How to deduce information about center of group if only order of centraliser is given?
$begingroup$
Suppose centraliser of group of element has order 4 . Then what information can we deduct about center
I know that $Z(G)subset Z(x)$ where $|Z(x)|=4$
Any group of order 4 is abelian $|Z(G)|=1,2,4$.
Now I got some examples of with above property
Like In $D_4$ or $Q_8$ $|Z(G)|=2$
In $Z_4$ ,$|Z(G)|=4$
But I could not able to find example with $|Z(G)|=1|
$ and $|Z(x)=4|$
What else info can we deduce?
Any Help will be appreciated
Edit:
As Derek says $A_4$ has $Z(G)=1$ there is (12)(43) which has 4 centraliser element
group-theory
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
$endgroup$
add a comment |
$begingroup$
Suppose centraliser of group of element has order 4 . Then what information can we deduct about center
I know that $Z(G)subset Z(x)$ where $|Z(x)|=4$
Any group of order 4 is abelian $|Z(G)|=1,2,4$.
Now I got some examples of with above property
Like In $D_4$ or $Q_8$ $|Z(G)|=2$
In $Z_4$ ,$|Z(G)|=4$
But I could not able to find example with $|Z(G)|=1|
$ and $|Z(x)=4|$
What else info can we deduce?
Any Help will be appreciated
Edit:
As Derek says $A_4$ has $Z(G)=1$ there is (12)(43) which has 4 centraliser element
group-theory
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
$endgroup$
1
$begingroup$
Try $G = A_4$ with $x$ an element of order $2$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:24
$begingroup$
I got now .But What else info can we deduct
$endgroup$
– MathLover
Dec 6 '18 at 12:30
1
$begingroup$
You have proved that $|Z(G)|=1,2,4$ are all possible. What else could there be to say?
$endgroup$
– Derek Holt
Dec 6 '18 at 12:33
1
$begingroup$
You can say that if $|Z(G)| = 4$, then $G = Z(G)$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:36
$begingroup$
If MathLover in the second line means strict inclusion (⊂), then x∉Z(G), so |Z(G)|=1 or |Z(G)|=2.
$endgroup$
– Nicky Hekster
Dec 6 '18 at 16:18
add a comment |
$begingroup$
Suppose centraliser of group of element has order 4 . Then what information can we deduct about center
I know that $Z(G)subset Z(x)$ where $|Z(x)|=4$
Any group of order 4 is abelian $|Z(G)|=1,2,4$.
Now I got some examples of with above property
Like In $D_4$ or $Q_8$ $|Z(G)|=2$
In $Z_4$ ,$|Z(G)|=4$
But I could not able to find example with $|Z(G)|=1|
$ and $|Z(x)=4|$
What else info can we deduce?
Any Help will be appreciated
Edit:
As Derek says $A_4$ has $Z(G)=1$ there is (12)(43) which has 4 centraliser element
group-theory
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
$endgroup$
Suppose centraliser of group of element has order 4 . Then what information can we deduct about center
I know that $Z(G)subset Z(x)$ where $|Z(x)|=4$
Any group of order 4 is abelian $|Z(G)|=1,2,4$.
Now I got some examples of with above property
Like In $D_4$ or $Q_8$ $|Z(G)|=2$
In $Z_4$ ,$|Z(G)|=4$
But I could not able to find example with $|Z(G)|=1|
$ and $|Z(x)=4|$
What else info can we deduce?
Any Help will be appreciated
Edit:
As Derek says $A_4$ has $Z(G)=1$ there is (12)(43) which has 4 centraliser element
group-theory
group-theory
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
edited Dec 6 '18 at 17:15
Arturo Magidin
261k34586906
261k34586906
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
asked Dec 6 '18 at 12:03
MathLoverMathLover
49310
49310
1
$begingroup$
Try $G = A_4$ with $x$ an element of order $2$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:24
$begingroup$
I got now .But What else info can we deduct
$endgroup$
– MathLover
Dec 6 '18 at 12:30
1
$begingroup$
You have proved that $|Z(G)|=1,2,4$ are all possible. What else could there be to say?
$endgroup$
– Derek Holt
Dec 6 '18 at 12:33
1
$begingroup$
You can say that if $|Z(G)| = 4$, then $G = Z(G)$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:36
$begingroup$
If MathLover in the second line means strict inclusion (⊂), then x∉Z(G), so |Z(G)|=1 or |Z(G)|=2.
$endgroup$
– Nicky Hekster
Dec 6 '18 at 16:18
add a comment |
1
$begingroup$
Try $G = A_4$ with $x$ an element of order $2$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:24
$begingroup$
I got now .But What else info can we deduct
$endgroup$
– MathLover
Dec 6 '18 at 12:30
1
$begingroup$
You have proved that $|Z(G)|=1,2,4$ are all possible. What else could there be to say?
$endgroup$
– Derek Holt
Dec 6 '18 at 12:33
1
$begingroup$
You can say that if $|Z(G)| = 4$, then $G = Z(G)$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:36
$begingroup$
If MathLover in the second line means strict inclusion (⊂), then x∉Z(G), so |Z(G)|=1 or |Z(G)|=2.
$endgroup$
– Nicky Hekster
Dec 6 '18 at 16:18
1
1
$begingroup$
Try $G = A_4$ with $x$ an element of order $2$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:24
$begingroup$
Try $G = A_4$ with $x$ an element of order $2$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:24
$begingroup$
I got now .But What else info can we deduct
$endgroup$
– MathLover
Dec 6 '18 at 12:30
$begingroup$
I got now .But What else info can we deduct
$endgroup$
– MathLover
Dec 6 '18 at 12:30
1
1
$begingroup$
You have proved that $|Z(G)|=1,2,4$ are all possible. What else could there be to say?
$endgroup$
– Derek Holt
Dec 6 '18 at 12:33
$begingroup$
You have proved that $|Z(G)|=1,2,4$ are all possible. What else could there be to say?
$endgroup$
– Derek Holt
Dec 6 '18 at 12:33
1
1
$begingroup$
You can say that if $|Z(G)| = 4$, then $G = Z(G)$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:36
$begingroup$
You can say that if $|Z(G)| = 4$, then $G = Z(G)$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:36
$begingroup$
If MathLover in the second line means strict inclusion (⊂), then x∉Z(G), so |Z(G)|=1 or |Z(G)|=2.
$endgroup$
– Nicky Hekster
Dec 6 '18 at 16:18
$begingroup$
If MathLover in the second line means strict inclusion (⊂), then x∉Z(G), so |Z(G)|=1 or |Z(G)|=2.
$endgroup$
– Nicky Hekster
Dec 6 '18 at 16:18
add a comment |
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1
$begingroup$
Try $G = A_4$ with $x$ an element of order $2$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:24
$begingroup$
I got now .But What else info can we deduct
$endgroup$
– MathLover
Dec 6 '18 at 12:30
1
$begingroup$
You have proved that $|Z(G)|=1,2,4$ are all possible. What else could there be to say?
$endgroup$
– Derek Holt
Dec 6 '18 at 12:33
1
$begingroup$
You can say that if $|Z(G)| = 4$, then $G = Z(G)$.
$endgroup$
– Derek Holt
Dec 6 '18 at 12:36
$begingroup$
If MathLover in the second line means strict inclusion (⊂), then x∉Z(G), so |Z(G)|=1 or |Z(G)|=2.
$endgroup$
– Nicky Hekster
Dec 6 '18 at 16:18