x,y,z]^y^-1$. I don't know how to continue.
group-theory finite-groups nilpotent-groups
group-theory finite-groups nilpotent-groups
asked Oct 31 '18 at 7:50
YasminYasmin
303111
asked Oct 31 '18 at 7:50
YasminYasmin
303111
edited Mar 23 at 5:36
Yasmin
asked Oct 31 '18 at 7:50
YasminYasmin
303111
asked Oct 31 '18 at 7:50
YasminYasmin
303111
asked Oct 31 '18 at 7:50
YasminYasmin
303111
303111
1
$begingroup$
You need to prove that $[x,y^-1,z] = [x,y,z]^-1$. More generally, in a group of class $3$, triple commutators $[x,y,z]$ are linear in all three variables i.e. $[wx,y,z] = [w,y,z][x,y,z]$, etc.
$endgroup$
– Derek Holt
Oct 31 '18 at 8:02
1
$begingroup$
This holds in any metabelian group, so in particular in a nilpotent group of class $3$, see Lemma 6.
$endgroup$
– Dietrich Burde
Oct 31 '18 at 9:59
1
$begingroup$
There are at least 4 conventions for the commutator ($[a,b]=aba^-1b^-1$, $bab^-1a^-1$, $a^-1b^-1ab$, $b^-1a^-1ba$), and for each of them there are two conventions for the double commutator: $[[a,b],c]$ and $[a,[b,c]]$. So it's always good to recall which definition you choose (even if if probably the given equality hold for each of these 8 conventions).
$endgroup$
– YCor
Nov 1 '18 at 22:26
1
$begingroup$
With the convention $[a,b]=a^-1b^-1ab$, there's a Hall identity $[a^b,[b,c]]cdot[b^c,[c,a]]cdot[c^a,[a,b]]$, which directly entails the desired equality in any metabelian group.
$endgroup$
– YCor
Nov 1 '18 at 23:19
1
$begingroup$
OK, so the Hall identity writes as $[a,b,c^a]cdot [c,a,b^c]cdot [b,c,a^b]=1$ (in any group).
$endgroup$
– YCor
Nov 2 '18 at 8:25
|
show 10 more comments
1
$begingroup$
You need to prove that $[x,y^-1,z] = [x,y,z]^-1$. More generally, in a group of class $3$, triple commutators $[x,y,z]$ are linear in all three variables i.e. $[wx,y,z] = [w,y,z][x,y,z]$, etc.
$endgroup$
– Derek Holt
Oct 31 '18 at 8:02
1
$begingroup$
This holds in any metabelian group, so in particular in a nilpotent group of class $3$, see Lemma 6.
$endgroup$
– Dietrich Burde
Oct 31 '18 at 9:59
1
$begingroup$
There are at least 4 conventions for the commutator ($[a,b]=aba^-1b^-1$, $bab^-1a^-1$, $a^-1b^-1ab$, $b^-1a^-1ba$), and for each of them there are two conventions for the double commutator: $[[a,b],c]$ and $[a,[b,c]]$. So it's always good to recall which definition you choose (even if if probably the given equality hold for each of these 8 conventions).
$endgroup$
– YCor
Nov 1 '18 at 22:26
1
$begingroup$
With the convention $[a,b]=a^-1b^-1ab$, there's a Hall identity $[a^b,[b,c]]cdot[b^c,[c,a]]cdot[c^a,[a,b]]$, which directly entails the desired equality in any metabelian group.
$endgroup$
– YCor
Nov 1 '18 at 23:19
1
$begingroup$
OK, so the Hall identity writes as $[a,b,c^a]cdot [c,a,b^c]cdot [b,c,a^b]=1$ (in any group).
$endgroup$
– YCor
Nov 2 '18 at 8:25
1
1
$begingroup$
You need to prove that $[x,y^-1,z] = [x,y,z]^-1$. More generally, in a group of class $3$, triple commutators $[x,y,z]$ are linear in all three variables i.e. $[wx,y,z] = [w,y,z][x,y,z]$, etc.
$endgroup$
– Derek Holt
Oct 31 '18 at 8:02
$begingroup$
You need to prove that $[x,y^-1,z] = [x,y,z]^-1$. More generally, in a group of class $3$, triple commutators $[x,y,z]$ are linear in all three variables i.e. $[wx,y,z] = [w,y,z][x,y,z]$, etc.
$endgroup$
– Derek Holt
Oct 31 '18 at 8:02
1
1
$begingroup$
This holds in any metabelian group, so in particular in a nilpotent group of class $3$, see Lemma 6.
$endgroup$
– Dietrich Burde
Oct 31 '18 at 9:59
$begingroup$
This holds in any metabelian group, so in particular in a nilpotent group of class $3$, see Lemma 6.
$endgroup$
– Dietrich Burde
Oct 31 '18 at 9:59
1
1
$begingroup$
There are at least 4 conventions for the commutator ($[a,b]=aba^-1b^-1$, $bab^-1a^-1$, $a^-1b^-1ab$, $b^-1a^-1ba$), and for each of them there are two conventions for the double commutator: $[[a,b],c]$ and $[a,[b,c]]$. So it's always good to recall which definition you choose (even if if probably the given equality hold for each of these 8 conventions).
$endgroup$
– YCor
Nov 1 '18 at 22:26
$begingroup$
There are at least 4 conventions for the commutator ($[a,b]=aba^-1b^-1$, $bab^-1a^-1$, $a^-1b^-1ab$, $b^-1a^-1ba$), and for each of them there are two conventions for the double commutator: $[[a,b],c]$ and $[a,[b,c]]$. So it's always good to recall which definition you choose (even if if probably the given equality hold for each of these 8 conventions).
$endgroup$
– YCor
Nov 1 '18 at 22:26
1
1
$begingroup$
With the convention $[a,b]=a^-1b^-1ab$, there's a Hall identity $[a^b,[b,c]]cdot[b^c,[c,a]]cdot[c^a,[a,b]]$, which directly entails the desired equality in any metabelian group.
$endgroup$
– YCor
Nov 1 '18 at 23:19
$begingroup$
With the convention $[a,b]=a^-1b^-1ab$, there's a Hall identity $[a^b,[b,c]]cdot[b^c,[c,a]]cdot[c^a,[a,b]]$, which directly entails the desired equality in any metabelian group.
$endgroup$
– YCor
Nov 1 '18 at 23:19
1
1
$begingroup$
OK, so the Hall identity writes as $[a,b,c^a]cdot [c,a,b^c]cdot [b,c,a^b]=1$ (in any group).
$endgroup$
– YCor
Nov 2 '18 at 8:25
$begingroup$
OK, so the Hall identity writes as $[a,b,c^a]cdot [c,a,b^c]cdot [b,c,a^b]=1$ (in any group).
$endgroup$
– YCor
Nov 2 '18 at 8:25
|
show 10 more comments
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$begingroup$
You need to prove that $[x,y^-1,z] = [x,y,z]^-1$. More generally, in a group of class $3$, triple commutators $[x,y,z]$ are linear in all three variables i.e. $[wx,y,z] = [w,y,z][x,y,z]$, etc.
$endgroup$
– Derek Holt
Oct 31 '18 at 8:02
1
$begingroup$
This holds in any metabelian group, so in particular in a nilpotent group of class $3$, see Lemma 6.
$endgroup$
– Dietrich Burde
Oct 31 '18 at 9:59
1
$begingroup$
There are at least 4 conventions for the commutator ($[a,b]=aba^-1b^-1$, $bab^-1a^-1$, $a^-1b^-1ab$, $b^-1a^-1ba$), and for each of them there are two conventions for the double commutator: $[[a,b],c]$ and $[a,[b,c]]$. So it's always good to recall which definition you choose (even if if probably the given equality hold for each of these 8 conventions).
$endgroup$
– YCor
Nov 1 '18 at 22:26
1
$begingroup$
With the convention $[a,b]=a^-1b^-1ab$, there's a Hall identity $[a^b,[b,c]]cdot[b^c,[c,a]]cdot[c^a,[a,b]]$, which directly entails the desired equality in any metabelian group.
$endgroup$
– YCor
Nov 1 '18 at 23:19
1
$begingroup$
OK, so the Hall identity writes as $[a,b,c^a]cdot [c,a,b^c]cdot [b,c,a^b]=1$ (in any group).
$endgroup$
– YCor
Nov 2 '18 at 8:25