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If a group $Gleq S_{13}$ has an element of order $40$, then $G$ has a normal non-trivial subgroup.

If $G$ is a group with order $364$, then it has a normal subgroup of order $13$Prove that a group of order 30 has at least three different normal subgroupsShowing There Is Only One Non-Trivial Normal Subgroup.Let $G$ be the group $S_4times S_3$ .Prove that $G$ has a normal subgroup of order $72$.Prove a group that has a normal subgroup isomorphic to $D_8$ has a non-trivial center“Non-trivial element” in factor group terminologyLet $G$ be a non-trivial group with no non-trivial proper subgroup. Prove that $G$ cannot be infinite group.Find Order and Normal Subgroup in a Matrix GroupProving a group is simple given that potential normal subgroups have order $2$ and $4$Show that a group of order 66 has a normal subgroup of order 33.

5

$begingroup$

I came across with the following question and I have no idea how to approach it.

Let $Gleq S_{13}$ be a subgroup with an element of order $40$.

Prove that $G$ has a normal proper non-trivial subgroup.

I thought of using the sylow theorem but I don't know order $G$. How should I prove it?

Also, if someone knows from which book the question was taken it will be great (Would like to practice with similar questions).

abstract-algebra group-theory permutations

share|cite|improve this question

edited Mar 12 at 23:22

Robert Chamberlain

4,4171621

asked Mar 12 at 22:54

vesiivesii

3878

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly you mean "nontrivial proper subgroup". Hint: signature.
  $endgroup$
  – YCor
  Mar 12 at 23:14
  

  
  
  
  

add a comment | 

5

$begingroup$

I came across with the following question and I have no idea how to approach it.

Let $Gleq S_{13}$ be a subgroup with an element of order $40$.

Prove that $G$ has a normal proper non-trivial subgroup.

I thought of using the sylow theorem but I don't know order $G$. How should I prove it?

Also, if someone knows from which book the question was taken it will be great (Would like to practice with similar questions).

abstract-algebra group-theory permutations

share|cite|improve this question

edited Mar 12 at 23:22

Robert Chamberlain

4,4171621

asked Mar 12 at 22:54

vesiivesii

3878

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly you mean "nontrivial proper subgroup". Hint: signature.
  $endgroup$
  – YCor
  Mar 12 at 23:14
  

  
  
  
  

add a comment | 

$begingroup$

I came across with the following question and I have no idea how to approach it.

Let $Gleq S_{13}$ be a subgroup with an element of order $40$.

Prove that $G$ has a normal proper non-trivial subgroup.

I thought of using the sylow theorem but I don't know order $G$. How should I prove it?

Also, if someone knows from which book the question was taken it will be great (Would like to practice with similar questions).

abstract-algebra group-theory permutations

share|cite|improve this question

edited Mar 12 at 23:22

Robert Chamberlain

4,4171621

asked Mar 12 at 22:54

vesiivesii

3878

$endgroup$

share|cite|improve this question

edited Mar 12 at 23:22

Robert Chamberlain

4,4171621

asked Mar 12 at 22:54

vesiivesii

3878

share|cite|improve this question

edited Mar 12 at 23:22

Robert Chamberlain

4,4171621

asked Mar 12 at 22:54

vesiivesii

3878

edited Mar 12 at 23:22

Robert Chamberlain

4,4171621

edited Mar 12 at 23:22

Robert Chamberlain

4,4171621

asked Mar 12 at 22:54

vesiivesii

3878

asked Mar 12 at 22:54

vesiivesii

3878

1 Answer
1

active

oldest

votes

7

$begingroup$

The only permutations in $S_{13}$ that have order $40$ are the product of a $5$-cycle and a disjoint $8$-cycle. Let's call this element $x$. Then $x^2 in A_{13}$ but $x notin A_{13}$, and $G cap A_{13}$ is a non-trivial proper normal subgroup of $G$.

Edited to add a discussion of motivation: By the way, this isn't something that I had "in the can," so to speak. But since we know that the only non-trivial proper normal subgroup of $S_{13}$ is $A_{13}$, it seemed natural to actually consider what $G$ has to look like when inserted into $S_{13}$, in the hope that we could force a non-trivial intersection with $A_{13}$. Once you ask yourself that question, the answer follows pretty quickly.

share|cite|improve this answer

edited Mar 13 at 0:55

answered Mar 12 at 23:37

Robert ShoreRobert Shore

3,235323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what does it mean "product of a 5-cycle"?
  $endgroup$
  – vesii
  Mar 12 at 23:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with cycle notation for elements of symmetric groups? So for example, we might write $(1~3~5~13~7)(2~4~8~6~12~10~11~9)$ to denote the permutation that sends $1$ to $3$, $3$ to $5$, $5$ to $13$, $13$ to $7$, and $7$ back to $1$ (the $5$-cycle; i.e., a cycle of length $5$) and also cycling the remaining eight numbers as indicated in the $8$-cycle.
  $endgroup$
  – Robert Shore
  Mar 12 at 23:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I am familiar. so if we have $sigma=alpha cdot beta cdot gamma$ we say that $sigma$ is a product of a $5-cycle$ if the order of $sigma$ is $5$?
  $endgroup$
  – vesii
  Mar 12 at 23:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not exactly. It could be that $o(sigma)=5$ because $sigma$ is the product of two distinct $5$-cycles. A $5$-cycle is a permutation that cyclically permutes $5$ elements of your set while leaving the remaining elements fixed.
  $endgroup$
  – Robert Shore
  Mar 12 at 23:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so if we have a group of order $x=acdot b$ and $a$ is prime and $b$ is not, so we should have one cycle with length $b$ and other cycles of length $a$? How many 5-cycles and 8-cycles should the subgroup contain? one 5-cycle and one 8-cycle?
  $endgroup$
  – vesii
  Mar 12 at 23:56
  
  
  

  
  
  
  

 | 
show 1 more comment

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7

$begingroup$

The only permutations in $S_{13}$ that have order $40$ are the product of a $5$-cycle and a disjoint $8$-cycle. Let's call this element $x$. Then $x^2 in A_{13}$ but $x notin A_{13}$, and $G cap A_{13}$ is a non-trivial proper normal subgroup of $G$.

Edited to add a discussion of motivation: By the way, this isn't something that I had "in the can," so to speak. But since we know that the only non-trivial proper normal subgroup of $S_{13}$ is $A_{13}$, it seemed natural to actually consider what $G$ has to look like when inserted into $S_{13}$, in the hope that we could force a non-trivial intersection with $A_{13}$. Once you ask yourself that question, the answer follows pretty quickly.

share|cite|improve this answer

edited Mar 13 at 0:55

answered Mar 12 at 23:37

Robert ShoreRobert Shore

3,235323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what does it mean "product of a 5-cycle"?
  $endgroup$
  – vesii
  Mar 12 at 23:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with cycle notation for elements of symmetric groups? So for example, we might write $(1~3~5~13~7)(2~4~8~6~12~10~11~9)$ to denote the permutation that sends $1$ to $3$, $3$ to $5$, $5$ to $13$, $13$ to $7$, and $7$ back to $1$ (the $5$-cycle; i.e., a cycle of length $5$) and also cycling the remaining eight numbers as indicated in the $8$-cycle.
  $endgroup$
  – Robert Shore
  Mar 12 at 23:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I am familiar. so if we have $sigma=alpha cdot beta cdot gamma$ we say that $sigma$ is a product of a $5-cycle$ if the order of $sigma$ is $5$?
  $endgroup$
  – vesii
  Mar 12 at 23:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not exactly. It could be that $o(sigma)=5$ because $sigma$ is the product of two distinct $5$-cycles. A $5$-cycle is a permutation that cyclically permutes $5$ elements of your set while leaving the remaining elements fixed.
  $endgroup$
  – Robert Shore
  Mar 12 at 23:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so if we have a group of order $x=acdot b$ and $a$ is prime and $b$ is not, so we should have one cycle with length $b$ and other cycles of length $a$? How many 5-cycles and 8-cycles should the subgroup contain? one 5-cycle and one 8-cycle?
  $endgroup$
  – vesii
  Mar 12 at 23:56
  
  
  

  
  
  
  

 | 
show 1 more comment

7

$begingroup$

The only permutations in $S_{13}$ that have order $40$ are the product of a $5$-cycle and a disjoint $8$-cycle. Let's call this element $x$. Then $x^2 in A_{13}$ but $x notin A_{13}$, and $G cap A_{13}$ is a non-trivial proper normal subgroup of $G$.

Edited to add a discussion of motivation: By the way, this isn't something that I had "in the can," so to speak. But since we know that the only non-trivial proper normal subgroup of $S_{13}$ is $A_{13}$, it seemed natural to actually consider what $G$ has to look like when inserted into $S_{13}$, in the hope that we could force a non-trivial intersection with $A_{13}$. Once you ask yourself that question, the answer follows pretty quickly.

share|cite|improve this answer

edited Mar 13 at 0:55

answered Mar 12 at 23:37

Robert ShoreRobert Shore

3,235323

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what does it mean "product of a 5-cycle"?
  $endgroup$
  – vesii
  Mar 12 at 23:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with cycle notation for elements of symmetric groups? So for example, we might write $(1~3~5~13~7)(2~4~8~6~12~10~11~9)$ to denote the permutation that sends $1$ to $3$, $3$ to $5$, $5$ to $13$, $13$ to $7$, and $7$ back to $1$ (the $5$-cycle; i.e., a cycle of length $5$) and also cycling the remaining eight numbers as indicated in the $8$-cycle.
  $endgroup$
  – Robert Shore
  Mar 12 at 23:50
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I am familiar. so if we have $sigma=alpha cdot beta cdot gamma$ we say that $sigma$ is a product of a $5-cycle$ if the order of $sigma$ is $5$?
  $endgroup$
  – vesii
  Mar 12 at 23:52
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not exactly. It could be that $o(sigma)=5$ because $sigma$ is the product of two distinct $5$-cycles. A $5$-cycle is a permutation that cyclically permutes $5$ elements of your set while leaving the remaining elements fixed.
  $endgroup$
  – Robert Shore
  Mar 12 at 23:53
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so if we have a group of order $x=acdot b$ and $a$ is prime and $b$ is not, so we should have one cycle with length $b$ and other cycles of length $a$? How many 5-cycles and 8-cycles should the subgroup contain? one 5-cycle and one 8-cycle?
  $endgroup$
  – vesii
  Mar 12 at 23:56
  
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

The only permutations in $S_{13}$ that have order $40$ are the product of a $5$-cycle and a disjoint $8$-cycle. Let's call this element $x$. Then $x^2 in A_{13}$ but $x notin A_{13}$, and $G cap A_{13}$ is a non-trivial proper normal subgroup of $G$.

Edited to add a discussion of motivation: By the way, this isn't something that I had "in the can," so to speak. But since we know that the only non-trivial proper normal subgroup of $S_{13}$ is $A_{13}$, it seemed natural to actually consider what $G$ has to look like when inserted into $S_{13}$, in the hope that we could force a non-trivial intersection with $A_{13}$. Once you ask yourself that question, the answer follows pretty quickly.

share|cite|improve this answer

edited Mar 13 at 0:55

answered Mar 12 at 23:37

Robert ShoreRobert Shore

3,235323

$endgroup$

share|cite|improve this answer

edited Mar 13 at 0:55

answered Mar 12 at 23:37

Robert ShoreRobert Shore

3,235323

answered Mar 12 at 23:37

Robert ShoreRobert Shore

3,235323

answered Mar 12 at 23:37

Robert ShoreRobert Shore

3,235323