Mixing
Information from index of center.
Let G be the group. Z is center of group with $|G:Z|=9$.then i have to show that $forall gin G$ ,$g^3in Z $ .
I know that if $G/Z(G)$ cyclic then abelian .But here I can only deduce that G/Z is abelian as any group of order $p^2$.I am not abel to deduce any thing further.
Any hint will be appreciated
group-theory
asked Nov 20 '18 at 16:17
Shubham
1,5951519
add a comment |
Let G be the group. Z is center of group with $|G:Z|=9$.then i have to show that $forall gin G$ ,$g^3in Z $ .
I know that if $G/Z(G)$ cyclic then abelian .But here I can only deduce that G/Z is abelian as any group of order $p^2$.I am not abel to deduce any thing further.
Any hint will be appreciated
group-theory
asked Nov 20 '18 at 16:17
Shubham
1,5951519
Look at the quotient G/Z. As you said, if it's cyclic then G is abelian, which is contradiction with |G:Z|=9. What about the order of the elements in G/Z, then?
– Pietro Gheri
Nov 20 '18 at 16:33
add a comment |
Let G be the group. Z is center of group with $|G:Z|=9$.then i have to show that $forall gin G$ ,$g^3in Z $ .
I know that if $G/Z(G)$ cyclic then abelian .But here I can only deduce that G/Z is abelian as any group of order $p^2$.I am not abel to deduce any thing further.
Any hint will be appreciated
group-theory
asked Nov 20 '18 at 16:17
Shubham
1,5951519
Let G be the group. Z is center of group with $|G:Z|=9$.then i have to show that $forall gin G$ ,$g^3in Z $ .
I know that if $G/Z(G)$ cyclic then abelian .But here I can only deduce that G/Z is abelian as any group of order $p^2$.I am not abel to deduce any thing further.
Any hint will be appreciated
group-theory
group-theory
asked Nov 20 '18 at 16:17
Shubham
1,5951519
asked Nov 20 '18 at 16:17
Shubham
1,5951519
edited Nov 20 '18 at 16:24
asked Nov 20 '18 at 16:17
Shubham
1,5951519
asked Nov 20 '18 at 16:17
Shubham
1,5951519
asked Nov 20 '18 at 16:17
Shubham
1,5951519
1,5951519
Look at the quotient G/Z. As you said, if it's cyclic then G is abelian, which is contradiction with |G:Z|=9. What about the order of the elements in G/Z, then?
– Pietro Gheri
Nov 20 '18 at 16:33
add a comment |
Look at the quotient G/Z. As you said, if it's cyclic then G is abelian, which is contradiction with |G:Z|=9. What about the order of the elements in G/Z, then?
– Pietro Gheri
Nov 20 '18 at 16:33
Look at the quotient G/Z. As you said, if it's cyclic then G is abelian, which is contradiction with |G:Z|=9. What about the order of the elements in G/Z, then?
– Pietro Gheri
Nov 20 '18 at 16:33
Look at the quotient G/Z. As you said, if it's cyclic then G is abelian, which is contradiction with |G:Z|=9. What about the order of the elements in G/Z, then?
– Pietro Gheri
Nov 20 '18 at 16:33
add a comment |
1 Answer
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Hint: $G/Z(G)$ is an abelian group of order $p^2=3^2$. So it is either isomorphic to $Bbb{Z}/3times Bbb{Z}/3$, or it is cyclic and isomorphic to $Bbb{Z}/9$. In the second case we have $G=Z(G)$ as you said, which is a contradiction to $(G:Z(G))=9$.
A related result (but different): Suppose that $N$ is a normal subgroup of $G$ and $|G/N|= m$. Then we have $g^min N$ for all $gin G$. The proof uses the order of elements in $G/N$.
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
add a comment |
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Hint: $G/Z(G)$ is an abelian group of order $p^2=3^2$. So it is either isomorphic to $Bbb{Z}/3times Bbb{Z}/3$, or it is cyclic and isomorphic to $Bbb{Z}/9$. In the second case we have $G=Z(G)$ as you said, which is a contradiction to $(G:Z(G))=9$.
A related result (but different): Suppose that $N$ is a normal subgroup of $G$ and $|G/N|= m$. Then we have $g^min N$ for all $gin G$. The proof uses the order of elements in $G/N$.
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
add a comment |
Hint: $G/Z(G)$ is an abelian group of order $p^2=3^2$. So it is either isomorphic to $Bbb{Z}/3times Bbb{Z}/3$, or it is cyclic and isomorphic to $Bbb{Z}/9$. In the second case we have $G=Z(G)$ as you said, which is a contradiction to $(G:Z(G))=9$.
A related result (but different): Suppose that $N$ is a normal subgroup of $G$ and $|G/N|= m$. Then we have $g^min N$ for all $gin G$. The proof uses the order of elements in $G/N$.
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
add a comment |
Hint: $G/Z(G)$ is an abelian group of order $p^2=3^2$. So it is either isomorphic to $Bbb{Z}/3times Bbb{Z}/3$, or it is cyclic and isomorphic to $Bbb{Z}/9$. In the second case we have $G=Z(G)$ as you said, which is a contradiction to $(G:Z(G))=9$.
A related result (but different): Suppose that $N$ is a normal subgroup of $G$ and $|G/N|= m$. Then we have $g^min N$ for all $gin G$. The proof uses the order of elements in $G/N$.
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
Hint: $G/Z(G)$ is an abelian group of order $p^2=3^2$. So it is either isomorphic to $Bbb{Z}/3times Bbb{Z}/3$, or it is cyclic and isomorphic to $Bbb{Z}/9$. In the second case we have $G=Z(G)$ as you said, which is a contradiction to $(G:Z(G))=9$.
A related result (but different): Suppose that $N$ is a normal subgroup of $G$ and $|G/N|= m$. Then we have $g^min N$ for all $gin G$. The proof uses the order of elements in $G/N$.
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
edited Nov 20 '18 at 16:37
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
answered Nov 20 '18 at 16:31
Dietrich Burde
77.8k64386
77.8k64386
add a comment |
add a comment |
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Look at the quotient G/Z. As you said, if it's cyclic then G is abelian, which is contradiction with |G:Z|=9. What about the order of the elements in G/Z, then?
– Pietro Gheri
Nov 20 '18 at 16:33