Notational confusion about HNN-extensions: $G=K ast_{H,t}$.
$begingroup$
Let $G=K ast_{H,t}$ denote an HNN-extension, i.e., $$H le K le G, H^t le K.$$
Is it true that ${K, t}$ is a generating system for $G$? In particular, $G/K$ is cyclic?
group-theory notation group-presentation quotient-group combinatorial-group-theory
edited Dec 24 '18 at 11:42
user1729
17.3k64193
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
$endgroup$
add a comment |
$begingroup$
Let $G=K ast_{H,t}$ denote an HNN-extension, i.e., $$H le K le G, H^t le K.$$
Is it true that ${K, t}$ is a generating system for $G$? In particular, $G/K$ is cyclic?
group-theory notation group-presentation quotient-group combinatorial-group-theory
edited Dec 24 '18 at 11:42
user1729
17.3k64193
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
$endgroup$
add a comment |
$begingroup$
Let $G=K ast_{H,t}$ denote an HNN-extension, i.e., $$H le K le G, H^t le K.$$
Is it true that ${K, t}$ is a generating system for $G$? In particular, $G/K$ is cyclic?
group-theory notation group-presentation quotient-group combinatorial-group-theory
edited Dec 24 '18 at 11:42
user1729
17.3k64193
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
$endgroup$
Let $G=K ast_{H,t}$ denote an HNN-extension, i.e., $$H le K le G, H^t le K.$$
Is it true that ${K, t}$ is a generating system for $G$? In particular, $G/K$ is cyclic?
group-theory notation group-presentation quotient-group combinatorial-group-theory
group-theory notation group-presentation quotient-group combinatorial-group-theory
edited Dec 24 '18 at 11:42
user1729
17.3k64193
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
edited Dec 24 '18 at 11:42
user1729
17.3k64193
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
edited Dec 24 '18 at 11:42
user1729
17.3k64193
edited Dec 24 '18 at 11:42
user1729
17.3k64193
edited Dec 24 '18 at 11:42
user1729
17.3k64193
17.3k64193
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
asked Mar 11 '18 at 18:20
JavelinaJavelina
153
153
add a comment |
add a comment |
1 Answer
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$begingroup$
An HNN-extension is not a triple $K_{H, t}sim(K, H, t)$, but instead a $5$-tuple $(K, H, H', phi, t)$ where $H, H'leq K$ and $phi$ is an isomorphism $phi: Hrightarrow H'$. HNN-extensions are usually given by the relative presentation:$$G=langle K, tmid t^{-1}ht=phi(h):forall hin Hrangle.$$ Sometimes the notation $G=K_{H^t=H'}$ is used (and we all assume that the isomorphism $phi$ exists but that its precise definition doesn't matter), and possibly sometimes $G=K_{H^t=phi(H)}$. See the book Lyndon and Schupp, Combinatorial group theory for more details.
So yes, ${K, t}$ is a generating system for $G$. This is basically by definition.
On the other hand, in general $G/K$ is a set of cosets without any group structure! Hence, it cannot be a cyclic group. This is because $K$ is not necessarily normal in $G$. Indeed, the follows holds:
Lemma 1. Let $G$ be an HNN-extension as above. Then $K$ is a normal subgroup of $G$ if and only if $K=H=H'$, whence $phiinoperatorname{Aut}(K)$ and $G$ is the mapping torus of $phi$, so $G=Krtimes_{phi}mathbb{Z}$.
The proof of this lemma is an easy consequence of Britton's lemma, or of Bass-Serre theory. The main step is showing that if $Klhd G$ then $K=H=H'$, but if $Klhd G$ then for all $kin K$ we have $t^{-epsilon}kt^{epsilon}in K$ for $epsilon=pm1$, so for all $kin K$ we have $t^{-epsilon}kt^{epsilon}k'=1$. Now apply your favourite machinery, whilst exploiting the phrase "for all".
On the other hand, the following lemma holds.
Lemma 2. Let $G$ be an HNN-extension as above, and let $langlelangle Kranglerangle$ denote the normal closure in $G$ of the subgroup $K$. Then $G/langlelangle Kranglerangle$ is infinite cyclic.
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
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add a comment |
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$begingroup$
An HNN-extension is not a triple $K_{H, t}sim(K, H, t)$, but instead a $5$-tuple $(K, H, H', phi, t)$ where $H, H'leq K$ and $phi$ is an isomorphism $phi: Hrightarrow H'$. HNN-extensions are usually given by the relative presentation:$$G=langle K, tmid t^{-1}ht=phi(h):forall hin Hrangle.$$ Sometimes the notation $G=K_{H^t=H'}$ is used (and we all assume that the isomorphism $phi$ exists but that its precise definition doesn't matter), and possibly sometimes $G=K_{H^t=phi(H)}$. See the book Lyndon and Schupp, Combinatorial group theory for more details.
So yes, ${K, t}$ is a generating system for $G$. This is basically by definition.
On the other hand, in general $G/K$ is a set of cosets without any group structure! Hence, it cannot be a cyclic group. This is because $K$ is not necessarily normal in $G$. Indeed, the follows holds:
Lemma 1. Let $G$ be an HNN-extension as above. Then $K$ is a normal subgroup of $G$ if and only if $K=H=H'$, whence $phiinoperatorname{Aut}(K)$ and $G$ is the mapping torus of $phi$, so $G=Krtimes_{phi}mathbb{Z}$.
The proof of this lemma is an easy consequence of Britton's lemma, or of Bass-Serre theory. The main step is showing that if $Klhd G$ then $K=H=H'$, but if $Klhd G$ then for all $kin K$ we have $t^{-epsilon}kt^{epsilon}in K$ for $epsilon=pm1$, so for all $kin K$ we have $t^{-epsilon}kt^{epsilon}k'=1$. Now apply your favourite machinery, whilst exploiting the phrase "for all".
On the other hand, the following lemma holds.
Lemma 2. Let $G$ be an HNN-extension as above, and let $langlelangle Kranglerangle$ denote the normal closure in $G$ of the subgroup $K$. Then $G/langlelangle Kranglerangle$ is infinite cyclic.
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
$endgroup$
add a comment |
$begingroup$
An HNN-extension is not a triple $K_{H, t}sim(K, H, t)$, but instead a $5$-tuple $(K, H, H', phi, t)$ where $H, H'leq K$ and $phi$ is an isomorphism $phi: Hrightarrow H'$. HNN-extensions are usually given by the relative presentation:$$G=langle K, tmid t^{-1}ht=phi(h):forall hin Hrangle.$$ Sometimes the notation $G=K_{H^t=H'}$ is used (and we all assume that the isomorphism $phi$ exists but that its precise definition doesn't matter), and possibly sometimes $G=K_{H^t=phi(H)}$. See the book Lyndon and Schupp, Combinatorial group theory for more details.
So yes, ${K, t}$ is a generating system for $G$. This is basically by definition.
On the other hand, in general $G/K$ is a set of cosets without any group structure! Hence, it cannot be a cyclic group. This is because $K$ is not necessarily normal in $G$. Indeed, the follows holds:
Lemma 1. Let $G$ be an HNN-extension as above. Then $K$ is a normal subgroup of $G$ if and only if $K=H=H'$, whence $phiinoperatorname{Aut}(K)$ and $G$ is the mapping torus of $phi$, so $G=Krtimes_{phi}mathbb{Z}$.
The proof of this lemma is an easy consequence of Britton's lemma, or of Bass-Serre theory. The main step is showing that if $Klhd G$ then $K=H=H'$, but if $Klhd G$ then for all $kin K$ we have $t^{-epsilon}kt^{epsilon}in K$ for $epsilon=pm1$, so for all $kin K$ we have $t^{-epsilon}kt^{epsilon}k'=1$. Now apply your favourite machinery, whilst exploiting the phrase "for all".
On the other hand, the following lemma holds.
Lemma 2. Let $G$ be an HNN-extension as above, and let $langlelangle Kranglerangle$ denote the normal closure in $G$ of the subgroup $K$. Then $G/langlelangle Kranglerangle$ is infinite cyclic.
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
$endgroup$
add a comment |
$begingroup$
An HNN-extension is not a triple $K_{H, t}sim(K, H, t)$, but instead a $5$-tuple $(K, H, H', phi, t)$ where $H, H'leq K$ and $phi$ is an isomorphism $phi: Hrightarrow H'$. HNN-extensions are usually given by the relative presentation:$$G=langle K, tmid t^{-1}ht=phi(h):forall hin Hrangle.$$ Sometimes the notation $G=K_{H^t=H'}$ is used (and we all assume that the isomorphism $phi$ exists but that its precise definition doesn't matter), and possibly sometimes $G=K_{H^t=phi(H)}$. See the book Lyndon and Schupp, Combinatorial group theory for more details.
So yes, ${K, t}$ is a generating system for $G$. This is basically by definition.
On the other hand, in general $G/K$ is a set of cosets without any group structure! Hence, it cannot be a cyclic group. This is because $K$ is not necessarily normal in $G$. Indeed, the follows holds:
Lemma 1. Let $G$ be an HNN-extension as above. Then $K$ is a normal subgroup of $G$ if and only if $K=H=H'$, whence $phiinoperatorname{Aut}(K)$ and $G$ is the mapping torus of $phi$, so $G=Krtimes_{phi}mathbb{Z}$.
The proof of this lemma is an easy consequence of Britton's lemma, or of Bass-Serre theory. The main step is showing that if $Klhd G$ then $K=H=H'$, but if $Klhd G$ then for all $kin K$ we have $t^{-epsilon}kt^{epsilon}in K$ for $epsilon=pm1$, so for all $kin K$ we have $t^{-epsilon}kt^{epsilon}k'=1$. Now apply your favourite machinery, whilst exploiting the phrase "for all".
On the other hand, the following lemma holds.
Lemma 2. Let $G$ be an HNN-extension as above, and let $langlelangle Kranglerangle$ denote the normal closure in $G$ of the subgroup $K$. Then $G/langlelangle Kranglerangle$ is infinite cyclic.
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
$endgroup$
An HNN-extension is not a triple $K_{H, t}sim(K, H, t)$, but instead a $5$-tuple $(K, H, H', phi, t)$ where $H, H'leq K$ and $phi$ is an isomorphism $phi: Hrightarrow H'$. HNN-extensions are usually given by the relative presentation:$$G=langle K, tmid t^{-1}ht=phi(h):forall hin Hrangle.$$ Sometimes the notation $G=K_{H^t=H'}$ is used (and we all assume that the isomorphism $phi$ exists but that its precise definition doesn't matter), and possibly sometimes $G=K_{H^t=phi(H)}$. See the book Lyndon and Schupp, Combinatorial group theory for more details.
So yes, ${K, t}$ is a generating system for $G$. This is basically by definition.
On the other hand, in general $G/K$ is a set of cosets without any group structure! Hence, it cannot be a cyclic group. This is because $K$ is not necessarily normal in $G$. Indeed, the follows holds:
Lemma 1. Let $G$ be an HNN-extension as above. Then $K$ is a normal subgroup of $G$ if and only if $K=H=H'$, whence $phiinoperatorname{Aut}(K)$ and $G$ is the mapping torus of $phi$, so $G=Krtimes_{phi}mathbb{Z}$.
The proof of this lemma is an easy consequence of Britton's lemma, or of Bass-Serre theory. The main step is showing that if $Klhd G$ then $K=H=H'$, but if $Klhd G$ then for all $kin K$ we have $t^{-epsilon}kt^{epsilon}in K$ for $epsilon=pm1$, so for all $kin K$ we have $t^{-epsilon}kt^{epsilon}k'=1$. Now apply your favourite machinery, whilst exploiting the phrase "for all".
On the other hand, the following lemma holds.
Lemma 2. Let $G$ be an HNN-extension as above, and let $langlelangle Kranglerangle$ denote the normal closure in $G$ of the subgroup $K$. Then $G/langlelangle Kranglerangle$ is infinite cyclic.
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
edited Dec 24 '18 at 11:36
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
answered Dec 24 '18 at 11:29
user1729user1729
17.3k64193
17.3k64193
add a comment |
add a comment |
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