Mixing
What are the semi-direct products of $mathbb{Z}$ with itself? (Check my work please)
$begingroup$
I am just starting out with semi-direct products. I would like to list and describe the semi-direct products of $mathbb{Z}$ with itself.
I first need to find the automorphisms $varphi$ from $mathbb{Z}$ to $mathbb{Z}$. These automorphisms are determined by $varphi(1)$ by the additive property of such morphisms : $varphi(n+m)=varphi(n)+varphi(m).$ Therefore, I only need to determine the possible values of $varphi(1)$. But I know that an automorphism must send generators on generators, and the generators of $mathbb{Z}$ are $pm1$, so there are at most two automorphisms : $Id:x mapsto x$ and $-Id:x mapsto -x$ (and they are automorphisms, so these are the only ones).
Therefore, a semi-direct product $mathbb{Z} rtimes_{psi}mathbb{Z}$ is given by a morphism $psi:mathbb{Z}rightarrow Aut({mathbb{Z}}) cong mathbb{Z}/2mathbb{Z}$. The only possibilites, if I didn't make any mistakes, are $psi:n rightarrow Id$ (constant morphism) and $psi:nrightarrow (-1)^nId$ ($psi$ is determined by $psi(1)$ which is either $Id$ yielding the constant morphism or $-Id$ which yields the second one.)
So there are only two semi-direct products possible, one of which is the direct product (I guess?).
$mathbb{Z} rtimes_{n rightarrow id}mathbb{Z}congmathbb{Z}^2$
and
$mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$
Did I miss any of them, and is there anything to say about this last semi-direct product? Is $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$ isomorphic to any known groups? I'm not sure what to say now, I should give it a brief description but I don't know what else there is to say than $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$. Its multiplicative law is $(x,y)*(z,t)=(x+psi(y)(z),z+t)=(x+(-1)^yz,z+t)$, so it kind of "oscillates" and it doesn't sound familiar to me... I don't recall any groups with such a weird multiplication law. It's not even commutative I guess, since $(1,1)*(2,2)=(1-2,4)=(-1,4)$ but $(2,2)*(1,1)=(2+1,2)=(3,2) neq(-1,4)...$
abstract-algebra group-theory proof-verification semidirect-product
edited Jan 30 at 22:52
Shaun
10.3k113686
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
$endgroup$
add a comment |
$begingroup$
I am just starting out with semi-direct products. I would like to list and describe the semi-direct products of $mathbb{Z}$ with itself.
I first need to find the automorphisms $varphi$ from $mathbb{Z}$ to $mathbb{Z}$. These automorphisms are determined by $varphi(1)$ by the additive property of such morphisms : $varphi(n+m)=varphi(n)+varphi(m).$ Therefore, I only need to determine the possible values of $varphi(1)$. But I know that an automorphism must send generators on generators, and the generators of $mathbb{Z}$ are $pm1$, so there are at most two automorphisms : $Id:x mapsto x$ and $-Id:x mapsto -x$ (and they are automorphisms, so these are the only ones).
Therefore, a semi-direct product $mathbb{Z} rtimes_{psi}mathbb{Z}$ is given by a morphism $psi:mathbb{Z}rightarrow Aut({mathbb{Z}}) cong mathbb{Z}/2mathbb{Z}$. The only possibilites, if I didn't make any mistakes, are $psi:n rightarrow Id$ (constant morphism) and $psi:nrightarrow (-1)^nId$ ($psi$ is determined by $psi(1)$ which is either $Id$ yielding the constant morphism or $-Id$ which yields the second one.)
So there are only two semi-direct products possible, one of which is the direct product (I guess?).
$mathbb{Z} rtimes_{n rightarrow id}mathbb{Z}congmathbb{Z}^2$
and
$mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$
Did I miss any of them, and is there anything to say about this last semi-direct product? Is $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$ isomorphic to any known groups? I'm not sure what to say now, I should give it a brief description but I don't know what else there is to say than $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$. Its multiplicative law is $(x,y)*(z,t)=(x+psi(y)(z),z+t)=(x+(-1)^yz,z+t)$, so it kind of "oscillates" and it doesn't sound familiar to me... I don't recall any groups with such a weird multiplication law. It's not even commutative I guess, since $(1,1)*(2,2)=(1-2,4)=(-1,4)$ but $(2,2)*(1,1)=(2+1,2)=(3,2) neq(-1,4)...$
abstract-algebra group-theory proof-verification semidirect-product
edited Jan 30 at 22:52
Shaun
10.3k113686
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
$endgroup$
add a comment |
$begingroup$
I am just starting out with semi-direct products. I would like to list and describe the semi-direct products of $mathbb{Z}$ with itself.
I first need to find the automorphisms $varphi$ from $mathbb{Z}$ to $mathbb{Z}$. These automorphisms are determined by $varphi(1)$ by the additive property of such morphisms : $varphi(n+m)=varphi(n)+varphi(m).$ Therefore, I only need to determine the possible values of $varphi(1)$. But I know that an automorphism must send generators on generators, and the generators of $mathbb{Z}$ are $pm1$, so there are at most two automorphisms : $Id:x mapsto x$ and $-Id:x mapsto -x$ (and they are automorphisms, so these are the only ones).
Therefore, a semi-direct product $mathbb{Z} rtimes_{psi}mathbb{Z}$ is given by a morphism $psi:mathbb{Z}rightarrow Aut({mathbb{Z}}) cong mathbb{Z}/2mathbb{Z}$. The only possibilites, if I didn't make any mistakes, are $psi:n rightarrow Id$ (constant morphism) and $psi:nrightarrow (-1)^nId$ ($psi$ is determined by $psi(1)$ which is either $Id$ yielding the constant morphism or $-Id$ which yields the second one.)
So there are only two semi-direct products possible, one of which is the direct product (I guess?).
$mathbb{Z} rtimes_{n rightarrow id}mathbb{Z}congmathbb{Z}^2$
and
$mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$
Did I miss any of them, and is there anything to say about this last semi-direct product? Is $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$ isomorphic to any known groups? I'm not sure what to say now, I should give it a brief description but I don't know what else there is to say than $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$. Its multiplicative law is $(x,y)*(z,t)=(x+psi(y)(z),z+t)=(x+(-1)^yz,z+t)$, so it kind of "oscillates" and it doesn't sound familiar to me... I don't recall any groups with such a weird multiplication law. It's not even commutative I guess, since $(1,1)*(2,2)=(1-2,4)=(-1,4)$ but $(2,2)*(1,1)=(2+1,2)=(3,2) neq(-1,4)...$
abstract-algebra group-theory proof-verification semidirect-product
edited Jan 30 at 22:52
Shaun
10.3k113686
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
$endgroup$
I am just starting out with semi-direct products. I would like to list and describe the semi-direct products of $mathbb{Z}$ with itself.
I first need to find the automorphisms $varphi$ from $mathbb{Z}$ to $mathbb{Z}$. These automorphisms are determined by $varphi(1)$ by the additive property of such morphisms : $varphi(n+m)=varphi(n)+varphi(m).$ Therefore, I only need to determine the possible values of $varphi(1)$. But I know that an automorphism must send generators on generators, and the generators of $mathbb{Z}$ are $pm1$, so there are at most two automorphisms : $Id:x mapsto x$ and $-Id:x mapsto -x$ (and they are automorphisms, so these are the only ones).
Therefore, a semi-direct product $mathbb{Z} rtimes_{psi}mathbb{Z}$ is given by a morphism $psi:mathbb{Z}rightarrow Aut({mathbb{Z}}) cong mathbb{Z}/2mathbb{Z}$. The only possibilites, if I didn't make any mistakes, are $psi:n rightarrow Id$ (constant morphism) and $psi:nrightarrow (-1)^nId$ ($psi$ is determined by $psi(1)$ which is either $Id$ yielding the constant morphism or $-Id$ which yields the second one.)
So there are only two semi-direct products possible, one of which is the direct product (I guess?).
$mathbb{Z} rtimes_{n rightarrow id}mathbb{Z}congmathbb{Z}^2$
and
$mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$
Did I miss any of them, and is there anything to say about this last semi-direct product? Is $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$ isomorphic to any known groups? I'm not sure what to say now, I should give it a brief description but I don't know what else there is to say than $mathbb{Z} rtimes_{n rightarrow (-1)^nId} mathbb{Z}$. Its multiplicative law is $(x,y)*(z,t)=(x+psi(y)(z),z+t)=(x+(-1)^yz,z+t)$, so it kind of "oscillates" and it doesn't sound familiar to me... I don't recall any groups with such a weird multiplication law. It's not even commutative I guess, since $(1,1)*(2,2)=(1-2,4)=(-1,4)$ but $(2,2)*(1,1)=(2+1,2)=(3,2) neq(-1,4)...$
abstract-algebra group-theory proof-verification semidirect-product
abstract-algebra group-theory proof-verification semidirect-product
edited Jan 30 at 22:52
Shaun
10.3k113686
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
edited Jan 30 at 22:52
Shaun
10.3k113686
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
edited Jan 30 at 22:52
Shaun
10.3k113686
edited Jan 30 at 22:52
Shaun
10.3k113686
edited Jan 30 at 22:52
Shaun
10.3k113686
10.3k113686
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
asked Jan 30 at 21:18
EvaristeEvariste
1,1862618
1,1862618
add a comment |
add a comment |
1 Answer
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$begingroup$
Your work is correct. The first semidirect product is indeed just the direct product: a semidirect product is just the direct product if the homomorphism used is trivial (since then the group operation on the semidirect product reduces to just the usual coordinatewise operation).
I don't really know a common name for the second semidirect product. Another way to think of it is via a presentation: it is generated by two elements $a$ and $b$ with the relation $bab^{-1}=a^{-1}$ (here $a=(1,0)$ and $b=(0,1)$). This group arises naturally in topology: it is isomorphic to the fundamental group of the Klein bottle.
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
$endgroup$
$begingroup$
The first part was just meant to determine $Aut(mathbb{Z})$. The homomorphism from $mathbb{Z}$ to $Aut(mathbb{Z})$ was meant to be $psi$. I agree this was confusing. Thank you very much for your insight, especially the last part!
$endgroup$
– Evariste
Jan 30 at 21:41
$begingroup$
Ah, that makes sense. Normally we don't talk about "automorphisms from ... to ..." but instead just say "automorphisms of ..." since "automorphism" already means the domain and codomain are the same.
$endgroup$
– Eric Wofsey
Jan 30 at 21:43
$begingroup$
For the reasons you say, topologists refer to the second semidirect product as the Klein bottle group. Crystallographers refer to it in their cryptic notation as pg.
$endgroup$
– Lee Mosher
Jan 31 at 0:58
add a comment |
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$begingroup$
Your work is correct. The first semidirect product is indeed just the direct product: a semidirect product is just the direct product if the homomorphism used is trivial (since then the group operation on the semidirect product reduces to just the usual coordinatewise operation).
I don't really know a common name for the second semidirect product. Another way to think of it is via a presentation: it is generated by two elements $a$ and $b$ with the relation $bab^{-1}=a^{-1}$ (here $a=(1,0)$ and $b=(0,1)$). This group arises naturally in topology: it is isomorphic to the fundamental group of the Klein bottle.
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
$endgroup$
$begingroup$
The first part was just meant to determine $Aut(mathbb{Z})$. The homomorphism from $mathbb{Z}$ to $Aut(mathbb{Z})$ was meant to be $psi$. I agree this was confusing. Thank you very much for your insight, especially the last part!
$endgroup$
– Evariste
Jan 30 at 21:41
$begingroup$
Ah, that makes sense. Normally we don't talk about "automorphisms from ... to ..." but instead just say "automorphisms of ..." since "automorphism" already means the domain and codomain are the same.
$endgroup$
– Eric Wofsey
Jan 30 at 21:43
$begingroup$
For the reasons you say, topologists refer to the second semidirect product as the Klein bottle group. Crystallographers refer to it in their cryptic notation as pg.
$endgroup$
– Lee Mosher
Jan 31 at 0:58
add a comment |
$begingroup$
Your work is correct. The first semidirect product is indeed just the direct product: a semidirect product is just the direct product if the homomorphism used is trivial (since then the group operation on the semidirect product reduces to just the usual coordinatewise operation).
I don't really know a common name for the second semidirect product. Another way to think of it is via a presentation: it is generated by two elements $a$ and $b$ with the relation $bab^{-1}=a^{-1}$ (here $a=(1,0)$ and $b=(0,1)$). This group arises naturally in topology: it is isomorphic to the fundamental group of the Klein bottle.
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
$endgroup$
$begingroup$
The first part was just meant to determine $Aut(mathbb{Z})$. The homomorphism from $mathbb{Z}$ to $Aut(mathbb{Z})$ was meant to be $psi$. I agree this was confusing. Thank you very much for your insight, especially the last part!
$endgroup$
– Evariste
Jan 30 at 21:41
$begingroup$
Ah, that makes sense. Normally we don't talk about "automorphisms from ... to ..." but instead just say "automorphisms of ..." since "automorphism" already means the domain and codomain are the same.
$endgroup$
– Eric Wofsey
Jan 30 at 21:43
$begingroup$
For the reasons you say, topologists refer to the second semidirect product as the Klein bottle group. Crystallographers refer to it in their cryptic notation as pg.
$endgroup$
– Lee Mosher
Jan 31 at 0:58
add a comment |
$begingroup$
Your work is correct. The first semidirect product is indeed just the direct product: a semidirect product is just the direct product if the homomorphism used is trivial (since then the group operation on the semidirect product reduces to just the usual coordinatewise operation).
I don't really know a common name for the second semidirect product. Another way to think of it is via a presentation: it is generated by two elements $a$ and $b$ with the relation $bab^{-1}=a^{-1}$ (here $a=(1,0)$ and $b=(0,1)$). This group arises naturally in topology: it is isomorphic to the fundamental group of the Klein bottle.
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
$endgroup$
Your work is correct. The first semidirect product is indeed just the direct product: a semidirect product is just the direct product if the homomorphism used is trivial (since then the group operation on the semidirect product reduces to just the usual coordinatewise operation).
I don't really know a common name for the second semidirect product. Another way to think of it is via a presentation: it is generated by two elements $a$ and $b$ with the relation $bab^{-1}=a^{-1}$ (here $a=(1,0)$ and $b=(0,1)$). This group arises naturally in topology: it is isomorphic to the fundamental group of the Klein bottle.
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
edited Jan 30 at 21:44
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
answered Jan 30 at 21:36
Eric WofseyEric Wofsey
192k14218351
192k14218351
$begingroup$
The first part was just meant to determine $Aut(mathbb{Z})$. The homomorphism from $mathbb{Z}$ to $Aut(mathbb{Z})$ was meant to be $psi$. I agree this was confusing. Thank you very much for your insight, especially the last part!
$endgroup$
– Evariste
Jan 30 at 21:41
$begingroup$
Ah, that makes sense. Normally we don't talk about "automorphisms from ... to ..." but instead just say "automorphisms of ..." since "automorphism" already means the domain and codomain are the same.
$endgroup$
– Eric Wofsey
Jan 30 at 21:43
$begingroup$
For the reasons you say, topologists refer to the second semidirect product as the Klein bottle group. Crystallographers refer to it in their cryptic notation as pg.
$endgroup$
– Lee Mosher
Jan 31 at 0:58
add a comment |
$begingroup$
The first part was just meant to determine $Aut(mathbb{Z})$. The homomorphism from $mathbb{Z}$ to $Aut(mathbb{Z})$ was meant to be $psi$. I agree this was confusing. Thank you very much for your insight, especially the last part!
$endgroup$
– Evariste
Jan 30 at 21:41
$begingroup$
Ah, that makes sense. Normally we don't talk about "automorphisms from ... to ..." but instead just say "automorphisms of ..." since "automorphism" already means the domain and codomain are the same.
$endgroup$
– Eric Wofsey
Jan 30 at 21:43
$begingroup$
For the reasons you say, topologists refer to the second semidirect product as the Klein bottle group. Crystallographers refer to it in their cryptic notation as pg.
$endgroup$
– Lee Mosher
Jan 31 at 0:58
$begingroup$
The first part was just meant to determine $Aut(mathbb{Z})$. The homomorphism from $mathbb{Z}$ to $Aut(mathbb{Z})$ was meant to be $psi$. I agree this was confusing. Thank you very much for your insight, especially the last part!
$endgroup$
– Evariste
Jan 30 at 21:41
$begingroup$
The first part was just meant to determine $Aut(mathbb{Z})$. The homomorphism from $mathbb{Z}$ to $Aut(mathbb{Z})$ was meant to be $psi$. I agree this was confusing. Thank you very much for your insight, especially the last part!
$endgroup$
– Evariste
Jan 30 at 21:41
$begingroup$
Ah, that makes sense. Normally we don't talk about "automorphisms from ... to ..." but instead just say "automorphisms of ..." since "automorphism" already means the domain and codomain are the same.
$endgroup$
– Eric Wofsey
Jan 30 at 21:43
$begingroup$
Ah, that makes sense. Normally we don't talk about "automorphisms from ... to ..." but instead just say "automorphisms of ..." since "automorphism" already means the domain and codomain are the same.
$endgroup$
– Eric Wofsey
Jan 30 at 21:43
$begingroup$
For the reasons you say, topologists refer to the second semidirect product as the Klein bottle group. Crystallographers refer to it in their cryptic notation as pg.
$endgroup$
– Lee Mosher
Jan 31 at 0:58
$begingroup$
For the reasons you say, topologists refer to the second semidirect product as the Klein bottle group. Crystallographers refer to it in their cryptic notation as pg.
$endgroup$
– Lee Mosher
Jan 31 at 0:58
add a comment |
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