Groups of order $2pqr$ [closed]

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What are the groups of order $2pqr$ up to isomorphism, where $p$, $q$, $r$ are distinct primes greater than $2$? How would you identify them?

Thanks a lot in advance.

group-theory

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edited Aug 1 at 15:52

Daniel Buck

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asked Aug 1 at 15:48

Buddhini Angelika

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closed as off-topic by Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde Aug 1 at 18:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde

If this question can be reworded to fit the rules in the help center, please edit the question.

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  Why not try some special cases first? If you tell me the groups of order $2.3.7.43$ that you can find then I'll check if you've found them all.
  – ancientmathematician
  Aug 1 at 16:27
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

What are the groups of order $2pqr$ up to isomorphism, where $p$, $q$, $r$ are distinct primes greater than $2$? How would you identify them?

Thanks a lot in advance.

group-theory

share|cite|improve this question

edited Aug 1 at 15:52

Daniel Buck

2,2641623

asked Aug 1 at 15:48

Buddhini Angelika

121

closed as off-topic by Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde Aug 1 at 18:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why not try some special cases first? If you tell me the groups of order $2.3.7.43$ that you can find then I'll check if you've found them all.
  – ancientmathematician
  Aug 1 at 16:27
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

What are the groups of order $2pqr$ up to isomorphism, where $p$, $q$, $r$ are distinct primes greater than $2$? How would you identify them?

Thanks a lot in advance.

group-theory

share|cite|improve this question

edited Aug 1 at 15:52

Daniel Buck

2,2641623

asked Aug 1 at 15:48

Buddhini Angelika

121

What are the groups of order $2pqr$ up to isomorphism, where $p$, $q$, $r$ are distinct primes greater than $2$? How would you identify them?

Thanks a lot in advance.

group-theory

share|cite|improve this question

edited Aug 1 at 15:52

Daniel Buck

2,2641623

asked Aug 1 at 15:48

Buddhini Angelika

121

share|cite|improve this question

share|cite|improve this question

edited Aug 1 at 15:52

Daniel Buck

2,2641623

edited Aug 1 at 15:52

Daniel Buck

2,2641623

edited Aug 1 at 15:52

Daniel Buck

2,2641623

2,2641623

asked Aug 1 at 15:48

Buddhini Angelika

121

asked Aug 1 at 15:48

Buddhini Angelika

121

asked Aug 1 at 15:48

Buddhini Angelika

121

121

closed as off-topic by Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde Aug 1 at 18:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde Aug 1 at 18:09

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, Alan Wang, John Ma, Isaac Browne, Dietrich Burde

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why not try some special cases first? If you tell me the groups of order $2.3.7.43$ that you can find then I'll check if you've found them all.
  – ancientmathematician
  Aug 1 at 16:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why not try some special cases first? If you tell me the groups of order $2.3.7.43$ that you can find then I'll check if you've found them all.
  – ancientmathematician
  Aug 1 at 16:27
  

  
  
  
  

Why not try some special cases first? If you tell me the groups of order $2.3.7.43$ that you can find then I'll check if you've found them all.
– ancientmathematician
Aug 1 at 16:27

Why not try some special cases first? If you tell me the groups of order $2.3.7.43$ that you can find then I'll check if you've found them all.
– ancientmathematician
Aug 1 at 16:27

add a comment | 

1 Answer
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1
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Finite group of squarefree order is solvable (it follows from the fact that Sylow subgroups are cyclic). As it's solvable, it has all Hall subgroups, in particular, maximal subgroup of odd order. It has index 2, therefore normal. (First part of this paragraph is probably an overkill reasoning, but the first one that popped in my head.)

So you have split extension
$$G = C_2 ltimes H$$

Denote odd primes dividing $|H|$ as $p < q < r$.

First, assume $(q, r-1) = 1$, so $p'$-Hall subgroup of $H$ is abelian.

1. $(p, q-1) = (p, r-1) = 1$, then $H = C_pqr$ (it's obvious)


2. $p | q-1$ or $p|r-1$, but not both. Then $H = C_pqr$, $H = (C_p ltimes C_q) times C_r$ or $H = (C_p ltimes C_r) times C_q$ (also pretty easy to see)


3. $p$ divides both $q-1$ an $r-1$. Then there are $p+2$ different semidirect products $C_p ltimes C_qr$ incluing trivial one $C_pqr$. Proof I'm aware of is elementary, but long and tedious, so I wouldn't reproduce it here. I suspect that some nifty number-theoretic argument makes it easy, but old case by case analysis still does the job.

Also you want classification when $q|r-1$. It's too long for an answer here, but it's done already by Hölder in 1893 — you can look up if you're not afraid of ancient notation. (There should be more modern exposition, but I do not remember any.)

---

Then, finally, after possible $H$'s are identified, you want to somehow understand which involutions in $Aut(H)$ will give same semidirect products. There's no general recipe to do this, but we're lucky enough to be in simple enough situation.

In case $H = C_pqr$ see answer to this https://math.stackexchange.com/q/571248 question. If $H$ is cyclic times nonabelian, then similarly to previous case we can treat those direct factors disjointly (because there are no nonzero homomorphisms between them), and both cases are already classified above (we haven't used that $p, q, r$ are odd in section about $H$).

There's also hard one when $H$ is indecomposable; I guess you can find full classification in some article, but it almost guaranteed to be just tedious analysis of multiple cases.

share|cite|improve this answer

answered Aug 1 at 18:02

xsnl

1,155418

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. Can you please tell me the link to Holders publication? I tried to find it out but can't find it. I want to find all the publications by Holder (There is a publication on groups of order $pq^2$ and $pqrs$ but it seems none are available).
  – Buddhini Angelika
  2 days ago
  
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Finite group of squarefree order is solvable (it follows from the fact that Sylow subgroups are cyclic). As it's solvable, it has all Hall subgroups, in particular, maximal subgroup of odd order. It has index 2, therefore normal. (First part of this paragraph is probably an overkill reasoning, but the first one that popped in my head.)

So you have split extension
$$G = C_2 ltimes H$$

Denote odd primes dividing $|H|$ as $p < q < r$.

First, assume $(q, r-1) = 1$, so $p'$-Hall subgroup of $H$ is abelian.

1. $(p, q-1) = (p, r-1) = 1$, then $H = C_pqr$ (it's obvious)


2. $p | q-1$ or $p|r-1$, but not both. Then $H = C_pqr$, $H = (C_p ltimes C_q) times C_r$ or $H = (C_p ltimes C_r) times C_q$ (also pretty easy to see)


3. $p$ divides both $q-1$ an $r-1$. Then there are $p+2$ different semidirect products $C_p ltimes C_qr$ incluing trivial one $C_pqr$. Proof I'm aware of is elementary, but long and tedious, so I wouldn't reproduce it here. I suspect that some nifty number-theoretic argument makes it easy, but old case by case analysis still does the job.

Also you want classification when $q|r-1$. It's too long for an answer here, but it's done already by Hölder in 1893 — you can look up if you're not afraid of ancient notation. (There should be more modern exposition, but I do not remember any.)

---

Then, finally, after possible $H$'s are identified, you want to somehow understand which involutions in $Aut(H)$ will give same semidirect products. There's no general recipe to do this, but we're lucky enough to be in simple enough situation.

In case $H = C_pqr$ see answer to this https://math.stackexchange.com/q/571248 question. If $H$ is cyclic times nonabelian, then similarly to previous case we can treat those direct factors disjointly (because there are no nonzero homomorphisms between them), and both cases are already classified above (we haven't used that $p, q, r$ are odd in section about $H$).

There's also hard one when $H$ is indecomposable; I guess you can find full classification in some article, but it almost guaranteed to be just tedious analysis of multiple cases.

share|cite|improve this answer

answered Aug 1 at 18:02

xsnl

1,155418

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. Can you please tell me the link to Holders publication? I tried to find it out but can't find it. I want to find all the publications by Holder (There is a publication on groups of order $pq^2$ and $pqrs$ but it seems none are available).
  – Buddhini Angelika
  2 days ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Finite group of squarefree order is solvable (it follows from the fact that Sylow subgroups are cyclic). As it's solvable, it has all Hall subgroups, in particular, maximal subgroup of odd order. It has index 2, therefore normal. (First part of this paragraph is probably an overkill reasoning, but the first one that popped in my head.)

So you have split extension
$$G = C_2 ltimes H$$

Denote odd primes dividing $|H|$ as $p < q < r$.

First, assume $(q, r-1) = 1$, so $p'$-Hall subgroup of $H$ is abelian.

1. $(p, q-1) = (p, r-1) = 1$, then $H = C_pqr$ (it's obvious)


2. $p | q-1$ or $p|r-1$, but not both. Then $H = C_pqr$, $H = (C_p ltimes C_q) times C_r$ or $H = (C_p ltimes C_r) times C_q$ (also pretty easy to see)


3. $p$ divides both $q-1$ an $r-1$. Then there are $p+2$ different semidirect products $C_p ltimes C_qr$ incluing trivial one $C_pqr$. Proof I'm aware of is elementary, but long and tedious, so I wouldn't reproduce it here. I suspect that some nifty number-theoretic argument makes it easy, but old case by case analysis still does the job.

Also you want classification when $q|r-1$. It's too long for an answer here, but it's done already by Hölder in 1893 — you can look up if you're not afraid of ancient notation. (There should be more modern exposition, but I do not remember any.)

---

Then, finally, after possible $H$'s are identified, you want to somehow understand which involutions in $Aut(H)$ will give same semidirect products. There's no general recipe to do this, but we're lucky enough to be in simple enough situation.

In case $H = C_pqr$ see answer to this https://math.stackexchange.com/q/571248 question. If $H$ is cyclic times nonabelian, then similarly to previous case we can treat those direct factors disjointly (because there are no nonzero homomorphisms between them), and both cases are already classified above (we haven't used that $p, q, r$ are odd in section about $H$).

There's also hard one when $H$ is indecomposable; I guess you can find full classification in some article, but it almost guaranteed to be just tedious analysis of multiple cases.

share|cite|improve this answer

answered Aug 1 at 18:02

xsnl

1,155418

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. Can you please tell me the link to Holders publication? I tried to find it out but can't find it. I want to find all the publications by Holder (There is a publication on groups of order $pq^2$ and $pqrs$ but it seems none are available).
  – Buddhini Angelika
  2 days ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Finite group of squarefree order is solvable (it follows from the fact that Sylow subgroups are cyclic). As it's solvable, it has all Hall subgroups, in particular, maximal subgroup of odd order. It has index 2, therefore normal. (First part of this paragraph is probably an overkill reasoning, but the first one that popped in my head.)

So you have split extension
$$G = C_2 ltimes H$$

Denote odd primes dividing $|H|$ as $p < q < r$.

First, assume $(q, r-1) = 1$, so $p'$-Hall subgroup of $H$ is abelian.

1. $(p, q-1) = (p, r-1) = 1$, then $H = C_pqr$ (it's obvious)


2. $p | q-1$ or $p|r-1$, but not both. Then $H = C_pqr$, $H = (C_p ltimes C_q) times C_r$ or $H = (C_p ltimes C_r) times C_q$ (also pretty easy to see)


3. $p$ divides both $q-1$ an $r-1$. Then there are $p+2$ different semidirect products $C_p ltimes C_qr$ incluing trivial one $C_pqr$. Proof I'm aware of is elementary, but long and tedious, so I wouldn't reproduce it here. I suspect that some nifty number-theoretic argument makes it easy, but old case by case analysis still does the job.

Also you want classification when $q|r-1$. It's too long for an answer here, but it's done already by Hölder in 1893 — you can look up if you're not afraid of ancient notation. (There should be more modern exposition, but I do not remember any.)

---

Then, finally, after possible $H$'s are identified, you want to somehow understand which involutions in $Aut(H)$ will give same semidirect products. There's no general recipe to do this, but we're lucky enough to be in simple enough situation.

In case $H = C_pqr$ see answer to this https://math.stackexchange.com/q/571248 question. If $H$ is cyclic times nonabelian, then similarly to previous case we can treat those direct factors disjointly (because there are no nonzero homomorphisms between them), and both cases are already classified above (we haven't used that $p, q, r$ are odd in section about $H$).

There's also hard one when $H$ is indecomposable; I guess you can find full classification in some article, but it almost guaranteed to be just tedious analysis of multiple cases.

share|cite|improve this answer

answered Aug 1 at 18:02

xsnl

1,155418

Finite group of squarefree order is solvable (it follows from the fact that Sylow subgroups are cyclic). As it's solvable, it has all Hall subgroups, in particular, maximal subgroup of odd order. It has index 2, therefore normal. (First part of this paragraph is probably an overkill reasoning, but the first one that popped in my head.)

So you have split extension
$$G = C_2 ltimes H$$

Denote odd primes dividing $|H|$ as $p < q < r$.

First, assume $(q, r-1) = 1$, so $p'$-Hall subgroup of $H$ is abelian.

1. $(p, q-1) = (p, r-1) = 1$, then $H = C_pqr$ (it's obvious)


2. $p | q-1$ or $p|r-1$, but not both. Then $H = C_pqr$, $H = (C_p ltimes C_q) times C_r$ or $H = (C_p ltimes C_r) times C_q$ (also pretty easy to see)


3. $p$ divides both $q-1$ an $r-1$. Then there are $p+2$ different semidirect products $C_p ltimes C_qr$ incluing trivial one $C_pqr$. Proof I'm aware of is elementary, but long and tedious, so I wouldn't reproduce it here. I suspect that some nifty number-theoretic argument makes it easy, but old case by case analysis still does the job.

Also you want classification when $q|r-1$. It's too long for an answer here, but it's done already by Hölder in 1893 — you can look up if you're not afraid of ancient notation. (There should be more modern exposition, but I do not remember any.)

---

Then, finally, after possible $H$'s are identified, you want to somehow understand which involutions in $Aut(H)$ will give same semidirect products. There's no general recipe to do this, but we're lucky enough to be in simple enough situation.

In case $H = C_pqr$ see answer to this https://math.stackexchange.com/q/571248 question. If $H$ is cyclic times nonabelian, then similarly to previous case we can treat those direct factors disjointly (because there are no nonzero homomorphisms between them), and both cases are already classified above (we haven't used that $p, q, r$ are odd in section about $H$).

There's also hard one when $H$ is indecomposable; I guess you can find full classification in some article, but it almost guaranteed to be just tedious analysis of multiple cases.

share|cite|improve this answer

answered Aug 1 at 18:02

xsnl

1,155418

share|cite|improve this answer

share|cite|improve this answer

answered Aug 1 at 18:02

xsnl

1,155418

answered Aug 1 at 18:02

xsnl

1,155418

answered Aug 1 at 18:02

xsnl

1,155418

1,155418

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. Can you please tell me the link to Holders publication? I tried to find it out but can't find it. I want to find all the publications by Holder (There is a publication on groups of order $pq^2$ and $pqrs$ but it seems none are available).
  – Buddhini Angelika
  2 days ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you very much. Can you please tell me the link to Holders publication? I tried to find it out but can't find it. I want to find all the publications by Holder (There is a publication on groups of order $pq^2$ and $pqrs$ but it seems none are available).
  – Buddhini Angelika
  2 days ago
  
  

  
  
  
  

Thank you very much. Can you please tell me the link to Holders publication? I tried to find it out but can't find it. I want to find all the publications by Holder (There is a publication on groups of order $pq^2$ and $pqrs$ but it seems none are available).
– Buddhini Angelika
2 days ago

Thank you very much. Can you please tell me the link to Holders publication? I tried to find it out but can't find it. I want to find all the publications by Holder (There is a publication on groups of order $pq^2$ and $pqrs$ but it seems none are available).
– Buddhini Angelika
2 days ago

add a comment |