$(mathbb Z/p mathbb Z rtimes mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z congmathbb Z/p mathbb Z rtimes...
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Given:
Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.
It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.
Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.
Note that composition of maps is evaluated from right to left.
Question:
If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?
Thoughts:
My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.
Makes this line of attack sense? Or is the required argument quite obvious?
Thank you very much for your insights!
Context:
I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.
group-theory finite-groups group-isomorphism semidirect-product
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up vote
4
down vote
favorite
Given:
Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.
It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.
Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.
Note that composition of maps is evaluated from right to left.
Question:
If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?
Thoughts:
My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.
Makes this line of attack sense? Or is the required argument quite obvious?
Thank you very much for your insights!
Context:
I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.
group-theory finite-groups group-isomorphism semidirect-product
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given:
Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.
It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.
Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.
Note that composition of maps is evaluated from right to left.
Question:
If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?
Thoughts:
My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.
Makes this line of attack sense? Or is the required argument quite obvious?
Thank you very much for your insights!
Context:
I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.
group-theory finite-groups group-isomorphism semidirect-product
Given:
Let $p$ and $q$ be prime numbers such that $q$ divides $p-1$.
It is well-know that there is a monomorphism $varphi: mathbb Z/q mathbb Z to Aut(mathbb Z/p mathbb Z)$.
Define homomorphisms $varsigma: (mathbb Z/q mathbb Z)^2 to mathbb Z/ q mathbb Z$ where $(a,b) mapsto a-b$ and $vartheta: (mathbb Z/q mathbb Z)^2 to Aut(mathbb Z/p mathbb Z)$ via $vartheta = varphi circ varsigma$.
Note that composition of maps is evaluated from right to left.
Question:
If we consider the semi-direct products $G := (mathbb Z/p mathbb Z rtimes_varphi mathbb Z/q mathbb Z) times mathbb Z/q mathbb Z$ and $H := mathbb Z/p mathbb Z rtimes_vartheta (mathbb Z/q mathbb Z)^2$, are these groups isomorphic?
Thoughts:
My intuition says: No, $G$ and $H$ are not isomorphic. But I am unsure how to prove this hypothesis. I tried to evaluate the centers $Z(G)$ and $Z(H)$ of $G$ and $H$, respectively, which gave me $Z(G) = {(0,0)} times mathbb Z/q mathbb Z$ and ${(0,r,r): r in mathbb Z/q mathbb Z} subseteq Z(H)$.
Makes this line of attack sense? Or is the required argument quite obvious?
Thank you very much for your insights!
Context:
I stumbled upon this question while reading a collection of problems about group theory which interested me as a layperson.
group-theory finite-groups group-isomorphism semidirect-product
group-theory finite-groups group-isomorphism semidirect-product
asked Nov 17 at 20:10
Moritz
1,1511622
1,1511622
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1 Answer
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I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
The second way is to show that $H$ "has the same structure as $G$".
Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!
Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
– Moritz
Nov 18 at 15:04
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
The second way is to show that $H$ "has the same structure as $G$".
Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!
Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
– Moritz
Nov 18 at 15:04
add a comment |
up vote
3
down vote
accepted
I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
The second way is to show that $H$ "has the same structure as $G$".
Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!
Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
– Moritz
Nov 18 at 15:04
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
The second way is to show that $H$ "has the same structure as $G$".
Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!
I think that your groups are isomorphic. You can prove it in two ways. The first one is to show an explicit isomorphism, which is not hard to do, but kind of boring.
The second way is to show that $H$ "has the same structure as $G$".
Namely you have to find two commuting normal subgroups of H, whose intersection is trivial, one isomorphic to $mathbb{Z}/pmathbb{Z} rtimes_varphi mathbb{Z}/qmathbb{Z}$ and the other isomorphic to $mathbb{Z}/qmathbb{Z}$. It's not hard to find the former. As for the latter, you have already found it!
answered Nov 17 at 21:30
Pietro Gheri
1362
1362
Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
– Moritz
Nov 18 at 15:04
add a comment |
Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
– Moritz
Nov 18 at 15:04
Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
– Moritz
Nov 18 at 15:04
Thank you! If your are right, I was looking in the wrong direction. I will try to your argument in the following days. Until then +1 from me.
– Moritz
Nov 18 at 15:04
add a comment |
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