Centre of non-abelian group of order $p^3$, where $p$ is prime
$begingroup$
If $p$ is a prime number and $G$ is a non-abelian group of order $p^3$, then what can we say about the number of elements in the centre of the group $G$, i.e. $Z(G)$?
abstract-algebra group-theory
edited Nov 25 '18 at 13:49
the_fox
2,58711533
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
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add a comment |
$begingroup$
If $p$ is a prime number and $G$ is a non-abelian group of order $p^3$, then what can we say about the number of elements in the centre of the group $G$, i.e. $Z(G)$?
abstract-algebra group-theory
edited Nov 25 '18 at 13:49
the_fox
2,58711533
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
$endgroup$
add a comment |
$begingroup$
If $p$ is a prime number and $G$ is a non-abelian group of order $p^3$, then what can we say about the number of elements in the centre of the group $G$, i.e. $Z(G)$?
abstract-algebra group-theory
edited Nov 25 '18 at 13:49
the_fox
2,58711533
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
$endgroup$
If $p$ is a prime number and $G$ is a non-abelian group of order $p^3$, then what can we say about the number of elements in the centre of the group $G$, i.e. $Z(G)$?
abstract-algebra group-theory
abstract-algebra group-theory
edited Nov 25 '18 at 13:49
the_fox
2,58711533
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
edited Nov 25 '18 at 13:49
the_fox
2,58711533
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
edited Nov 25 '18 at 13:49
the_fox
2,58711533
edited Nov 25 '18 at 13:49
the_fox
2,58711533
edited Nov 25 '18 at 13:49
the_fox
2,58711533
2,58711533
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
asked Nov 25 '18 at 13:16
kapil pundirkapil pundir
574
574
add a comment |
add a comment |
1 Answer
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Hint:By class equation, it can be shown that $Z(G)$ is non-trivial. Now if $O(Z(G)) =p^2 $,we will have order of $frac{G}{Z(G)}=p $,which will imply $frac{G}{Z(G)}$ is cyclic. But that will imply $G$ is abelian(Verify),leading to a contradiction. Then the only other possibility is $p$,since it is already given that $G$ is non-abelian.
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
$begingroup$
Do we have some method other than 'class equation' to show that Z(G) is non trivial?
$endgroup$
– kapil pundir
Nov 25 '18 at 14:52
$begingroup$
I'm sure there are, but it's the easiest way. To see that counting is likely to be involved in some essential way, note that there are infinite p-groups with trivial center. But all finite p-groups have non-trivial center (by the class equation argument).
$endgroup$
– C Monsour
Nov 25 '18 at 15:15
1
$begingroup$
Note that the class equation argument is elementary. All conjugacy classes have order a power of p, and the group is divisible by p. Thus, if one conjugacy class has size 1, at least p-1 more must also be size 1 so that the sum is divisible by p.
$endgroup$
– C Monsour
Nov 25 '18 at 15:18
add a comment |
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$begingroup$
Hint:By class equation, it can be shown that $Z(G)$ is non-trivial. Now if $O(Z(G)) =p^2 $,we will have order of $frac{G}{Z(G)}=p $,which will imply $frac{G}{Z(G)}$ is cyclic. But that will imply $G$ is abelian(Verify),leading to a contradiction. Then the only other possibility is $p$,since it is already given that $G$ is non-abelian.
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
$begingroup$
Do we have some method other than 'class equation' to show that Z(G) is non trivial?
$endgroup$
– kapil pundir
Nov 25 '18 at 14:52
$begingroup$
I'm sure there are, but it's the easiest way. To see that counting is likely to be involved in some essential way, note that there are infinite p-groups with trivial center. But all finite p-groups have non-trivial center (by the class equation argument).
$endgroup$
– C Monsour
Nov 25 '18 at 15:15
1
$begingroup$
Note that the class equation argument is elementary. All conjugacy classes have order a power of p, and the group is divisible by p. Thus, if one conjugacy class has size 1, at least p-1 more must also be size 1 so that the sum is divisible by p.
$endgroup$
– C Monsour
Nov 25 '18 at 15:18
add a comment |
$begingroup$
Hint:By class equation, it can be shown that $Z(G)$ is non-trivial. Now if $O(Z(G)) =p^2 $,we will have order of $frac{G}{Z(G)}=p $,which will imply $frac{G}{Z(G)}$ is cyclic. But that will imply $G$ is abelian(Verify),leading to a contradiction. Then the only other possibility is $p$,since it is already given that $G$ is non-abelian.
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
$begingroup$
Do we have some method other than 'class equation' to show that Z(G) is non trivial?
$endgroup$
– kapil pundir
Nov 25 '18 at 14:52
$begingroup$
I'm sure there are, but it's the easiest way. To see that counting is likely to be involved in some essential way, note that there are infinite p-groups with trivial center. But all finite p-groups have non-trivial center (by the class equation argument).
$endgroup$
– C Monsour
Nov 25 '18 at 15:15
1
$begingroup$
Note that the class equation argument is elementary. All conjugacy classes have order a power of p, and the group is divisible by p. Thus, if one conjugacy class has size 1, at least p-1 more must also be size 1 so that the sum is divisible by p.
$endgroup$
– C Monsour
Nov 25 '18 at 15:18
add a comment |
$begingroup$
Hint:By class equation, it can be shown that $Z(G)$ is non-trivial. Now if $O(Z(G)) =p^2 $,we will have order of $frac{G}{Z(G)}=p $,which will imply $frac{G}{Z(G)}$ is cyclic. But that will imply $G$ is abelian(Verify),leading to a contradiction. Then the only other possibility is $p$,since it is already given that $G$ is non-abelian.
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
$endgroup$
Hint:By class equation, it can be shown that $Z(G)$ is non-trivial. Now if $O(Z(G)) =p^2 $,we will have order of $frac{G}{Z(G)}=p $,which will imply $frac{G}{Z(G)}$ is cyclic. But that will imply $G$ is abelian(Verify),leading to a contradiction. Then the only other possibility is $p$,since it is already given that $G$ is non-abelian.
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
answered Nov 25 '18 at 13:29
Thomas ShelbyThomas Shelby
2,250220
2,250220
$begingroup$
Do we have some method other than 'class equation' to show that Z(G) is non trivial?
$endgroup$
– kapil pundir
Nov 25 '18 at 14:52
$begingroup$
I'm sure there are, but it's the easiest way. To see that counting is likely to be involved in some essential way, note that there are infinite p-groups with trivial center. But all finite p-groups have non-trivial center (by the class equation argument).
$endgroup$
– C Monsour
Nov 25 '18 at 15:15
1
$begingroup$
Note that the class equation argument is elementary. All conjugacy classes have order a power of p, and the group is divisible by p. Thus, if one conjugacy class has size 1, at least p-1 more must also be size 1 so that the sum is divisible by p.
$endgroup$
– C Monsour
Nov 25 '18 at 15:18
add a comment |
$begingroup$
Do we have some method other than 'class equation' to show that Z(G) is non trivial?
$endgroup$
– kapil pundir
Nov 25 '18 at 14:52
$begingroup$
I'm sure there are, but it's the easiest way. To see that counting is likely to be involved in some essential way, note that there are infinite p-groups with trivial center. But all finite p-groups have non-trivial center (by the class equation argument).
$endgroup$
– C Monsour
Nov 25 '18 at 15:15
1
$begingroup$
Note that the class equation argument is elementary. All conjugacy classes have order a power of p, and the group is divisible by p. Thus, if one conjugacy class has size 1, at least p-1 more must also be size 1 so that the sum is divisible by p.
$endgroup$
– C Monsour
Nov 25 '18 at 15:18
$begingroup$
Do we have some method other than 'class equation' to show that Z(G) is non trivial?
$endgroup$
– kapil pundir
Nov 25 '18 at 14:52
$begingroup$
Do we have some method other than 'class equation' to show that Z(G) is non trivial?
$endgroup$
– kapil pundir
Nov 25 '18 at 14:52
$begingroup$
I'm sure there are, but it's the easiest way. To see that counting is likely to be involved in some essential way, note that there are infinite p-groups with trivial center. But all finite p-groups have non-trivial center (by the class equation argument).
$endgroup$
– C Monsour
Nov 25 '18 at 15:15
$begingroup$
I'm sure there are, but it's the easiest way. To see that counting is likely to be involved in some essential way, note that there are infinite p-groups with trivial center. But all finite p-groups have non-trivial center (by the class equation argument).
$endgroup$
– C Monsour
Nov 25 '18 at 15:15
1
1
$begingroup$
Note that the class equation argument is elementary. All conjugacy classes have order a power of p, and the group is divisible by p. Thus, if one conjugacy class has size 1, at least p-1 more must also be size 1 so that the sum is divisible by p.
$endgroup$
– C Monsour
Nov 25 '18 at 15:18
$begingroup$
Note that the class equation argument is elementary. All conjugacy classes have order a power of p, and the group is divisible by p. Thus, if one conjugacy class has size 1, at least p-1 more must also be size 1 so that the sum is divisible by p.
$endgroup$
– C Monsour
Nov 25 '18 at 15:18
add a comment |
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