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Let $G$ have exactly $24$ elements of order $6$. Why are there $12$ cyclic subgroups of $G$ of order $6$?

Elements and cyclic subgroups of order $15$ in $Bbb Z_30times Bbb Z_20.$How to find subgroups of $ ;;Bbb Z_2times Bbb Z_6$Subgroups of a cyclic group and their order.$l$-Sylow subgroups of $SL_3(BbbF_p)$ are cyclic for $l|p^2 + p + 1$Determine the number of subgroups of $Bbb Z_p times Bbb Z_p$, where p is prime.How do we know all Sylow $p$-subgroups of a given order are distinct?Why does the fact that $nexists Hle A_4$ with $lvert Hrvert=6$ imply $lvert Z(A_4)rvert=1$?Is $|z|=18?$Elements and cyclic subgroups of order $15$ in $Bbb Z_30times Bbb Z_20.$A formula for the number of order $2$ elements of $D_mtimes D_n$ for even $m>2$ and odd $n>2$. (Gallian 8.24.)Find a subgroup of $Bbb Z_4oplusBbb Z_2$ not of the form $Hoplus K$ for some $Hle Bbb Z_4, Kle Bbb Z_2$.

3

2

$begingroup$

This Exercise 8.47 of Gallian's "Contemporary Abstract Algebra".

Answers that use material from the textbook prior to the question are preferred.

Let $G$ be a group with exactly $24$ elements of order $6$. Why are there $12$ cyclic subgroups of $G$ of order $6$?

The question is phrased differently in the textbook. It asks for the number of such cyclic subgroups and provides $12$ as an answer in the back.

Context:

I've been thinking about this on & off for about a week but have got nowhere. This is particularly frustrating because I think I should be able to do it myself.

Here is a similar question of mine:

Elements and cyclic subgroups of order $15$ in $Bbb Z_30times Bbb Z_20.$

The kind of answer I'm looking for is a general approach (with the preference above). The reason why is that it's evident that my understanding is lacking here. Perhaps some easier exercises to this end might help too.

Please help :)

group-theory finite-groups cyclic-groups

share|cite|improve this question

edited Mar 11 at 19:45

Shaun

asked Mar 11 at 19:13

ShaunShaun

9,442113684

$endgroup$

add a comment | 

3

2

$begingroup$

This Exercise 8.47 of Gallian's "Contemporary Abstract Algebra".

Answers that use material from the textbook prior to the question are preferred.

Let $G$ be a group with exactly $24$ elements of order $6$. Why are there $12$ cyclic subgroups of $G$ of order $6$?

The question is phrased differently in the textbook. It asks for the number of such cyclic subgroups and provides $12$ as an answer in the back.

Context:

I've been thinking about this on & off for about a week but have got nowhere. This is particularly frustrating because I think I should be able to do it myself.

Here is a similar question of mine:

Elements and cyclic subgroups of order $15$ in $Bbb Z_30times Bbb Z_20.$

The kind of answer I'm looking for is a general approach (with the preference above). The reason why is that it's evident that my understanding is lacking here. Perhaps some easier exercises to this end might help too.

Please help :)

group-theory finite-groups cyclic-groups

share|cite|improve this question

edited Mar 11 at 19:45

Shaun

asked Mar 11 at 19:13

ShaunShaun

9,442113684

$endgroup$

add a comment | 

$begingroup$

This Exercise 8.47 of Gallian's "Contemporary Abstract Algebra".

Answers that use material from the textbook prior to the question are preferred.

Let $G$ be a group with exactly $24$ elements of order $6$. Why are there $12$ cyclic subgroups of $G$ of order $6$?

The question is phrased differently in the textbook. It asks for the number of such cyclic subgroups and provides $12$ as an answer in the back.

Context:

I've been thinking about this on & off for about a week but have got nowhere. This is particularly frustrating because I think I should be able to do it myself.

Here is a similar question of mine:

Elements and cyclic subgroups of order $15$ in $Bbb Z_30times Bbb Z_20.$

The kind of answer I'm looking for is a general approach (with the preference above). The reason why is that it's evident that my understanding is lacking here. Perhaps some easier exercises to this end might help too.

Please help :)

group-theory finite-groups cyclic-groups

share|cite|improve this question

edited Mar 11 at 19:45

Shaun

asked Mar 11 at 19:13

ShaunShaun

9,442113684

$endgroup$

share|cite|improve this question

edited Mar 11 at 19:45

Shaun

asked Mar 11 at 19:13

ShaunShaun

9,442113684

share|cite|improve this question

edited Mar 11 at 19:45

Shaun

asked Mar 11 at 19:13

ShaunShaun

9,442113684

asked Mar 11 at 19:13

ShaunShaun

9,442113684

asked Mar 11 at 19:13

ShaunShaun

9,442113684

1 Answer
1

active

oldest

votes

5

$begingroup$

I'll make a try:

We have $24$ elements of order $6$. Every cyclic group of order $6$ has $2$ generators. So we have at least $dfrac242=12$ cyclic groups of order $6$.

share|cite|improve this answer

edited Mar 11 at 19:45

Shaun

9,442113684

answered Mar 11 at 19:27

giannispapavgiannispapav

1,794324

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can the down voter please explain? It's not obvious to me that this answer is wrong. Obviously the elements of order 6 can't belong to an intersection of two different subgroups and there are two per subgroup so...
  $endgroup$
  – Evariste
  Mar 11 at 19:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  by your reasoning there should be exactly $13$ cyclic groups of order $6$. and I'm quite convinced that it's correct :) Luckily there is only abelian groups of order $6$. The downvotes make me think about Abel and Galois... whose intuition went far beyond their judgers :)
  $endgroup$
  – Leaning
  Mar 11 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I copied the question down wrong! It's $24$, not $26$.
  $endgroup$
  – Shaun
  Mar 11 at 19:44
  

  
  
  
  

add a comment | 

Your Answer

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5

$begingroup$

I'll make a try:

We have $24$ elements of order $6$. Every cyclic group of order $6$ has $2$ generators. So we have at least $dfrac242=12$ cyclic groups of order $6$.

share|cite|improve this answer

edited Mar 11 at 19:45

Shaun

9,442113684

answered Mar 11 at 19:27

giannispapavgiannispapav

1,794324

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can the down voter please explain? It's not obvious to me that this answer is wrong. Obviously the elements of order 6 can't belong to an intersection of two different subgroups and there are two per subgroup so...
  $endgroup$
  – Evariste
  Mar 11 at 19:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  by your reasoning there should be exactly $13$ cyclic groups of order $6$. and I'm quite convinced that it's correct :) Luckily there is only abelian groups of order $6$. The downvotes make me think about Abel and Galois... whose intuition went far beyond their judgers :)
  $endgroup$
  – Leaning
  Mar 11 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I copied the question down wrong! It's $24$, not $26$.
  $endgroup$
  – Shaun
  Mar 11 at 19:44
  

  
  
  
  

add a comment | 

5

$begingroup$

I'll make a try:

We have $24$ elements of order $6$. Every cyclic group of order $6$ has $2$ generators. So we have at least $dfrac242=12$ cyclic groups of order $6$.

share|cite|improve this answer

edited Mar 11 at 19:45

Shaun

9,442113684

answered Mar 11 at 19:27

giannispapavgiannispapav

1,794324

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can the down voter please explain? It's not obvious to me that this answer is wrong. Obviously the elements of order 6 can't belong to an intersection of two different subgroups and there are two per subgroup so...
  $endgroup$
  – Evariste
  Mar 11 at 19:34
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  by your reasoning there should be exactly $13$ cyclic groups of order $6$. and I'm quite convinced that it's correct :) Luckily there is only abelian groups of order $6$. The downvotes make me think about Abel and Galois... whose intuition went far beyond their judgers :)
  $endgroup$
  – Leaning
  Mar 11 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I copied the question down wrong! It's $24$, not $26$.
  $endgroup$
  – Shaun
  Mar 11 at 19:44
  

  
  
  
  

add a comment | 

$begingroup$

I'll make a try:

We have $24$ elements of order $6$. Every cyclic group of order $6$ has $2$ generators. So we have at least $dfrac242=12$ cyclic groups of order $6$.

share|cite|improve this answer

edited Mar 11 at 19:45

Shaun

9,442113684

answered Mar 11 at 19:27

giannispapavgiannispapav

1,794324

$endgroup$

share|cite|improve this answer

edited Mar 11 at 19:45

Shaun

9,442113684

answered Mar 11 at 19:27

giannispapavgiannispapav

1,794324

edited Mar 11 at 19:45

Shaun

9,442113684

edited Mar 11 at 19:45

Shaun

9,442113684

answered Mar 11 at 19:27

giannispapavgiannispapav

1,794324

answered Mar 11 at 19:27

giannispapavgiannispapav

1,794324