Mixing
$G$ with a central series that is different from the upper and the lower central series
$begingroup$
Let
$$1=G_0leq G_1leq ... leq G_{n-1} leq G_n = G$$
be a central series of the group $G$. That is, $G_{i-1}/G_ileq Z(G/G_i)$ for all $i$.
Let
$$1=Z_0(G)leq Z_1(G)leq ... $$
$$... leq gamma_2(G)leq gamma_1(G)=G$$
be the upper central series and the lower central series of a group $G$, respectively. I have proved $$G_ileq Z_i(G), gamma_{i+1}(G)leq G_{n-i}$$
I was wondering if there is any example of a group such that has a central series which is different from the upper and the lower central series.
Also I would like to prove that for a nilpotent group of nilpotency class $c$, $$gamma_{c+1-i}(G)leq Z_i(G)$$ but I do not see it. (For this, I have been thinking on using that a nilpotent group of class $c$ satisfies $$Z_c(G)=G,gamma_{c+1}(G)=1$$ would that be useful?)
Any help?
group-theory nilpotent-groups
asked Jan 20 at 17:07
idriskameniidriskameni
753321
$endgroup$
|
show 2 more comments
$begingroup$
Let
$$1=G_0leq G_1leq ... leq G_{n-1} leq G_n = G$$
be a central series of the group $G$. That is, $G_{i-1}/G_ileq Z(G/G_i)$ for all $i$.
Let
$$1=Z_0(G)leq Z_1(G)leq ... $$
$$... leq gamma_2(G)leq gamma_1(G)=G$$
be the upper central series and the lower central series of a group $G$, respectively. I have proved $$G_ileq Z_i(G), gamma_{i+1}(G)leq G_{n-i}$$
I was wondering if there is any example of a group such that has a central series which is different from the upper and the lower central series.
Also I would like to prove that for a nilpotent group of nilpotency class $c$, $$gamma_{c+1-i}(G)leq Z_i(G)$$ but I do not see it. (For this, I have been thinking on using that a nilpotent group of class $c$ satisfies $$Z_c(G)=G,gamma_{c+1}(G)=1$$ would that be useful?)
Any help?
group-theory nilpotent-groups
asked Jan 20 at 17:07
idriskameniidriskameni
753321
$endgroup$
$begingroup$
Probably the reason that nobody has answered is that it is unclear exactly what you are asking. You ask for an example in which "these three series" are all different. Which three series exactly? We have the upper central series, and the lower central series, but what exactly is your third series? You could ask for an example with at least three distinct central series - that would make sense.
$endgroup$
– Derek Holt
Jan 23 at 8:54
$begingroup$
I think that it is super clear. Let (---) be the central series (1), the upper central series (2) and the lower central series (3) of a group $G$.
$endgroup$
– idriskameni
Jan 23 at 9:01
$begingroup$
I agree that (2) and (3) are clear, but what do you mean by "the central series (1)"?
$endgroup$
– Derek Holt
Jan 23 at 9:13
$begingroup$
The definition of central series of a group is clear I think. Anyway, I have updated the post.
$endgroup$
– idriskameni
Jan 23 at 9:27
2
$begingroup$
Yes, the definition of a central series of a group is clear. But the group may have many different central series (or it may have none at all). So it makes no sense to write "the central series". How would anyone know which of the many diffetrent central series you meant?
$endgroup$
– Derek Holt
Jan 23 at 9:58
|
show 2 more comments
$begingroup$
Let
$$1=G_0leq G_1leq ... leq G_{n-1} leq G_n = G$$
be a central series of the group $G$. That is, $G_{i-1}/G_ileq Z(G/G_i)$ for all $i$.
Let
$$1=Z_0(G)leq Z_1(G)leq ... $$
$$... leq gamma_2(G)leq gamma_1(G)=G$$
be the upper central series and the lower central series of a group $G$, respectively. I have proved $$G_ileq Z_i(G), gamma_{i+1}(G)leq G_{n-i}$$
I was wondering if there is any example of a group such that has a central series which is different from the upper and the lower central series.
Also I would like to prove that for a nilpotent group of nilpotency class $c$, $$gamma_{c+1-i}(G)leq Z_i(G)$$ but I do not see it. (For this, I have been thinking on using that a nilpotent group of class $c$ satisfies $$Z_c(G)=G,gamma_{c+1}(G)=1$$ would that be useful?)
Any help?
group-theory nilpotent-groups
asked Jan 20 at 17:07
idriskameniidriskameni
753321
$endgroup$
Let
$$1=G_0leq G_1leq ... leq G_{n-1} leq G_n = G$$
be a central series of the group $G$. That is, $G_{i-1}/G_ileq Z(G/G_i)$ for all $i$.
Let
$$1=Z_0(G)leq Z_1(G)leq ... $$
$$... leq gamma_2(G)leq gamma_1(G)=G$$
be the upper central series and the lower central series of a group $G$, respectively. I have proved $$G_ileq Z_i(G), gamma_{i+1}(G)leq G_{n-i}$$
I was wondering if there is any example of a group such that has a central series which is different from the upper and the lower central series.
Also I would like to prove that for a nilpotent group of nilpotency class $c$, $$gamma_{c+1-i}(G)leq Z_i(G)$$ but I do not see it. (For this, I have been thinking on using that a nilpotent group of class $c$ satisfies $$Z_c(G)=G,gamma_{c+1}(G)=1$$ would that be useful?)
Any help?
group-theory nilpotent-groups
group-theory nilpotent-groups
asked Jan 20 at 17:07
idriskameniidriskameni
753321
asked Jan 20 at 17:07
idriskameniidriskameni
753321
edited Jan 23 at 16:47
idriskameni
asked Jan 20 at 17:07
idriskameniidriskameni
753321
asked Jan 20 at 17:07
idriskameniidriskameni
753321
asked Jan 20 at 17:07
idriskameniidriskameni
753321
753321
$begingroup$
Probably the reason that nobody has answered is that it is unclear exactly what you are asking. You ask for an example in which "these three series" are all different. Which three series exactly? We have the upper central series, and the lower central series, but what exactly is your third series? You could ask for an example with at least three distinct central series - that would make sense.
$endgroup$
– Derek Holt
Jan 23 at 8:54
$begingroup$
I think that it is super clear. Let (---) be the central series (1), the upper central series (2) and the lower central series (3) of a group $G$.
$endgroup$
– idriskameni
Jan 23 at 9:01
$begingroup$
I agree that (2) and (3) are clear, but what do you mean by "the central series (1)"?
$endgroup$
– Derek Holt
Jan 23 at 9:13
$begingroup$
The definition of central series of a group is clear I think. Anyway, I have updated the post.
$endgroup$
– idriskameni
Jan 23 at 9:27
2
$begingroup$
Yes, the definition of a central series of a group is clear. But the group may have many different central series (or it may have none at all). So it makes no sense to write "the central series". How would anyone know which of the many diffetrent central series you meant?
$endgroup$
– Derek Holt
Jan 23 at 9:58
|
show 2 more comments
$begingroup$
Probably the reason that nobody has answered is that it is unclear exactly what you are asking. You ask for an example in which "these three series" are all different. Which three series exactly? We have the upper central series, and the lower central series, but what exactly is your third series? You could ask for an example with at least three distinct central series - that would make sense.
$endgroup$
– Derek Holt
Jan 23 at 8:54
$begingroup$
I think that it is super clear. Let (---) be the central series (1), the upper central series (2) and the lower central series (3) of a group $G$.
$endgroup$
– idriskameni
Jan 23 at 9:01
$begingroup$
I agree that (2) and (3) are clear, but what do you mean by "the central series (1)"?
$endgroup$
– Derek Holt
Jan 23 at 9:13
$begingroup$
The definition of central series of a group is clear I think. Anyway, I have updated the post.
$endgroup$
– idriskameni
Jan 23 at 9:27
2
$begingroup$
Yes, the definition of a central series of a group is clear. But the group may have many different central series (or it may have none at all). So it makes no sense to write "the central series". How would anyone know which of the many diffetrent central series you meant?
$endgroup$
– Derek Holt
Jan 23 at 9:58
$begingroup$
Probably the reason that nobody has answered is that it is unclear exactly what you are asking. You ask for an example in which "these three series" are all different. Which three series exactly? We have the upper central series, and the lower central series, but what exactly is your third series? You could ask for an example with at least three distinct central series - that would make sense.
$endgroup$
– Derek Holt
Jan 23 at 8:54
$begingroup$
Probably the reason that nobody has answered is that it is unclear exactly what you are asking. You ask for an example in which "these three series" are all different. Which three series exactly? We have the upper central series, and the lower central series, but what exactly is your third series? You could ask for an example with at least three distinct central series - that would make sense.
$endgroup$
– Derek Holt
Jan 23 at 8:54
$begingroup$
I think that it is super clear. Let (---) be the central series (1), the upper central series (2) and the lower central series (3) of a group $G$.
$endgroup$
– idriskameni
Jan 23 at 9:01
$begingroup$
I think that it is super clear. Let (---) be the central series (1), the upper central series (2) and the lower central series (3) of a group $G$.
$endgroup$
– idriskameni
Jan 23 at 9:01
$begingroup$
I agree that (2) and (3) are clear, but what do you mean by "the central series (1)"?
$endgroup$
– Derek Holt
Jan 23 at 9:13
$begingroup$
I agree that (2) and (3) are clear, but what do you mean by "the central series (1)"?
$endgroup$
– Derek Holt
Jan 23 at 9:13
$begingroup$
The definition of central series of a group is clear I think. Anyway, I have updated the post.
$endgroup$
– idriskameni
Jan 23 at 9:27
$begingroup$
The definition of central series of a group is clear I think. Anyway, I have updated the post.
$endgroup$
– idriskameni
Jan 23 at 9:27
2
2
$begingroup$
Yes, the definition of a central series of a group is clear. But the group may have many different central series (or it may have none at all). So it makes no sense to write "the central series". How would anyone know which of the many diffetrent central series you meant?
$endgroup$
– Derek Holt
Jan 23 at 9:58
$begingroup$
Yes, the definition of a central series of a group is clear. But the group may have many different central series (or it may have none at all). So it makes no sense to write "the central series". How would anyone know which of the many diffetrent central series you meant?
$endgroup$
– Derek Holt
Jan 23 at 9:58
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Consider $G=Vtimes D_8$ (direct product of Klein four group with the group of the square) of order 32. The lower central series is $1<1times Z(D_8)<G$. The upper series is $1<Vtimes Z(D_8)<G$. And three additional central series are given by $1<Htimes Z(D_8)<G$ where $H$ is any of the three order 2 subgroups of $V$. Thus $Vtimes D_8$ is an example of what you are looking for.
Note that I have answered the strict version of this problem, where the additional central series have the same lengths as the upper and lower series. If they are allowed to be longer, the problem is even easier.
As for why the upper series' terms always contain the corresponding lower series, this is a standard result you can find in many textbooks. The essential idea is to start at the top, noting that $G'=gamma_2(G)$ is the smallest normal subgroup with abelian quotient, so that $Z_{c-1}(G)ge gamma_2(G)$. And continue by an easy induction.
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
$endgroup$
$begingroup$
There are also plenty of longer central series.
$endgroup$
– Derek Holt
Jan 23 at 18:51
$begingroup$
Can that be done with $G=D_4 times mathbb{Z}_2$? I have been trying but I can not figure it out. I have seen on internet that this is the easier counterexample. Can you help me with that one? Anyway, brilliant contribution. Thank you very much.
$endgroup$
– idriskameni
Jan 23 at 19:13
$begingroup$
Let me ask you some questions. To be a central series, it should satisfy $G_{i+1}/G_i leq Z(G/G_i)$. I think that your example, does not satisfy that for the three central series. Does it?
$endgroup$
– idriskameni
Jan 23 at 21:30
1
$begingroup$
@idriskameni All the series I gave are central series. Since they have length 2, we just need $G_1=G_1/G_0le Z(G/G_0)=Z(G)$ and $G/G_1le Z(G/G_1)$. The first condition just says $G_1le Z(G)$. The second condition just says $G/G_1$ i abelian, which is the same as $G_1ge G'$ Since $Z(Vtimes D_8)=Vtimes Z(D_8)$ and $(Vtimes D_8)'=1times Z(D_8)$, you can select $G_1$ to be any group containing $1times Z(D_8)$ and contained in $Vtimes Z(D_8)$. Those are my examples. Nowhere do I mention the subgroup ${1,a}times D_4$ (it would be $Htimes D_8$ in my notation). Only you do that.
$endgroup$
– C Monsour
Jan 24 at 1:42
1
$begingroup$
As for $2 times D_8$ (the other group you mention), the lower and upper central series are different, but the group is class 2 and $|Z(G):G'|=|(2times 2)/(1times 2)|=2$ there are no groups in between $Z(G)$ and $G'$ to use for $G_1$, so those are the only minimal length central series and you won't find a third one. (Note that $p$ is standard notation for the cyclic group of order $p$. $Bbb{Z}_2$ denotes the 2-adic integers.. The cyclic group of order 2 is $2$, or $C_2$, or $Bbb{Z}/langle 2rangle$
$endgroup$
– C Monsour
Jan 24 at 1:48
|
show 2 more comments
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Consider $G=Vtimes D_8$ (direct product of Klein four group with the group of the square) of order 32. The lower central series is $1<1times Z(D_8)<G$. The upper series is $1<Vtimes Z(D_8)<G$. And three additional central series are given by $1<Htimes Z(D_8)<G$ where $H$ is any of the three order 2 subgroups of $V$. Thus $Vtimes D_8$ is an example of what you are looking for.
Note that I have answered the strict version of this problem, where the additional central series have the same lengths as the upper and lower series. If they are allowed to be longer, the problem is even easier.
As for why the upper series' terms always contain the corresponding lower series, this is a standard result you can find in many textbooks. The essential idea is to start at the top, noting that $G'=gamma_2(G)$ is the smallest normal subgroup with abelian quotient, so that $Z_{c-1}(G)ge gamma_2(G)$. And continue by an easy induction.
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
$endgroup$
$begingroup$
There are also plenty of longer central series.
$endgroup$
– Derek Holt
Jan 23 at 18:51
$begingroup$
Can that be done with $G=D_4 times mathbb{Z}_2$? I have been trying but I can not figure it out. I have seen on internet that this is the easier counterexample. Can you help me with that one? Anyway, brilliant contribution. Thank you very much.
$endgroup$
– idriskameni
Jan 23 at 19:13
$begingroup$
Let me ask you some questions. To be a central series, it should satisfy $G_{i+1}/G_i leq Z(G/G_i)$. I think that your example, does not satisfy that for the three central series. Does it?
$endgroup$
– idriskameni
Jan 23 at 21:30
1
$begingroup$
@idriskameni All the series I gave are central series. Since they have length 2, we just need $G_1=G_1/G_0le Z(G/G_0)=Z(G)$ and $G/G_1le Z(G/G_1)$. The first condition just says $G_1le Z(G)$. The second condition just says $G/G_1$ i abelian, which is the same as $G_1ge G'$ Since $Z(Vtimes D_8)=Vtimes Z(D_8)$ and $(Vtimes D_8)'=1times Z(D_8)$, you can select $G_1$ to be any group containing $1times Z(D_8)$ and contained in $Vtimes Z(D_8)$. Those are my examples. Nowhere do I mention the subgroup ${1,a}times D_4$ (it would be $Htimes D_8$ in my notation). Only you do that.
$endgroup$
– C Monsour
Jan 24 at 1:42
1
$begingroup$
As for $2 times D_8$ (the other group you mention), the lower and upper central series are different, but the group is class 2 and $|Z(G):G'|=|(2times 2)/(1times 2)|=2$ there are no groups in between $Z(G)$ and $G'$ to use for $G_1$, so those are the only minimal length central series and you won't find a third one. (Note that $p$ is standard notation for the cyclic group of order $p$. $Bbb{Z}_2$ denotes the 2-adic integers.. The cyclic group of order 2 is $2$, or $C_2$, or $Bbb{Z}/langle 2rangle$
$endgroup$
– C Monsour
Jan 24 at 1:48
|
show 2 more comments
$begingroup$
Consider $G=Vtimes D_8$ (direct product of Klein four group with the group of the square) of order 32. The lower central series is $1<1times Z(D_8)<G$. The upper series is $1<Vtimes Z(D_8)<G$. And three additional central series are given by $1<Htimes Z(D_8)<G$ where $H$ is any of the three order 2 subgroups of $V$. Thus $Vtimes D_8$ is an example of what you are looking for.
Note that I have answered the strict version of this problem, where the additional central series have the same lengths as the upper and lower series. If they are allowed to be longer, the problem is even easier.
As for why the upper series' terms always contain the corresponding lower series, this is a standard result you can find in many textbooks. The essential idea is to start at the top, noting that $G'=gamma_2(G)$ is the smallest normal subgroup with abelian quotient, so that $Z_{c-1}(G)ge gamma_2(G)$. And continue by an easy induction.
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
$endgroup$
$begingroup$
There are also plenty of longer central series.
$endgroup$
– Derek Holt
Jan 23 at 18:51
$begingroup$
Can that be done with $G=D_4 times mathbb{Z}_2$? I have been trying but I can not figure it out. I have seen on internet that this is the easier counterexample. Can you help me with that one? Anyway, brilliant contribution. Thank you very much.
$endgroup$
– idriskameni
Jan 23 at 19:13
$begingroup$
Let me ask you some questions. To be a central series, it should satisfy $G_{i+1}/G_i leq Z(G/G_i)$. I think that your example, does not satisfy that for the three central series. Does it?
$endgroup$
– idriskameni
Jan 23 at 21:30
1
$begingroup$
@idriskameni All the series I gave are central series. Since they have length 2, we just need $G_1=G_1/G_0le Z(G/G_0)=Z(G)$ and $G/G_1le Z(G/G_1)$. The first condition just says $G_1le Z(G)$. The second condition just says $G/G_1$ i abelian, which is the same as $G_1ge G'$ Since $Z(Vtimes D_8)=Vtimes Z(D_8)$ and $(Vtimes D_8)'=1times Z(D_8)$, you can select $G_1$ to be any group containing $1times Z(D_8)$ and contained in $Vtimes Z(D_8)$. Those are my examples. Nowhere do I mention the subgroup ${1,a}times D_4$ (it would be $Htimes D_8$ in my notation). Only you do that.
$endgroup$
– C Monsour
Jan 24 at 1:42
1
$begingroup$
As for $2 times D_8$ (the other group you mention), the lower and upper central series are different, but the group is class 2 and $|Z(G):G'|=|(2times 2)/(1times 2)|=2$ there are no groups in between $Z(G)$ and $G'$ to use for $G_1$, so those are the only minimal length central series and you won't find a third one. (Note that $p$ is standard notation for the cyclic group of order $p$. $Bbb{Z}_2$ denotes the 2-adic integers.. The cyclic group of order 2 is $2$, or $C_2$, or $Bbb{Z}/langle 2rangle$
$endgroup$
– C Monsour
Jan 24 at 1:48
|
show 2 more comments
$begingroup$
Consider $G=Vtimes D_8$ (direct product of Klein four group with the group of the square) of order 32. The lower central series is $1<1times Z(D_8)<G$. The upper series is $1<Vtimes Z(D_8)<G$. And three additional central series are given by $1<Htimes Z(D_8)<G$ where $H$ is any of the three order 2 subgroups of $V$. Thus $Vtimes D_8$ is an example of what you are looking for.
Note that I have answered the strict version of this problem, where the additional central series have the same lengths as the upper and lower series. If they are allowed to be longer, the problem is even easier.
As for why the upper series' terms always contain the corresponding lower series, this is a standard result you can find in many textbooks. The essential idea is to start at the top, noting that $G'=gamma_2(G)$ is the smallest normal subgroup with abelian quotient, so that $Z_{c-1}(G)ge gamma_2(G)$. And continue by an easy induction.
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
$endgroup$
Consider $G=Vtimes D_8$ (direct product of Klein four group with the group of the square) of order 32. The lower central series is $1<1times Z(D_8)<G$. The upper series is $1<Vtimes Z(D_8)<G$. And three additional central series are given by $1<Htimes Z(D_8)<G$ where $H$ is any of the three order 2 subgroups of $V$. Thus $Vtimes D_8$ is an example of what you are looking for.
Note that I have answered the strict version of this problem, where the additional central series have the same lengths as the upper and lower series. If they are allowed to be longer, the problem is even easier.
As for why the upper series' terms always contain the corresponding lower series, this is a standard result you can find in many textbooks. The essential idea is to start at the top, noting that $G'=gamma_2(G)$ is the smallest normal subgroup with abelian quotient, so that $Z_{c-1}(G)ge gamma_2(G)$. And continue by an easy induction.
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
edited Jan 23 at 18:54
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
answered Jan 23 at 18:38
C MonsourC Monsour
6,2541325
6,2541325
$begingroup$
There are also plenty of longer central series.
$endgroup$
– Derek Holt
Jan 23 at 18:51
$begingroup$
Can that be done with $G=D_4 times mathbb{Z}_2$? I have been trying but I can not figure it out. I have seen on internet that this is the easier counterexample. Can you help me with that one? Anyway, brilliant contribution. Thank you very much.
$endgroup$
– idriskameni
Jan 23 at 19:13
$begingroup$
Let me ask you some questions. To be a central series, it should satisfy $G_{i+1}/G_i leq Z(G/G_i)$. I think that your example, does not satisfy that for the three central series. Does it?
$endgroup$
– idriskameni
Jan 23 at 21:30
1
$begingroup$
@idriskameni All the series I gave are central series. Since they have length 2, we just need $G_1=G_1/G_0le Z(G/G_0)=Z(G)$ and $G/G_1le Z(G/G_1)$. The first condition just says $G_1le Z(G)$. The second condition just says $G/G_1$ i abelian, which is the same as $G_1ge G'$ Since $Z(Vtimes D_8)=Vtimes Z(D_8)$ and $(Vtimes D_8)'=1times Z(D_8)$, you can select $G_1$ to be any group containing $1times Z(D_8)$ and contained in $Vtimes Z(D_8)$. Those are my examples. Nowhere do I mention the subgroup ${1,a}times D_4$ (it would be $Htimes D_8$ in my notation). Only you do that.
$endgroup$
– C Monsour
Jan 24 at 1:42
1
$begingroup$
As for $2 times D_8$ (the other group you mention), the lower and upper central series are different, but the group is class 2 and $|Z(G):G'|=|(2times 2)/(1times 2)|=2$ there are no groups in between $Z(G)$ and $G'$ to use for $G_1$, so those are the only minimal length central series and you won't find a third one. (Note that $p$ is standard notation for the cyclic group of order $p$. $Bbb{Z}_2$ denotes the 2-adic integers.. The cyclic group of order 2 is $2$, or $C_2$, or $Bbb{Z}/langle 2rangle$
$endgroup$
– C Monsour
Jan 24 at 1:48
|
show 2 more comments
$begingroup$
There are also plenty of longer central series.
$endgroup$
– Derek Holt
Jan 23 at 18:51
$begingroup$
Can that be done with $G=D_4 times mathbb{Z}_2$? I have been trying but I can not figure it out. I have seen on internet that this is the easier counterexample. Can you help me with that one? Anyway, brilliant contribution. Thank you very much.
$endgroup$
– idriskameni
Jan 23 at 19:13
$begingroup$
Let me ask you some questions. To be a central series, it should satisfy $G_{i+1}/G_i leq Z(G/G_i)$. I think that your example, does not satisfy that for the three central series. Does it?
$endgroup$
– idriskameni
Jan 23 at 21:30
1
$begingroup$
@idriskameni All the series I gave are central series. Since they have length 2, we just need $G_1=G_1/G_0le Z(G/G_0)=Z(G)$ and $G/G_1le Z(G/G_1)$. The first condition just says $G_1le Z(G)$. The second condition just says $G/G_1$ i abelian, which is the same as $G_1ge G'$ Since $Z(Vtimes D_8)=Vtimes Z(D_8)$ and $(Vtimes D_8)'=1times Z(D_8)$, you can select $G_1$ to be any group containing $1times Z(D_8)$ and contained in $Vtimes Z(D_8)$. Those are my examples. Nowhere do I mention the subgroup ${1,a}times D_4$ (it would be $Htimes D_8$ in my notation). Only you do that.
$endgroup$
– C Monsour
Jan 24 at 1:42
1
$begingroup$
As for $2 times D_8$ (the other group you mention), the lower and upper central series are different, but the group is class 2 and $|Z(G):G'|=|(2times 2)/(1times 2)|=2$ there are no groups in between $Z(G)$ and $G'$ to use for $G_1$, so those are the only minimal length central series and you won't find a third one. (Note that $p$ is standard notation for the cyclic group of order $p$. $Bbb{Z}_2$ denotes the 2-adic integers.. The cyclic group of order 2 is $2$, or $C_2$, or $Bbb{Z}/langle 2rangle$
$endgroup$
– C Monsour
Jan 24 at 1:48
$begingroup$
There are also plenty of longer central series.
$endgroup$
– Derek Holt
Jan 23 at 18:51
$begingroup$
There are also plenty of longer central series.
$endgroup$
– Derek Holt
Jan 23 at 18:51
$begingroup$
Can that be done with $G=D_4 times mathbb{Z}_2$? I have been trying but I can not figure it out. I have seen on internet that this is the easier counterexample. Can you help me with that one? Anyway, brilliant contribution. Thank you very much.
$endgroup$
– idriskameni
Jan 23 at 19:13
$begingroup$
Can that be done with $G=D_4 times mathbb{Z}_2$? I have been trying but I can not figure it out. I have seen on internet that this is the easier counterexample. Can you help me with that one? Anyway, brilliant contribution. Thank you very much.
$endgroup$
– idriskameni
Jan 23 at 19:13
$begingroup$
Let me ask you some questions. To be a central series, it should satisfy $G_{i+1}/G_i leq Z(G/G_i)$. I think that your example, does not satisfy that for the three central series. Does it?
$endgroup$
– idriskameni
Jan 23 at 21:30
$begingroup$
Let me ask you some questions. To be a central series, it should satisfy $G_{i+1}/G_i leq Z(G/G_i)$. I think that your example, does not satisfy that for the three central series. Does it?
$endgroup$
– idriskameni
Jan 23 at 21:30
1
1
$begingroup$
@idriskameni All the series I gave are central series. Since they have length 2, we just need $G_1=G_1/G_0le Z(G/G_0)=Z(G)$ and $G/G_1le Z(G/G_1)$. The first condition just says $G_1le Z(G)$. The second condition just says $G/G_1$ i abelian, which is the same as $G_1ge G'$ Since $Z(Vtimes D_8)=Vtimes Z(D_8)$ and $(Vtimes D_8)'=1times Z(D_8)$, you can select $G_1$ to be any group containing $1times Z(D_8)$ and contained in $Vtimes Z(D_8)$. Those are my examples. Nowhere do I mention the subgroup ${1,a}times D_4$ (it would be $Htimes D_8$ in my notation). Only you do that.
$endgroup$
– C Monsour
Jan 24 at 1:42
$begingroup$
@idriskameni All the series I gave are central series. Since they have length 2, we just need $G_1=G_1/G_0le Z(G/G_0)=Z(G)$ and $G/G_1le Z(G/G_1)$. The first condition just says $G_1le Z(G)$. The second condition just says $G/G_1$ i abelian, which is the same as $G_1ge G'$ Since $Z(Vtimes D_8)=Vtimes Z(D_8)$ and $(Vtimes D_8)'=1times Z(D_8)$, you can select $G_1$ to be any group containing $1times Z(D_8)$ and contained in $Vtimes Z(D_8)$. Those are my examples. Nowhere do I mention the subgroup ${1,a}times D_4$ (it would be $Htimes D_8$ in my notation). Only you do that.
$endgroup$
– C Monsour
Jan 24 at 1:42
1
1
$begingroup$
As for $2 times D_8$ (the other group you mention), the lower and upper central series are different, but the group is class 2 and $|Z(G):G'|=|(2times 2)/(1times 2)|=2$ there are no groups in between $Z(G)$ and $G'$ to use for $G_1$, so those are the only minimal length central series and you won't find a third one. (Note that $p$ is standard notation for the cyclic group of order $p$. $Bbb{Z}_2$ denotes the 2-adic integers.. The cyclic group of order 2 is $2$, or $C_2$, or $Bbb{Z}/langle 2rangle$
$endgroup$
– C Monsour
Jan 24 at 1:48
$begingroup$
As for $2 times D_8$ (the other group you mention), the lower and upper central series are different, but the group is class 2 and $|Z(G):G'|=|(2times 2)/(1times 2)|=2$ there are no groups in between $Z(G)$ and $G'$ to use for $G_1$, so those are the only minimal length central series and you won't find a third one. (Note that $p$ is standard notation for the cyclic group of order $p$. $Bbb{Z}_2$ denotes the 2-adic integers.. The cyclic group of order 2 is $2$, or $C_2$, or $Bbb{Z}/langle 2rangle$
$endgroup$
– C Monsour
Jan 24 at 1:48
|
show 2 more comments
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$begingroup$
Probably the reason that nobody has answered is that it is unclear exactly what you are asking. You ask for an example in which "these three series" are all different. Which three series exactly? We have the upper central series, and the lower central series, but what exactly is your third series? You could ask for an example with at least three distinct central series - that would make sense.
$endgroup$
– Derek Holt
Jan 23 at 8:54
$begingroup$
I think that it is super clear. Let (---) be the central series (1), the upper central series (2) and the lower central series (3) of a group $G$.
$endgroup$
– idriskameni
Jan 23 at 9:01
$begingroup$
I agree that (2) and (3) are clear, but what do you mean by "the central series (1)"?
$endgroup$
– Derek Holt
Jan 23 at 9:13
$begingroup$
The definition of central series of a group is clear I think. Anyway, I have updated the post.
$endgroup$
– idriskameni
Jan 23 at 9:27
2
$begingroup$
Yes, the definition of a central series of a group is clear. But the group may have many different central series (or it may have none at all). So it makes no sense to write "the central series". How would anyone know which of the many diffetrent central series you meant?
$endgroup$
– Derek Holt
Jan 23 at 9:58