Mixing
The relation between commutator series and the center
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Let $G$ be a group and $R$ be a maximal solvable subgroup in $G$. Consider the commutator series $R^(n)=[R^(n-1),R^(n-1)]$. Assume that $R^(k)$ intersects the center $Z$ of $G$ in a non-trivial subgroup. Now since the last non-trivial term of the series is abelain, does that mean it equals the center $Z$?
abstract-algebra group-theory
asked Jul 18 at 3:57
Ronald
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Let $G$ be a group and $R$ be a maximal solvable subgroup in $G$. Consider the commutator series $R^(n)=[R^(n-1),R^(n-1)]$. Assume that $R^(k)$ intersects the center $Z$ of $G$ in a non-trivial subgroup. Now since the last non-trivial term of the series is abelain, does that mean it equals the center $Z$?
abstract-algebra group-theory
asked Jul 18 at 3:57
Ronald
1,5391821
Perhaps I have misunderstood the question, but I believe the answer is "no", as your suggestion that some $R^(k)$ intersects the centre of $G$ seems to be incorrect. For instance, take $G$ to be solvable with trivial centre (e.g., a dihedral group of order $10$). Then $R=G$, and no term of the derived series meets the trivial centre in a non-trivial subgroup.
– James
Jul 18 at 6:27
2
@James I would guess that rather than "We know that" he actually means "Assume that". I am also wondering whether we are supposed to assume that $G$ itself if not solvable. But the answer to the question is no however you interpret it. The last nontrivial term in the series could be a proper subgroup of $Z$.
– Derek Holt
Jul 18 at 7:36
Sorry I actually meant that what Derek said. Assume that!
– Ronald
Jul 18 at 7:47
add a comment |Â
up vote
1
down vote
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up vote
1
down vote
favorite
Let $G$ be a group and $R$ be a maximal solvable subgroup in $G$. Consider the commutator series $R^(n)=[R^(n-1),R^(n-1)]$. Assume that $R^(k)$ intersects the center $Z$ of $G$ in a non-trivial subgroup. Now since the last non-trivial term of the series is abelain, does that mean it equals the center $Z$?
abstract-algebra group-theory
asked Jul 18 at 3:57
Ronald
1,5391821
Let $G$ be a group and $R$ be a maximal solvable subgroup in $G$. Consider the commutator series $R^(n)=[R^(n-1),R^(n-1)]$. Assume that $R^(k)$ intersects the center $Z$ of $G$ in a non-trivial subgroup. Now since the last non-trivial term of the series is abelain, does that mean it equals the center $Z$?
abstract-algebra group-theory
asked Jul 18 at 3:57
Ronald
1,5391821
edited Jul 18 at 7:47
asked Jul 18 at 3:57
Ronald
1,5391821
asked Jul 18 at 3:57
Ronald
1,5391821
asked Jul 18 at 3:57
Ronald
1,5391821
1,5391821
Perhaps I have misunderstood the question, but I believe the answer is "no", as your suggestion that some $R^(k)$ intersects the centre of $G$ seems to be incorrect. For instance, take $G$ to be solvable with trivial centre (e.g., a dihedral group of order $10$). Then $R=G$, and no term of the derived series meets the trivial centre in a non-trivial subgroup.
– James
Jul 18 at 6:27
2
@James I would guess that rather than "We know that" he actually means "Assume that". I am also wondering whether we are supposed to assume that $G$ itself if not solvable. But the answer to the question is no however you interpret it. The last nontrivial term in the series could be a proper subgroup of $Z$.
– Derek Holt
Jul 18 at 7:36
Sorry I actually meant that what Derek said. Assume that!
– Ronald
Jul 18 at 7:47
add a comment |Â
Perhaps I have misunderstood the question, but I believe the answer is "no", as your suggestion that some $R^(k)$ intersects the centre of $G$ seems to be incorrect. For instance, take $G$ to be solvable with trivial centre (e.g., a dihedral group of order $10$). Then $R=G$, and no term of the derived series meets the trivial centre in a non-trivial subgroup.
– James
Jul 18 at 6:27
2
@James I would guess that rather than "We know that" he actually means "Assume that". I am also wondering whether we are supposed to assume that $G$ itself if not solvable. But the answer to the question is no however you interpret it. The last nontrivial term in the series could be a proper subgroup of $Z$.
– Derek Holt
Jul 18 at 7:36
Sorry I actually meant that what Derek said. Assume that!
– Ronald
Jul 18 at 7:47
Perhaps I have misunderstood the question, but I believe the answer is "no", as your suggestion that some $R^(k)$ intersects the centre of $G$ seems to be incorrect. For instance, take $G$ to be solvable with trivial centre (e.g., a dihedral group of order $10$). Then $R=G$, and no term of the derived series meets the trivial centre in a non-trivial subgroup.
– James
Jul 18 at 6:27
Perhaps I have misunderstood the question, but I believe the answer is "no", as your suggestion that some $R^(k)$ intersects the centre of $G$ seems to be incorrect. For instance, take $G$ to be solvable with trivial centre (e.g., a dihedral group of order $10$). Then $R=G$, and no term of the derived series meets the trivial centre in a non-trivial subgroup.
– James
Jul 18 at 6:27
2
2
@James I would guess that rather than "We know that" he actually means "Assume that". I am also wondering whether we are supposed to assume that $G$ itself if not solvable. But the answer to the question is no however you interpret it. The last nontrivial term in the series could be a proper subgroup of $Z$.
– Derek Holt
Jul 18 at 7:36
@James I would guess that rather than "We know that" he actually means "Assume that". I am also wondering whether we are supposed to assume that $G$ itself if not solvable. But the answer to the question is no however you interpret it. The last nontrivial term in the series could be a proper subgroup of $Z$.
– Derek Holt
Jul 18 at 7:36
Sorry I actually meant that what Derek said. Assume that!
– Ronald
Jul 18 at 7:47
Sorry I actually meant that what Derek said. Assume that!
– Ronald
Jul 18 at 7:47
add a comment |Â
1 Answer
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If we take $G = R = C_2 times S_3$, then $R^(0)$ intersects $Z(G)$ non-trivially, but $R^(1) = (0, (123))$ which is abelian but not central.
answered Jul 18 at 12:34
Christopher
4,8251224
1
(tbh, I'm a bit worried I've misunderstood the question, since I can't see any reason to believe the question to be true if I have not misunderstood it)
– Christopher
Jul 18 at 12:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If we take $G = R = C_2 times S_3$, then $R^(0)$ intersects $Z(G)$ non-trivially, but $R^(1) = (0, (123))$ which is abelian but not central.
answered Jul 18 at 12:34
Christopher
4,8251224
1
(tbh, I'm a bit worried I've misunderstood the question, since I can't see any reason to believe the question to be true if I have not misunderstood it)
– Christopher
Jul 18 at 12:35
add a comment |Â
up vote
2
down vote
accepted
If we take $G = R = C_2 times S_3$, then $R^(0)$ intersects $Z(G)$ non-trivially, but $R^(1) = (0, (123))$ which is abelian but not central.
answered Jul 18 at 12:34
Christopher
4,8251224
1
(tbh, I'm a bit worried I've misunderstood the question, since I can't see any reason to believe the question to be true if I have not misunderstood it)
– Christopher
Jul 18 at 12:35
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If we take $G = R = C_2 times S_3$, then $R^(0)$ intersects $Z(G)$ non-trivially, but $R^(1) = (0, (123))$ which is abelian but not central.
answered Jul 18 at 12:34
Christopher
4,8251224
If we take $G = R = C_2 times S_3$, then $R^(0)$ intersects $Z(G)$ non-trivially, but $R^(1) = (0, (123))$ which is abelian but not central.
answered Jul 18 at 12:34
Christopher
4,8251224
answered Jul 18 at 12:34
Christopher
4,8251224
answered Jul 18 at 12:34
Christopher
4,8251224
answered Jul 18 at 12:34
Christopher
4,8251224
4,8251224
1
(tbh, I'm a bit worried I've misunderstood the question, since I can't see any reason to believe the question to be true if I have not misunderstood it)
– Christopher
Jul 18 at 12:35
add a comment |Â
1
(tbh, I'm a bit worried I've misunderstood the question, since I can't see any reason to believe the question to be true if I have not misunderstood it)
– Christopher
Jul 18 at 12:35
1
1
(tbh, I'm a bit worried I've misunderstood the question, since I can't see any reason to believe the question to be true if I have not misunderstood it)
– Christopher
Jul 18 at 12:35
(tbh, I'm a bit worried I've misunderstood the question, since I can't see any reason to believe the question to be true if I have not misunderstood it)
– Christopher
Jul 18 at 12:35
add a comment |Â
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Perhaps I have misunderstood the question, but I believe the answer is "no", as your suggestion that some $R^(k)$ intersects the centre of $G$ seems to be incorrect. For instance, take $G$ to be solvable with trivial centre (e.g., a dihedral group of order $10$). Then $R=G$, and no term of the derived series meets the trivial centre in a non-trivial subgroup.
– James
Jul 18 at 6:27
2
@James I would guess that rather than "We know that" he actually means "Assume that". I am also wondering whether we are supposed to assume that $G$ itself if not solvable. But the answer to the question is no however you interpret it. The last nontrivial term in the series could be a proper subgroup of $Z$.
– Derek Holt
Jul 18 at 7:36
Sorry I actually meant that what Derek said. Assume that!
– Ronald
Jul 18 at 7:47