Why does this cyclic subgroup have only 4 subgroups?
$begingroup$
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
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
$endgroup$
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
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
$endgroup$
$begingroup$
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
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
$endgroup$
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
abstract-algebra group-theory
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
edited Apr 3 at 20:02
J. W. Tanner
5,0551520
5,0551520
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
asked Apr 3 at 19:57
Evan KimEvan Kim
70219
70219
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
why? Could you help me understand how you got to that conclusion?
$endgroup$
– Evan Kim
Apr 3 at 20:00
2
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
$begingroup$
${1,a^5}$ 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
$begingroup$
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 |
4 Answers
4
active
oldest
votes
$begingroup$
$[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.
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
$endgroup$
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.
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
$endgroup$
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 $a$.
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
$endgroup$
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 $H$.
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
$endgroup$
1
$begingroup$
Why the downvote?
$endgroup$
– Shaun
Apr 3 at 20:46
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$[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.
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
$endgroup$
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.
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
$endgroup$
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.
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
$endgroup$
$[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.
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
answered Apr 3 at 20:01
PeterPeter
49.3k1240138
49.3k1240138
add a comment |
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.
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
$endgroup$
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.
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
$endgroup$
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.
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
$endgroup$
$lbrace 1, a^5 rbrace$ 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.
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
answered Apr 3 at 20:00
TheSilverDoeTheSilverDoe
5,608316
5,608316
add a comment |
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 $a$.
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
$endgroup$
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 $a$.
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
$endgroup$
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 $a$.
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
$endgroup$
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 $a$.
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
answered Apr 3 at 21:06
Jack PfaffingerJack Pfaffinger
3941212
3941212
add a comment |
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 $H$.
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
$endgroup$
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 $H$.
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
$endgroup$
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 $H$.
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
$endgroup$
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 $H$.
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
answered Apr 3 at 20:27
ShaunShaun
10.9k113687
10.9k113687
1
$begingroup$
Why the downvote?
$endgroup$
– Shaun
Apr 3 at 20:46
add a comment |
1
$begingroup$
Why the downvote?
$endgroup$
– Shaun
Apr 3 at 20:46
1
1
$begingroup$
Why the downvote?
$endgroup$
– Shaun
Apr 3 at 20:46
$begingroup$
Why the downvote?
$endgroup$
– Shaun
Apr 3 at 20:46
add a comment |
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$begingroup$
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