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2

$begingroup$

Let the cyclic group have 6 elements and be denoted as $G = 1, a, a^2, a^3, a^4, a^5$ where $a^6 = 1$.

Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, $1, a^2, a^4$ and $1, a^3$.

Why isnt $1, a^5$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?

---

There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.

If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^-5$ or $a^10$, where it is explicitly stated that $a^10 = 1$ as well.

Is my thought process correct?

abstract-algebra group-theory

share|cite|improve this question

edited Apr 3 at 20:02

J. W. Tanner

4,5141320

asked Apr 3 at 19:57

Evan KimEvan Kim

67819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
  $endgroup$
  – Minus One-Twelfth
  Apr 3 at 19:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why? Could you help me understand how you got to that conclusion?
  $endgroup$
  – Evan Kim
  Apr 3 at 20:00
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $1,a^5$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^10=a^4$
  $endgroup$
  – J. W. Tanner
  Apr 3 at 20:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
  $endgroup$
  – Minus One-Twelfth
  Apr 3 at 21:44
  

  
  
  
  

add a comment | 

2

$begingroup$

Let the cyclic group have 6 elements and be denoted as $G = 1, a, a^2, a^3, a^4, a^5$ where $a^6 = 1$.

Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, $1, a^2, a^4$ and $1, a^3$.

Why isnt $1, a^5$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?

---

There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.

If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^-5$ or $a^10$, where it is explicitly stated that $a^10 = 1$ as well.

Is my thought process correct?

abstract-algebra group-theory

share|cite|improve this question

edited Apr 3 at 20:02

J. W. Tanner

4,5141320

asked Apr 3 at 19:57

Evan KimEvan Kim

67819

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^3$ is itself ($a^3$). The inverse of $a^5$ is $a$.
  $endgroup$
  – Minus One-Twelfth
  Apr 3 at 19:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why? Could you help me understand how you got to that conclusion?
  $endgroup$
  – Evan Kim
  Apr 3 at 20:00
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $1,a^5$ is not a subgroup because it is not closed; it does not contain $a^5a^5=a^10=a^4$
  $endgroup$
  – J. W. Tanner
  Apr 3 at 20:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The inverse of $a^5$ is $a$ because $a^5cdot a = 1$ (since $a^5cdot a = a^6$, which we are told is $1$).
  $endgroup$
  – Minus One-Twelfth
  Apr 3 at 21:44
  

  
  
  
  

add a comment | 

$begingroup$

Let the cyclic group have 6 elements and be denoted as $G = 1, a, a^2, a^3, a^4, a^5$ where $a^6 = 1$.

Besides the trivial subgroup 1 and the entire subgroup G, my textbook says there are only two other subgroups, $1, a^2, a^4$ and $1, a^3$.

Why isnt $1, a^5$ a subgroup? Is it because $a^5$ has no inverse? If so, then what is the inverse of $a^3$?

---

There should be an element, $b$ such that $a^3 cdot b = 1$. The only reasoning I can think of is that if $b = a^3$, then $a^3 cdot a^3 = a^6 = 1$ only because $a^6 =1$ was explicitly stated.

If $a^5 cdot b = 1$ is true, then $b$ would have to be $a^-5$ or $a^10$, where it is explicitly stated that $a^10 = 1$ as well.

Is my thought process correct?

abstract-algebra group-theory

share|cite|improve this question

edited Apr 3 at 20:02

J. W. Tanner

4,5141320

asked Apr 3 at 19:57

Evan KimEvan Kim

67819

$endgroup$

share|cite|improve this question

edited Apr 3 at 20:02

J. W. Tanner

4,5141320

asked Apr 3 at 19:57

Evan KimEvan Kim

67819

share|cite|improve this question

edited Apr 3 at 20:02

J. W. Tanner

4,5141320

asked Apr 3 at 19:57

Evan KimEvan Kim

67819

edited Apr 3 at 20:02

J. W. Tanner

4,5141320

edited Apr 3 at 20:02

J. W. Tanner

4,5141320

asked Apr 3 at 19:57

Evan KimEvan Kim

67819

asked Apr 3 at 19:57

Evan KimEvan Kim

67819

4 Answers
4

active

oldest

votes

4

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered Apr 3 at 20:01

PeterPeter

49.1k1240138

$endgroup$

add a comment | 

3

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered Apr 3 at 20:00

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

add a comment | 

2

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered Apr 3 at 21:06

Jack PfaffingerJack Pfaffinger

3961112

$endgroup$

add a comment | 

0

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered Apr 3 at 20:27

ShaunShaun

10.3k113686

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote?
  $endgroup$
  – Shaun
  Apr 3 at 20:46
  

  
  
  
  

add a comment | 

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4

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered Apr 3 at 20:01

PeterPeter

49.1k1240138

$endgroup$

add a comment | 

4

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered Apr 3 at 20:01

PeterPeter

49.1k1240138

$endgroup$

add a comment | 

$begingroup$

$[1,a^5] $ is not a subgroup because $a^5cdot a^5=a^4$ which is not in the set $[1,a^5]$

But in a subgroup , with two elements $a,b$ , the product $ab$ must be in the subgroup as well.

share|cite|improve this answer

answered Apr 3 at 20:01

PeterPeter

49.1k1240138

$endgroup$

share|cite|improve this answer

answered Apr 3 at 20:01

PeterPeter

49.1k1240138

answered Apr 3 at 20:01

PeterPeter

49.1k1240138

answered Apr 3 at 20:01

PeterPeter

49.1k1240138

3

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered Apr 3 at 20:00

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

add a comment | 

3

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered Apr 3 at 20:00

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

add a comment | 

$begingroup$

$lbrace 1, a^5 rbrace$ is not a subgroup because
$$a^5 . a^5 = a^4$$
is not an element of $lbrace 1, a^5 rbrace$. So $lbrace 1, a^5 rbrace$ is not stable for the intern law of the group.

share|cite|improve this answer

answered Apr 3 at 20:00

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

share|cite|improve this answer

answered Apr 3 at 20:00

TheSilverDoeTheSilverDoe

5,433216

answered Apr 3 at 20:00

TheSilverDoeTheSilverDoe

5,433216

answered Apr 3 at 20:00

TheSilverDoeTheSilverDoe

5,433216

2

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered Apr 3 at 21:06

Jack PfaffingerJack Pfaffinger

3961112

$endgroup$

add a comment | 

2

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered Apr 3 at 21:06

Jack PfaffingerJack Pfaffinger

3961112

$endgroup$

add a comment | 

$begingroup$

Since nobody said it I'll also add that we know from the Fundamental Theorem of Cyclic Groups that for a finite cyclic group of order $n$, every subgroup's order is a divisor of $n$, and there is exactly one subgroup for each divisor. So to find the number of cyclic groups for a group of order $n$, just count the divisors of $n$. Here there are $4$ divisors of $6$, and so these must be all the subgroups.

It is also true that if $a$ is an element of order $n$ in a group and $k$ is a positive integer. Then $langle a^k rangle = langle a^gcd(n,k) rangle$. Where $langle a rangle$ denotes the group generated by $a$. Since $gcd(5,6) = 1$, we know that the group generated by $a^5$ is the same as the group generated by $a$.

share|cite|improve this answer

answered Apr 3 at 21:06

Jack PfaffingerJack Pfaffinger

3961112

$endgroup$

share|cite|improve this answer

answered Apr 3 at 21:06

Jack PfaffingerJack Pfaffinger

3961112

answered Apr 3 at 21:06

Jack PfaffingerJack Pfaffinger

3961112

answered Apr 3 at 21:06

Jack PfaffingerJack Pfaffinger

3961112

0

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered Apr 3 at 20:27

ShaunShaun

10.3k113686

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote?
  $endgroup$
  – Shaun
  Apr 3 at 20:46
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered Apr 3 at 20:27

ShaunShaun

10.3k113686

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why the downvote?
  $endgroup$
  – Shaun
  Apr 3 at 20:46
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Prove that subgroups of cyclic groups are themselves cyclic. Then use Lagrange's Theorem.

---

To address your misunderstanding: if $gin H$ for some $Hle G$, then all powers of $g$ are in $H$.

share|cite|improve this answer

answered Apr 3 at 20:27

ShaunShaun

10.3k113686

$endgroup$

share|cite|improve this answer

answered Apr 3 at 20:27

ShaunShaun

10.3k113686

answered Apr 3 at 20:27

ShaunShaun

10.3k113686

answered Apr 3 at 20:27

ShaunShaun

10.3k113686