Commutator subgroup of Heisenberg group.Commutator subgroup problemIs there a nonvirtually abelian group whose commutator subgroup is finite?Subgroup of the quotient group of the commutator G'.Commutator Subgroup is finitely generatedCommutator subgroup is the minimal normal subgroup such that quotient group is abelianIs the commutator subgroup of a commutator subgroup itself?When is a quotient group abelian?Problem on commutator subgroupLet $H$ be the Heisenberg group. Determine the center $Z(H)$ of $H$. Show that the quotient group $H/Z(H)$ is abelian.Proving that a quotient is virtually a nilpotent group
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Commutator subgroup of Heisenberg group.
Commutator subgroup problemIs there a nonvirtually abelian group whose commutator subgroup is finite?Subgroup of the quotient group of the commutator G'.Commutator Subgroup is finitely generatedCommutator subgroup is the minimal normal subgroup such that quotient group is abelianIs the commutator subgroup of a commutator subgroup itself?When is a quotient group abelian?Problem on commutator subgroupLet $H$ be the Heisenberg group. Determine the center $Z(H)$ of $H$. Show that the quotient group $H/Z(H)$ is abelian.Proving that a quotient is virtually a nilpotent group
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Dears,
Let $H$ be Heisenberg group, a group of $3times 3$ matrices with $1$ on the main diagonal, zeros below, and elements of $Bbb R$ above the main diagonal.
Its center is the subgroup of all matrices with $0$ on the first diagonal above the main diagonal.
My question is - is it also a commutator subgroup of that group?
The quotient group $H/Z(H)$ is abelian (this group is nilpotent of class two), so commutator subgroup must be inside the center. I can't imagine other subgroups being CS, but I want someone smarter to let me know.
Have a nice day.
abstract-algebra group-theory quotient-group nilpotent-groups heisenberg-group
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
$endgroup$
add a comment |
$begingroup$
Dears,
Let $H$ be Heisenberg group, a group of $3times 3$ matrices with $1$ on the main diagonal, zeros below, and elements of $Bbb R$ above the main diagonal.
Its center is the subgroup of all matrices with $0$ on the first diagonal above the main diagonal.
My question is - is it also a commutator subgroup of that group?
The quotient group $H/Z(H)$ is abelian (this group is nilpotent of class two), so commutator subgroup must be inside the center. I can't imagine other subgroups being CS, but I want someone smarter to let me know.
Have a nice day.
abstract-algebra group-theory quotient-group nilpotent-groups heisenberg-group
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
$endgroup$
add a comment |
$begingroup$
Dears,
Let $H$ be Heisenberg group, a group of $3times 3$ matrices with $1$ on the main diagonal, zeros below, and elements of $Bbb R$ above the main diagonal.
Its center is the subgroup of all matrices with $0$ on the first diagonal above the main diagonal.
My question is - is it also a commutator subgroup of that group?
The quotient group $H/Z(H)$ is abelian (this group is nilpotent of class two), so commutator subgroup must be inside the center. I can't imagine other subgroups being CS, but I want someone smarter to let me know.
Have a nice day.
abstract-algebra group-theory quotient-group nilpotent-groups heisenberg-group
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
$endgroup$
Dears,
Let $H$ be Heisenberg group, a group of $3times 3$ matrices with $1$ on the main diagonal, zeros below, and elements of $Bbb R$ above the main diagonal.
Its center is the subgroup of all matrices with $0$ on the first diagonal above the main diagonal.
My question is - is it also a commutator subgroup of that group?
The quotient group $H/Z(H)$ is abelian (this group is nilpotent of class two), so commutator subgroup must be inside the center. I can't imagine other subgroups being CS, but I want someone smarter to let me know.
Have a nice day.
abstract-algebra group-theory quotient-group nilpotent-groups heisenberg-group
abstract-algebra group-theory quotient-group nilpotent-groups heisenberg-group
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
edited 8 hours ago
Maciej Ficek
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
asked 9 hours ago
Maciej FicekMaciej Ficek
276 bronze badges
276 bronze badges
add a comment |
add a comment |
2 Answers
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$begingroup$
$$beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrix^-1=beginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrix$$
Therefore beginalignABA^-1B^-1&=beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrixbeginpmatrix1&s&t\ 0&1&u\ 0&0&1endpmatrixbeginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrixbeginpmatrix1&-s&su-t\ 0&1&-u\ 0&0&1endpmatrix=\&=beginpmatrix1&s+x&t+ux+y\ 0&1&u+z\ 0&0&1endpmatrixbeginpmatrix1&-x-s&ux+xz-y+su-t\ 0&1&-z-u\ 0&0&1endpmatrix=\&=beginpmatrix1&0&t+ux+y+ux+xz-y+su-t-(z+u)(s+x)\ 0&1&0\ 0&0&1endpmatrix=\&=beginpmatrix1&0&ux-zs\ 0&1&0\ 0&0&1endpmatrixendalign
It is therefore apparent that commmutators are exactly the elements in the form $beginpmatrix1&0&alpha\0&1&0\ 0&0&1endpmatrix$, which incidentally form a subgroup.
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
Since$$beginbmatrix0&a&c\0&1&b\0&0&1endbmatrix.beginbmatrix1&d&f\0&1&e\0&0&1endbmatrixbeginbmatrix0&a&c\0&1&b\0&0&1endbmatrix^-1beginbmatrix1&d&f\0&1&e\0&0&1endbmatrix^-1=beginbmatrix0 & 0 & a e-b d \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix,$$yes, the commutator group is the center.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
add a comment |
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$begingroup$
$$beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrix^-1=beginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrix$$
Therefore beginalignABA^-1B^-1&=beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrixbeginpmatrix1&s&t\ 0&1&u\ 0&0&1endpmatrixbeginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrixbeginpmatrix1&-s&su-t\ 0&1&-u\ 0&0&1endpmatrix=\&=beginpmatrix1&s+x&t+ux+y\ 0&1&u+z\ 0&0&1endpmatrixbeginpmatrix1&-x-s&ux+xz-y+su-t\ 0&1&-z-u\ 0&0&1endpmatrix=\&=beginpmatrix1&0&t+ux+y+ux+xz-y+su-t-(z+u)(s+x)\ 0&1&0\ 0&0&1endpmatrix=\&=beginpmatrix1&0&ux-zs\ 0&1&0\ 0&0&1endpmatrixendalign
It is therefore apparent that commmutators are exactly the elements in the form $beginpmatrix1&0&alpha\0&1&0\ 0&0&1endpmatrix$, which incidentally form a subgroup.
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
$$beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrix^-1=beginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrix$$
Therefore beginalignABA^-1B^-1&=beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrixbeginpmatrix1&s&t\ 0&1&u\ 0&0&1endpmatrixbeginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrixbeginpmatrix1&-s&su-t\ 0&1&-u\ 0&0&1endpmatrix=\&=beginpmatrix1&s+x&t+ux+y\ 0&1&u+z\ 0&0&1endpmatrixbeginpmatrix1&-x-s&ux+xz-y+su-t\ 0&1&-z-u\ 0&0&1endpmatrix=\&=beginpmatrix1&0&t+ux+y+ux+xz-y+su-t-(z+u)(s+x)\ 0&1&0\ 0&0&1endpmatrix=\&=beginpmatrix1&0&ux-zs\ 0&1&0\ 0&0&1endpmatrixendalign
It is therefore apparent that commmutators are exactly the elements in the form $beginpmatrix1&0&alpha\0&1&0\ 0&0&1endpmatrix$, which incidentally form a subgroup.
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
$endgroup$
add a comment |
$begingroup$
$$beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrix^-1=beginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrix$$
Therefore beginalignABA^-1B^-1&=beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrixbeginpmatrix1&s&t\ 0&1&u\ 0&0&1endpmatrixbeginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrixbeginpmatrix1&-s&su-t\ 0&1&-u\ 0&0&1endpmatrix=\&=beginpmatrix1&s+x&t+ux+y\ 0&1&u+z\ 0&0&1endpmatrixbeginpmatrix1&-x-s&ux+xz-y+su-t\ 0&1&-z-u\ 0&0&1endpmatrix=\&=beginpmatrix1&0&t+ux+y+ux+xz-y+su-t-(z+u)(s+x)\ 0&1&0\ 0&0&1endpmatrix=\&=beginpmatrix1&0&ux-zs\ 0&1&0\ 0&0&1endpmatrixendalign
It is therefore apparent that commmutators are exactly the elements in the form $beginpmatrix1&0&alpha\0&1&0\ 0&0&1endpmatrix$, which incidentally form a subgroup.
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
$endgroup$
$$beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrix^-1=beginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrix$$
Therefore beginalignABA^-1B^-1&=beginpmatrix1&x&y\ 0&1&z\ 0&0&1endpmatrixbeginpmatrix1&s&t\ 0&1&u\ 0&0&1endpmatrixbeginpmatrix1&-x&xz-y\ 0&1&-z\ 0&0&1endpmatrixbeginpmatrix1&-s&su-t\ 0&1&-u\ 0&0&1endpmatrix=\&=beginpmatrix1&s+x&t+ux+y\ 0&1&u+z\ 0&0&1endpmatrixbeginpmatrix1&-x-s&ux+xz-y+su-t\ 0&1&-z-u\ 0&0&1endpmatrix=\&=beginpmatrix1&0&t+ux+y+ux+xz-y+su-t-(z+u)(s+x)\ 0&1&0\ 0&0&1endpmatrix=\&=beginpmatrix1&0&ux-zs\ 0&1&0\ 0&0&1endpmatrixendalign
It is therefore apparent that commmutators are exactly the elements in the form $beginpmatrix1&0&alpha\0&1&0\ 0&0&1endpmatrix$, which incidentally form a subgroup.
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
edited 6 hours ago
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
answered 8 hours ago
Gae. S.Gae. S.
6073 silver badges14 bronze badges
6073 silver badges14 bronze badges
add a comment |
add a comment |
$begingroup$
Since$$beginbmatrix0&a&c\0&1&b\0&0&1endbmatrix.beginbmatrix1&d&f\0&1&e\0&0&1endbmatrixbeginbmatrix0&a&c\0&1&b\0&0&1endbmatrix^-1beginbmatrix1&d&f\0&1&e\0&0&1endbmatrix^-1=beginbmatrix0 & 0 & a e-b d \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix,$$yes, the commutator group is the center.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
add a comment |
$begingroup$
Since$$beginbmatrix0&a&c\0&1&b\0&0&1endbmatrix.beginbmatrix1&d&f\0&1&e\0&0&1endbmatrixbeginbmatrix0&a&c\0&1&b\0&0&1endbmatrix^-1beginbmatrix1&d&f\0&1&e\0&0&1endbmatrix^-1=beginbmatrix0 & 0 & a e-b d \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix,$$yes, the commutator group is the center.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
add a comment |
$begingroup$
Since$$beginbmatrix0&a&c\0&1&b\0&0&1endbmatrix.beginbmatrix1&d&f\0&1&e\0&0&1endbmatrixbeginbmatrix0&a&c\0&1&b\0&0&1endbmatrix^-1beginbmatrix1&d&f\0&1&e\0&0&1endbmatrix^-1=beginbmatrix0 & 0 & a e-b d \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix,$$yes, the commutator group is the center.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
$endgroup$
Since$$beginbmatrix0&a&c\0&1&b\0&0&1endbmatrix.beginbmatrix1&d&f\0&1&e\0&0&1endbmatrixbeginbmatrix0&a&c\0&1&b\0&0&1endbmatrix^-1beginbmatrix1&d&f\0&1&e\0&0&1endbmatrix^-1=beginbmatrix0 & 0 & a e-b d \ 0 & 0 & 0 \ 0 & 0 & 0endbmatrix,$$yes, the commutator group is the center.
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
edited 8 hours ago
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
answered 8 hours ago
José Carlos SantosJosé Carlos Santos
199k25 gold badges158 silver badges276 bronze badges
199k25 gold badges158 silver badges276 bronze badges
add a comment |
add a comment |
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