Normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially; What is wrong with the following trivial...
$begingroup$
I'm trying to show that a normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially (please don't tell how to show it), but it seem it is quite a trivial question considering the following argument:
Consider $Z(N)$, a nontrivial subgroup of $N$. Since any element in
$Z(G)$ also commutes with the elements of $N$, $Z(G) subseteq Z(N)$,
but we know that $Z(N) leq N$, so $ {e} not = Z(G) subseteq N.$
If the answer was this, I don't think my Algebra professor would ask it, so what is wrong with the above argument ?
group-theory normal-subgroups p-groups
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
$endgroup$
|
show 1 more comment
$begingroup$
I'm trying to show that a normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially (please don't tell how to show it), but it seem it is quite a trivial question considering the following argument:
Consider $Z(N)$, a nontrivial subgroup of $N$. Since any element in
$Z(G)$ also commutes with the elements of $N$, $Z(G) subseteq Z(N)$,
but we know that $Z(N) leq N$, so $ {e} not = Z(G) subseteq N.$
If the answer was this, I don't think my Algebra professor would ask it, so what is wrong with the above argument ?
group-theory normal-subgroups p-groups
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
$endgroup$
$begingroup$
It's certainly not the case that $Z(G)subseteq Z(N)$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown Why ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:09
1
$begingroup$
What if $N$ is a proper subgroup of $Z(G)$?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:13
1
$begingroup$
An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g in C_G(N)$ which is in general bigger than $Z(N)$.
$endgroup$
– Pietro Gheri
Dec 15 '18 at 12:40
|
show 1 more comment
$begingroup$
I'm trying to show that a normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially (please don't tell how to show it), but it seem it is quite a trivial question considering the following argument:
Consider $Z(N)$, a nontrivial subgroup of $N$. Since any element in
$Z(G)$ also commutes with the elements of $N$, $Z(G) subseteq Z(N)$,
but we know that $Z(N) leq N$, so $ {e} not = Z(G) subseteq N.$
If the answer was this, I don't think my Algebra professor would ask it, so what is wrong with the above argument ?
group-theory normal-subgroups p-groups
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
$endgroup$
I'm trying to show that a normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially (please don't tell how to show it), but it seem it is quite a trivial question considering the following argument:
Consider $Z(N)$, a nontrivial subgroup of $N$. Since any element in
$Z(G)$ also commutes with the elements of $N$, $Z(G) subseteq Z(N)$,
but we know that $Z(N) leq N$, so $ {e} not = Z(G) subseteq N.$
If the answer was this, I don't think my Algebra professor would ask it, so what is wrong with the above argument ?
group-theory normal-subgroups p-groups
group-theory normal-subgroups p-groups
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
asked Dec 15 '18 at 6:03
onurcanbektasonurcanbektas
3,40211036
3,40211036
$begingroup$
It's certainly not the case that $Z(G)subseteq Z(N)$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown Why ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:09
1
$begingroup$
What if $N$ is a proper subgroup of $Z(G)$?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:13
1
$begingroup$
An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g in C_G(N)$ which is in general bigger than $Z(N)$.
$endgroup$
– Pietro Gheri
Dec 15 '18 at 12:40
|
show 1 more comment
$begingroup$
It's certainly not the case that $Z(G)subseteq Z(N)$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown Why ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:09
1
$begingroup$
What if $N$ is a proper subgroup of $Z(G)$?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:13
1
$begingroup$
An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g in C_G(N)$ which is in general bigger than $Z(N)$.
$endgroup$
– Pietro Gheri
Dec 15 '18 at 12:40
$begingroup$
It's certainly not the case that $Z(G)subseteq Z(N)$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:08
$begingroup$
It's certainly not the case that $Z(G)subseteq Z(N)$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown Why ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown Why ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:09
$begingroup$
@LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:09
1
1
$begingroup$
What if $N$ is a proper subgroup of $Z(G)$?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:13
$begingroup$
What if $N$ is a proper subgroup of $Z(G)$?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:13
1
1
$begingroup$
An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g in C_G(N)$ which is in general bigger than $Z(N)$.
$endgroup$
– Pietro Gheri
Dec 15 '18 at 12:40
$begingroup$
An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g in C_G(N)$ which is in general bigger than $Z(N)$.
$endgroup$
– Pietro Gheri
Dec 15 '18 at 12:40
|
show 1 more comment
1 Answer
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$begingroup$
$Z(N)$ and $Z(G)$ need not be at all related. If $N$ is abelian, then $Z(N) =N$, but the center of $G$ might intersect $N$ trivially. For example, the center of $S_3times mathbb Z_3$ intersects a subgroup of order $2$ in the first factor trivially.
In general, $Z(G) $ is a subgroup of the centralizer of $N$, but not necessarily the center.
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
$endgroup$
$begingroup$
In my notation, $Z(G):= C(G)$.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:49
$begingroup$
@onur Yes, but $Z(N) neq C(N) $ in general, unless you're using highly unusual notation.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:51
$begingroup$
what is $Z(N)$ in your notation ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:53
$begingroup$
@onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:56
$begingroup$
Oh, I see. Thanks both for the answer and the replies.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:57
|
show 1 more comment
Your Answer
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$begingroup$
$Z(N)$ and $Z(G)$ need not be at all related. If $N$ is abelian, then $Z(N) =N$, but the center of $G$ might intersect $N$ trivially. For example, the center of $S_3times mathbb Z_3$ intersects a subgroup of order $2$ in the first factor trivially.
In general, $Z(G) $ is a subgroup of the centralizer of $N$, but not necessarily the center.
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
$endgroup$
$begingroup$
In my notation, $Z(G):= C(G)$.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:49
$begingroup$
@onur Yes, but $Z(N) neq C(N) $ in general, unless you're using highly unusual notation.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:51
$begingroup$
what is $Z(N)$ in your notation ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:53
$begingroup$
@onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:56
$begingroup$
Oh, I see. Thanks both for the answer and the replies.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:57
|
show 1 more comment
$begingroup$
$Z(N)$ and $Z(G)$ need not be at all related. If $N$ is abelian, then $Z(N) =N$, but the center of $G$ might intersect $N$ trivially. For example, the center of $S_3times mathbb Z_3$ intersects a subgroup of order $2$ in the first factor trivially.
In general, $Z(G) $ is a subgroup of the centralizer of $N$, but not necessarily the center.
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
$endgroup$
$begingroup$
In my notation, $Z(G):= C(G)$.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:49
$begingroup$
@onur Yes, but $Z(N) neq C(N) $ in general, unless you're using highly unusual notation.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:51
$begingroup$
what is $Z(N)$ in your notation ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:53
$begingroup$
@onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:56
$begingroup$
Oh, I see. Thanks both for the answer and the replies.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:57
|
show 1 more comment
$begingroup$
$Z(N)$ and $Z(G)$ need not be at all related. If $N$ is abelian, then $Z(N) =N$, but the center of $G$ might intersect $N$ trivially. For example, the center of $S_3times mathbb Z_3$ intersects a subgroup of order $2$ in the first factor trivially.
In general, $Z(G) $ is a subgroup of the centralizer of $N$, but not necessarily the center.
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
$endgroup$
$Z(N)$ and $Z(G)$ need not be at all related. If $N$ is abelian, then $Z(N) =N$, but the center of $G$ might intersect $N$ trivially. For example, the center of $S_3times mathbb Z_3$ intersects a subgroup of order $2$ in the first factor trivially.
In general, $Z(G) $ is a subgroup of the centralizer of $N$, but not necessarily the center.
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
edited Dec 15 '18 at 12:42
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
answered Dec 15 '18 at 12:36
Matt SamuelMatt Samuel
38.5k63768
38.5k63768
$begingroup$
In my notation, $Z(G):= C(G)$.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:49
$begingroup$
@onur Yes, but $Z(N) neq C(N) $ in general, unless you're using highly unusual notation.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:51
$begingroup$
what is $Z(N)$ in your notation ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:53
$begingroup$
@onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:56
$begingroup$
Oh, I see. Thanks both for the answer and the replies.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:57
|
show 1 more comment
$begingroup$
In my notation, $Z(G):= C(G)$.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:49
$begingroup$
@onur Yes, but $Z(N) neq C(N) $ in general, unless you're using highly unusual notation.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:51
$begingroup$
what is $Z(N)$ in your notation ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:53
$begingroup$
@onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:56
$begingroup$
Oh, I see. Thanks both for the answer and the replies.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:57
$begingroup$
In my notation, $Z(G):= C(G)$.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:49
$begingroup$
In my notation, $Z(G):= C(G)$.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:49
$begingroup$
@onur Yes, but $Z(N) neq C(N) $ in general, unless you're using highly unusual notation.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:51
$begingroup$
@onur Yes, but $Z(N) neq C(N) $ in general, unless you're using highly unusual notation.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:51
$begingroup$
what is $Z(N)$ in your notation ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:53
$begingroup$
what is $Z(N)$ in your notation ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:53
$begingroup$
@onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:56
$begingroup$
@onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$.
$endgroup$
– Matt Samuel
Dec 15 '18 at 14:56
$begingroup$
Oh, I see. Thanks both for the answer and the replies.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:57
$begingroup$
Oh, I see. Thanks both for the answer and the replies.
$endgroup$
– onurcanbektas
Dec 15 '18 at 14:57
|
show 1 more comment
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$begingroup$
It's certainly not the case that $Z(G)subseteq Z(N)$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown Why ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:08
$begingroup$
@LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ?
$endgroup$
– onurcanbektas
Dec 15 '18 at 6:09
1
$begingroup$
What if $N$ is a proper subgroup of $Z(G)$?
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 6:13
1
$begingroup$
An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g in C_G(N)$ which is in general bigger than $Z(N)$.
$endgroup$
– Pietro Gheri
Dec 15 '18 at 12:40