Mixing
Finding cyclic subgroups of a non-cyclic group
$begingroup$
An example would be $U(16) = {1,3,5,7,9,11,13,15}$. I know $U(16)$ is not cyclic because the order of $7$ and $15$ is $2$. But how do you go about finding the cyclic subgroups in cases like this?
abstract-algebra group-theory cyclic-groups
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
asked Oct 8 '17 at 0:46
VG3VG3
163
$endgroup$
add a comment |
$begingroup$
An example would be $U(16) = {1,3,5,7,9,11,13,15}$. I know $U(16)$ is not cyclic because the order of $7$ and $15$ is $2$. But how do you go about finding the cyclic subgroups in cases like this?
abstract-algebra group-theory cyclic-groups
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
asked Oct 8 '17 at 0:46
VG3VG3
163
$endgroup$
$begingroup$
A group is cyclic if and only if it is generated by one element. What are the subgroups of a group generated by one element?
$endgroup$
– leibnewtz
Oct 8 '17 at 18:45
add a comment |
$begingroup$
An example would be $U(16) = {1,3,5,7,9,11,13,15}$. I know $U(16)$ is not cyclic because the order of $7$ and $15$ is $2$. But how do you go about finding the cyclic subgroups in cases like this?
abstract-algebra group-theory cyclic-groups
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
asked Oct 8 '17 at 0:46
VG3VG3
163
$endgroup$
An example would be $U(16) = {1,3,5,7,9,11,13,15}$. I know $U(16)$ is not cyclic because the order of $7$ and $15$ is $2$. But how do you go about finding the cyclic subgroups in cases like this?
abstract-algebra group-theory cyclic-groups
abstract-algebra group-theory cyclic-groups
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
asked Oct 8 '17 at 0:46
VG3VG3
163
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
asked Oct 8 '17 at 0:46
VG3VG3
163
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
edited Oct 8 '17 at 18:52
leibnewtz
2,4491617
2,4491617
asked Oct 8 '17 at 0:46
VG3VG3
163
asked Oct 8 '17 at 0:46
VG3VG3
163
asked Oct 8 '17 at 0:46
VG3VG3
163
163
$begingroup$
A group is cyclic if and only if it is generated by one element. What are the subgroups of a group generated by one element?
$endgroup$
– leibnewtz
Oct 8 '17 at 18:45
add a comment |
$begingroup$
A group is cyclic if and only if it is generated by one element. What are the subgroups of a group generated by one element?
$endgroup$
– leibnewtz
Oct 8 '17 at 18:45
$begingroup$
A group is cyclic if and only if it is generated by one element. What are the subgroups of a group generated by one element?
$endgroup$
– leibnewtz
Oct 8 '17 at 18:45
$begingroup$
A group is cyclic if and only if it is generated by one element. What are the subgroups of a group generated by one element?
$endgroup$
– leibnewtz
Oct 8 '17 at 18:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This case is specialy simply: $U(16)$ is a small group $$ 16 = 2^4 quad therefore quad |U(16)| = varphi(16) = (2-1) 2^{4-1} = 8$$
and you can calculate $langle g rangle$ for all $g in U(16)$. So
$$langle 3 rangle = {1, 3, 9, 27 equiv 11}\
langle 5 rangle = {1, 5, 25 equiv 9, 45 equiv 13}\
langle 7 rangle = {1, 7} $$
and so on.
Generally, do this is not possible (because your group has infinity order or it's too big or it's boring and we have no pacient). There are some results that can be usefull:
Thm.: Let $G$ be a finite group of order $n$. $G$ is a cyclic iff for any divisor $d$ of $n$ there is at most one cyclic subgroup $H$ of $G$ with order $d$.
The proof is simple, you only use the classical result $n=sum_{d|n}varphi(d)$.
We have another interesting result if you are looking to $mathbb{Z}_p$
Thm.: Let $mathbb{K}$ a field and $mathbb{K}^{times}$ it's multiplicative group. Let $G$ a subgroup of $mathbb{K}^{times}$. If $G$ has finite order, then it's cyclic. In particular, $mathbb{F}_{p^n}^{times}$ is cyclic.
Proof: Let $n = |G|$ and $d$ a divisor of $n$. We will use the equivalence of the previous result. Suppose there are $H$ and $H'$ distinct subgroups of order $d$. By Lagrange Theorem, $h^d = 1$ for all $h in H cap H'$. So, the elements of $H cap H'$ are roots of $f(x) = x^d - 1 in mathbb{K}[x]$, but $$|H cap H'| leq {k in mathbb{K} mid f(k) = 0} leq deg f(x) = d < |H cap H'|$$ Contradiction. Therefore, there is at most one subgroup of order $d$ and $G$ is cyclic.
I think I should mention the Cauchy's Theorem:
Thm (Cauchy).: Let $G$ be a finite group and $p$ a prime. If $p$ divides $|G|$, there is $gin G$ with order $p$, i.e. , there is a cyclic subgroup $H = langle p rangle$ with order p.
(See here for details and proofs). Note, Cauchy's result don't tell you how to find some cyclic subgroup, but it say that there is at least one cyclic group with order $p$.
Edit: Probably, I should do a example. Let's try $U(140)$. We know that $$ 140 = 2^2 cdot 5 cdot 7 \ therefore |U(140)| = varphi(140) = (2-1)2^{2-1} (5-1) 5^{1-1} (7-1)7^{1-1} = 48 = 2^4 cdot 3$$
Then there is at last one cyclic subgroup of $U(140)$ with order 2 and 3. If you show that there are more then one subgroup woth order 2 or 3, then the first theorem says that $U(140)$ is not cyclic. Of course $-1$ have order 2 $$langle -1 rangle = {1, -1}$$ and we only need to find another element with order $
2$ (I guess this is what you did in your example). Let's try some numbers not divisible by 2, 5 and 7, maybe prime numbers are a good idea. I tried some random numbers with Wolfram's help and $$langle 29 rangle = {1, 29} \ langle 41 rangle = {1, 41}\ langle 99 rangle = {1, 99} \ langle 111 rangle = {1, 111}$$
Yeah, it's enought. So $U(140)$ is not cyclic. On particular there is no field $mathbb{K}$ such that $U(140)$ is a subgroup of $mathbb{K}^{times}$.
But how many cyclic subgroup we can find? Here is a partial answer: We could use Sylow Theorems.
Thm (3º Sylow) Let $G$ a finite group with order $n$ and p be a prime factor such that $n = p^m b$ with $gcd(p,b)=1$ Let $n_p$ be the number of Sylow p-subgroups of G (subgroup with order $p^m$). Then the following hold:
1) $n_p$ divides the index of the Sylow p-subgroup in G;
2) $n_p equiv 1 (mod p)$;
3) $n_p = |G : N(P)|$, where $P$ is any Sylow p-subgroup of $n$ and $N$ denotes the normalizer.
Using it in our example, we know that $$n_3 mid |U(140)/3| = 2^4 Rightarrow n_3 in {1, 2, 4, 8, 16}$$ and $$n_3 equiv 1 mod(3) Rightarrow n_3 in {1, 4, 16}$$
Ok, we know (without boring computations) that $U(140)$ is not cyclic; ok, we don't know how many subgroups with order 3, but we know that there are 1, 4 or 16. In this example, with a computer we can test some numbers and do a full list of cyclic subgroups because it's finite. For example, I discovered that is just one subgroup with order 3 $$langle 81 rangle = langle 121 rangle = {1, 81, 121}$$
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
$endgroup$
$begingroup$
Is this kind of result are you looking for?
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 1:29
$begingroup$
The results you've posted are mostly (partial) converses to Lagrange's theorem. The OP's question has a much simpler answer: The cyclic subgroups of a group $G$ are those generated by one of its elements.
$endgroup$
– leibnewtz
Oct 8 '17 at 18:50
$begingroup$
Yes, it's the definition and it's easy to calculate all cyclic subgroup in $U(16)$, but it's not always the case (and it's really boring). I interpreted the question as "how can we find cyclic groups in a real world problem?".
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 19:20
1
$begingroup$
Knowing something exists doesn't necessarily help you find it. Even if we know the order of a group, and we know that it has to contain an element of order $5$, we still have to go through and compute the order of elements by hand to find which one that is. Sometimes we can use other facts about the group, but order of elements is an elusive thing
$endgroup$
– leibnewtz
Oct 9 '17 at 2:42
1
$begingroup$
Although I do think OP's question is kind of ambiguous. I'm not sure if they want to find cyclic subgroups up to isomorphism, or if they want to explicitly write out the subgroups. In the former case, theorems like Cauchy's theorem would give you at least a partial answer
$endgroup$
– leibnewtz
Oct 9 '17 at 2:44
|
show 3 more comments
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$begingroup$
This case is specialy simply: $U(16)$ is a small group $$ 16 = 2^4 quad therefore quad |U(16)| = varphi(16) = (2-1) 2^{4-1} = 8$$
and you can calculate $langle g rangle$ for all $g in U(16)$. So
$$langle 3 rangle = {1, 3, 9, 27 equiv 11}\
langle 5 rangle = {1, 5, 25 equiv 9, 45 equiv 13}\
langle 7 rangle = {1, 7} $$
and so on.
Generally, do this is not possible (because your group has infinity order or it's too big or it's boring and we have no pacient). There are some results that can be usefull:
Thm.: Let $G$ be a finite group of order $n$. $G$ is a cyclic iff for any divisor $d$ of $n$ there is at most one cyclic subgroup $H$ of $G$ with order $d$.
The proof is simple, you only use the classical result $n=sum_{d|n}varphi(d)$.
We have another interesting result if you are looking to $mathbb{Z}_p$
Thm.: Let $mathbb{K}$ a field and $mathbb{K}^{times}$ it's multiplicative group. Let $G$ a subgroup of $mathbb{K}^{times}$. If $G$ has finite order, then it's cyclic. In particular, $mathbb{F}_{p^n}^{times}$ is cyclic.
Proof: Let $n = |G|$ and $d$ a divisor of $n$. We will use the equivalence of the previous result. Suppose there are $H$ and $H'$ distinct subgroups of order $d$. By Lagrange Theorem, $h^d = 1$ for all $h in H cap H'$. So, the elements of $H cap H'$ are roots of $f(x) = x^d - 1 in mathbb{K}[x]$, but $$|H cap H'| leq {k in mathbb{K} mid f(k) = 0} leq deg f(x) = d < |H cap H'|$$ Contradiction. Therefore, there is at most one subgroup of order $d$ and $G$ is cyclic.
I think I should mention the Cauchy's Theorem:
Thm (Cauchy).: Let $G$ be a finite group and $p$ a prime. If $p$ divides $|G|$, there is $gin G$ with order $p$, i.e. , there is a cyclic subgroup $H = langle p rangle$ with order p.
(See here for details and proofs). Note, Cauchy's result don't tell you how to find some cyclic subgroup, but it say that there is at least one cyclic group with order $p$.
Edit: Probably, I should do a example. Let's try $U(140)$. We know that $$ 140 = 2^2 cdot 5 cdot 7 \ therefore |U(140)| = varphi(140) = (2-1)2^{2-1} (5-1) 5^{1-1} (7-1)7^{1-1} = 48 = 2^4 cdot 3$$
Then there is at last one cyclic subgroup of $U(140)$ with order 2 and 3. If you show that there are more then one subgroup woth order 2 or 3, then the first theorem says that $U(140)$ is not cyclic. Of course $-1$ have order 2 $$langle -1 rangle = {1, -1}$$ and we only need to find another element with order $
2$ (I guess this is what you did in your example). Let's try some numbers not divisible by 2, 5 and 7, maybe prime numbers are a good idea. I tried some random numbers with Wolfram's help and $$langle 29 rangle = {1, 29} \ langle 41 rangle = {1, 41}\ langle 99 rangle = {1, 99} \ langle 111 rangle = {1, 111}$$
Yeah, it's enought. So $U(140)$ is not cyclic. On particular there is no field $mathbb{K}$ such that $U(140)$ is a subgroup of $mathbb{K}^{times}$.
But how many cyclic subgroup we can find? Here is a partial answer: We could use Sylow Theorems.
Thm (3º Sylow) Let $G$ a finite group with order $n$ and p be a prime factor such that $n = p^m b$ with $gcd(p,b)=1$ Let $n_p$ be the number of Sylow p-subgroups of G (subgroup with order $p^m$). Then the following hold:
1) $n_p$ divides the index of the Sylow p-subgroup in G;
2) $n_p equiv 1 (mod p)$;
3) $n_p = |G : N(P)|$, where $P$ is any Sylow p-subgroup of $n$ and $N$ denotes the normalizer.
Using it in our example, we know that $$n_3 mid |U(140)/3| = 2^4 Rightarrow n_3 in {1, 2, 4, 8, 16}$$ and $$n_3 equiv 1 mod(3) Rightarrow n_3 in {1, 4, 16}$$
Ok, we know (without boring computations) that $U(140)$ is not cyclic; ok, we don't know how many subgroups with order 3, but we know that there are 1, 4 or 16. In this example, with a computer we can test some numbers and do a full list of cyclic subgroups because it's finite. For example, I discovered that is just one subgroup with order 3 $$langle 81 rangle = langle 121 rangle = {1, 81, 121}$$
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
$endgroup$
$begingroup$
Is this kind of result are you looking for?
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 1:29
$begingroup$
The results you've posted are mostly (partial) converses to Lagrange's theorem. The OP's question has a much simpler answer: The cyclic subgroups of a group $G$ are those generated by one of its elements.
$endgroup$
– leibnewtz
Oct 8 '17 at 18:50
$begingroup$
Yes, it's the definition and it's easy to calculate all cyclic subgroup in $U(16)$, but it's not always the case (and it's really boring). I interpreted the question as "how can we find cyclic groups in a real world problem?".
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 19:20
1
$begingroup$
Knowing something exists doesn't necessarily help you find it. Even if we know the order of a group, and we know that it has to contain an element of order $5$, we still have to go through and compute the order of elements by hand to find which one that is. Sometimes we can use other facts about the group, but order of elements is an elusive thing
$endgroup$
– leibnewtz
Oct 9 '17 at 2:42
1
$begingroup$
Although I do think OP's question is kind of ambiguous. I'm not sure if they want to find cyclic subgroups up to isomorphism, or if they want to explicitly write out the subgroups. In the former case, theorems like Cauchy's theorem would give you at least a partial answer
$endgroup$
– leibnewtz
Oct 9 '17 at 2:44
|
show 3 more comments
$begingroup$
This case is specialy simply: $U(16)$ is a small group $$ 16 = 2^4 quad therefore quad |U(16)| = varphi(16) = (2-1) 2^{4-1} = 8$$
and you can calculate $langle g rangle$ for all $g in U(16)$. So
$$langle 3 rangle = {1, 3, 9, 27 equiv 11}\
langle 5 rangle = {1, 5, 25 equiv 9, 45 equiv 13}\
langle 7 rangle = {1, 7} $$
and so on.
Generally, do this is not possible (because your group has infinity order or it's too big or it's boring and we have no pacient). There are some results that can be usefull:
Thm.: Let $G$ be a finite group of order $n$. $G$ is a cyclic iff for any divisor $d$ of $n$ there is at most one cyclic subgroup $H$ of $G$ with order $d$.
The proof is simple, you only use the classical result $n=sum_{d|n}varphi(d)$.
We have another interesting result if you are looking to $mathbb{Z}_p$
Thm.: Let $mathbb{K}$ a field and $mathbb{K}^{times}$ it's multiplicative group. Let $G$ a subgroup of $mathbb{K}^{times}$. If $G$ has finite order, then it's cyclic. In particular, $mathbb{F}_{p^n}^{times}$ is cyclic.
Proof: Let $n = |G|$ and $d$ a divisor of $n$. We will use the equivalence of the previous result. Suppose there are $H$ and $H'$ distinct subgroups of order $d$. By Lagrange Theorem, $h^d = 1$ for all $h in H cap H'$. So, the elements of $H cap H'$ are roots of $f(x) = x^d - 1 in mathbb{K}[x]$, but $$|H cap H'| leq {k in mathbb{K} mid f(k) = 0} leq deg f(x) = d < |H cap H'|$$ Contradiction. Therefore, there is at most one subgroup of order $d$ and $G$ is cyclic.
I think I should mention the Cauchy's Theorem:
Thm (Cauchy).: Let $G$ be a finite group and $p$ a prime. If $p$ divides $|G|$, there is $gin G$ with order $p$, i.e. , there is a cyclic subgroup $H = langle p rangle$ with order p.
(See here for details and proofs). Note, Cauchy's result don't tell you how to find some cyclic subgroup, but it say that there is at least one cyclic group with order $p$.
Edit: Probably, I should do a example. Let's try $U(140)$. We know that $$ 140 = 2^2 cdot 5 cdot 7 \ therefore |U(140)| = varphi(140) = (2-1)2^{2-1} (5-1) 5^{1-1} (7-1)7^{1-1} = 48 = 2^4 cdot 3$$
Then there is at last one cyclic subgroup of $U(140)$ with order 2 and 3. If you show that there are more then one subgroup woth order 2 or 3, then the first theorem says that $U(140)$ is not cyclic. Of course $-1$ have order 2 $$langle -1 rangle = {1, -1}$$ and we only need to find another element with order $
2$ (I guess this is what you did in your example). Let's try some numbers not divisible by 2, 5 and 7, maybe prime numbers are a good idea. I tried some random numbers with Wolfram's help and $$langle 29 rangle = {1, 29} \ langle 41 rangle = {1, 41}\ langle 99 rangle = {1, 99} \ langle 111 rangle = {1, 111}$$
Yeah, it's enought. So $U(140)$ is not cyclic. On particular there is no field $mathbb{K}$ such that $U(140)$ is a subgroup of $mathbb{K}^{times}$.
But how many cyclic subgroup we can find? Here is a partial answer: We could use Sylow Theorems.
Thm (3º Sylow) Let $G$ a finite group with order $n$ and p be a prime factor such that $n = p^m b$ with $gcd(p,b)=1$ Let $n_p$ be the number of Sylow p-subgroups of G (subgroup with order $p^m$). Then the following hold:
1) $n_p$ divides the index of the Sylow p-subgroup in G;
2) $n_p equiv 1 (mod p)$;
3) $n_p = |G : N(P)|$, where $P$ is any Sylow p-subgroup of $n$ and $N$ denotes the normalizer.
Using it in our example, we know that $$n_3 mid |U(140)/3| = 2^4 Rightarrow n_3 in {1, 2, 4, 8, 16}$$ and $$n_3 equiv 1 mod(3) Rightarrow n_3 in {1, 4, 16}$$
Ok, we know (without boring computations) that $U(140)$ is not cyclic; ok, we don't know how many subgroups with order 3, but we know that there are 1, 4 or 16. In this example, with a computer we can test some numbers and do a full list of cyclic subgroups because it's finite. For example, I discovered that is just one subgroup with order 3 $$langle 81 rangle = langle 121 rangle = {1, 81, 121}$$
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
$endgroup$
$begingroup$
Is this kind of result are you looking for?
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 1:29
$begingroup$
The results you've posted are mostly (partial) converses to Lagrange's theorem. The OP's question has a much simpler answer: The cyclic subgroups of a group $G$ are those generated by one of its elements.
$endgroup$
– leibnewtz
Oct 8 '17 at 18:50
$begingroup$
Yes, it's the definition and it's easy to calculate all cyclic subgroup in $U(16)$, but it's not always the case (and it's really boring). I interpreted the question as "how can we find cyclic groups in a real world problem?".
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 19:20
1
$begingroup$
Knowing something exists doesn't necessarily help you find it. Even if we know the order of a group, and we know that it has to contain an element of order $5$, we still have to go through and compute the order of elements by hand to find which one that is. Sometimes we can use other facts about the group, but order of elements is an elusive thing
$endgroup$
– leibnewtz
Oct 9 '17 at 2:42
1
$begingroup$
Although I do think OP's question is kind of ambiguous. I'm not sure if they want to find cyclic subgroups up to isomorphism, or if they want to explicitly write out the subgroups. In the former case, theorems like Cauchy's theorem would give you at least a partial answer
$endgroup$
– leibnewtz
Oct 9 '17 at 2:44
|
show 3 more comments
$begingroup$
This case is specialy simply: $U(16)$ is a small group $$ 16 = 2^4 quad therefore quad |U(16)| = varphi(16) = (2-1) 2^{4-1} = 8$$
and you can calculate $langle g rangle$ for all $g in U(16)$. So
$$langle 3 rangle = {1, 3, 9, 27 equiv 11}\
langle 5 rangle = {1, 5, 25 equiv 9, 45 equiv 13}\
langle 7 rangle = {1, 7} $$
and so on.
Generally, do this is not possible (because your group has infinity order or it's too big or it's boring and we have no pacient). There are some results that can be usefull:
Thm.: Let $G$ be a finite group of order $n$. $G$ is a cyclic iff for any divisor $d$ of $n$ there is at most one cyclic subgroup $H$ of $G$ with order $d$.
The proof is simple, you only use the classical result $n=sum_{d|n}varphi(d)$.
We have another interesting result if you are looking to $mathbb{Z}_p$
Thm.: Let $mathbb{K}$ a field and $mathbb{K}^{times}$ it's multiplicative group. Let $G$ a subgroup of $mathbb{K}^{times}$. If $G$ has finite order, then it's cyclic. In particular, $mathbb{F}_{p^n}^{times}$ is cyclic.
Proof: Let $n = |G|$ and $d$ a divisor of $n$. We will use the equivalence of the previous result. Suppose there are $H$ and $H'$ distinct subgroups of order $d$. By Lagrange Theorem, $h^d = 1$ for all $h in H cap H'$. So, the elements of $H cap H'$ are roots of $f(x) = x^d - 1 in mathbb{K}[x]$, but $$|H cap H'| leq {k in mathbb{K} mid f(k) = 0} leq deg f(x) = d < |H cap H'|$$ Contradiction. Therefore, there is at most one subgroup of order $d$ and $G$ is cyclic.
I think I should mention the Cauchy's Theorem:
Thm (Cauchy).: Let $G$ be a finite group and $p$ a prime. If $p$ divides $|G|$, there is $gin G$ with order $p$, i.e. , there is a cyclic subgroup $H = langle p rangle$ with order p.
(See here for details and proofs). Note, Cauchy's result don't tell you how to find some cyclic subgroup, but it say that there is at least one cyclic group with order $p$.
Edit: Probably, I should do a example. Let's try $U(140)$. We know that $$ 140 = 2^2 cdot 5 cdot 7 \ therefore |U(140)| = varphi(140) = (2-1)2^{2-1} (5-1) 5^{1-1} (7-1)7^{1-1} = 48 = 2^4 cdot 3$$
Then there is at last one cyclic subgroup of $U(140)$ with order 2 and 3. If you show that there are more then one subgroup woth order 2 or 3, then the first theorem says that $U(140)$ is not cyclic. Of course $-1$ have order 2 $$langle -1 rangle = {1, -1}$$ and we only need to find another element with order $
2$ (I guess this is what you did in your example). Let's try some numbers not divisible by 2, 5 and 7, maybe prime numbers are a good idea. I tried some random numbers with Wolfram's help and $$langle 29 rangle = {1, 29} \ langle 41 rangle = {1, 41}\ langle 99 rangle = {1, 99} \ langle 111 rangle = {1, 111}$$
Yeah, it's enought. So $U(140)$ is not cyclic. On particular there is no field $mathbb{K}$ such that $U(140)$ is a subgroup of $mathbb{K}^{times}$.
But how many cyclic subgroup we can find? Here is a partial answer: We could use Sylow Theorems.
Thm (3º Sylow) Let $G$ a finite group with order $n$ and p be a prime factor such that $n = p^m b$ with $gcd(p,b)=1$ Let $n_p$ be the number of Sylow p-subgroups of G (subgroup with order $p^m$). Then the following hold:
1) $n_p$ divides the index of the Sylow p-subgroup in G;
2) $n_p equiv 1 (mod p)$;
3) $n_p = |G : N(P)|$, where $P$ is any Sylow p-subgroup of $n$ and $N$ denotes the normalizer.
Using it in our example, we know that $$n_3 mid |U(140)/3| = 2^4 Rightarrow n_3 in {1, 2, 4, 8, 16}$$ and $$n_3 equiv 1 mod(3) Rightarrow n_3 in {1, 4, 16}$$
Ok, we know (without boring computations) that $U(140)$ is not cyclic; ok, we don't know how many subgroups with order 3, but we know that there are 1, 4 or 16. In this example, with a computer we can test some numbers and do a full list of cyclic subgroups because it's finite. For example, I discovered that is just one subgroup with order 3 $$langle 81 rangle = langle 121 rangle = {1, 81, 121}$$
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
$endgroup$
This case is specialy simply: $U(16)$ is a small group $$ 16 = 2^4 quad therefore quad |U(16)| = varphi(16) = (2-1) 2^{4-1} = 8$$
and you can calculate $langle g rangle$ for all $g in U(16)$. So
$$langle 3 rangle = {1, 3, 9, 27 equiv 11}\
langle 5 rangle = {1, 5, 25 equiv 9, 45 equiv 13}\
langle 7 rangle = {1, 7} $$
and so on.
Generally, do this is not possible (because your group has infinity order or it's too big or it's boring and we have no pacient). There are some results that can be usefull:
Thm.: Let $G$ be a finite group of order $n$. $G$ is a cyclic iff for any divisor $d$ of $n$ there is at most one cyclic subgroup $H$ of $G$ with order $d$.
The proof is simple, you only use the classical result $n=sum_{d|n}varphi(d)$.
We have another interesting result if you are looking to $mathbb{Z}_p$
Thm.: Let $mathbb{K}$ a field and $mathbb{K}^{times}$ it's multiplicative group. Let $G$ a subgroup of $mathbb{K}^{times}$. If $G$ has finite order, then it's cyclic. In particular, $mathbb{F}_{p^n}^{times}$ is cyclic.
Proof: Let $n = |G|$ and $d$ a divisor of $n$. We will use the equivalence of the previous result. Suppose there are $H$ and $H'$ distinct subgroups of order $d$. By Lagrange Theorem, $h^d = 1$ for all $h in H cap H'$. So, the elements of $H cap H'$ are roots of $f(x) = x^d - 1 in mathbb{K}[x]$, but $$|H cap H'| leq {k in mathbb{K} mid f(k) = 0} leq deg f(x) = d < |H cap H'|$$ Contradiction. Therefore, there is at most one subgroup of order $d$ and $G$ is cyclic.
I think I should mention the Cauchy's Theorem:
Thm (Cauchy).: Let $G$ be a finite group and $p$ a prime. If $p$ divides $|G|$, there is $gin G$ with order $p$, i.e. , there is a cyclic subgroup $H = langle p rangle$ with order p.
(See here for details and proofs). Note, Cauchy's result don't tell you how to find some cyclic subgroup, but it say that there is at least one cyclic group with order $p$.
Edit: Probably, I should do a example. Let's try $U(140)$. We know that $$ 140 = 2^2 cdot 5 cdot 7 \ therefore |U(140)| = varphi(140) = (2-1)2^{2-1} (5-1) 5^{1-1} (7-1)7^{1-1} = 48 = 2^4 cdot 3$$
Then there is at last one cyclic subgroup of $U(140)$ with order 2 and 3. If you show that there are more then one subgroup woth order 2 or 3, then the first theorem says that $U(140)$ is not cyclic. Of course $-1$ have order 2 $$langle -1 rangle = {1, -1}$$ and we only need to find another element with order $
2$ (I guess this is what you did in your example). Let's try some numbers not divisible by 2, 5 and 7, maybe prime numbers are a good idea. I tried some random numbers with Wolfram's help and $$langle 29 rangle = {1, 29} \ langle 41 rangle = {1, 41}\ langle 99 rangle = {1, 99} \ langle 111 rangle = {1, 111}$$
Yeah, it's enought. So $U(140)$ is not cyclic. On particular there is no field $mathbb{K}$ such that $U(140)$ is a subgroup of $mathbb{K}^{times}$.
But how many cyclic subgroup we can find? Here is a partial answer: We could use Sylow Theorems.
Thm (3º Sylow) Let $G$ a finite group with order $n$ and p be a prime factor such that $n = p^m b$ with $gcd(p,b)=1$ Let $n_p$ be the number of Sylow p-subgroups of G (subgroup with order $p^m$). Then the following hold:
1) $n_p$ divides the index of the Sylow p-subgroup in G;
2) $n_p equiv 1 (mod p)$;
3) $n_p = |G : N(P)|$, where $P$ is any Sylow p-subgroup of $n$ and $N$ denotes the normalizer.
Using it in our example, we know that $$n_3 mid |U(140)/3| = 2^4 Rightarrow n_3 in {1, 2, 4, 8, 16}$$ and $$n_3 equiv 1 mod(3) Rightarrow n_3 in {1, 4, 16}$$
Ok, we know (without boring computations) that $U(140)$ is not cyclic; ok, we don't know how many subgroups with order 3, but we know that there are 1, 4 or 16. In this example, with a computer we can test some numbers and do a full list of cyclic subgroups because it's finite. For example, I discovered that is just one subgroup with order 3 $$langle 81 rangle = langle 121 rangle = {1, 81, 121}$$
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
edited Oct 8 '17 at 18:39
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
answered Oct 8 '17 at 1:28
Thadeu Henrique CostaThadeu Henrique Costa
397213
397213
$begingroup$
Is this kind of result are you looking for?
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 1:29
$begingroup$
The results you've posted are mostly (partial) converses to Lagrange's theorem. The OP's question has a much simpler answer: The cyclic subgroups of a group $G$ are those generated by one of its elements.
$endgroup$
– leibnewtz
Oct 8 '17 at 18:50
$begingroup$
Yes, it's the definition and it's easy to calculate all cyclic subgroup in $U(16)$, but it's not always the case (and it's really boring). I interpreted the question as "how can we find cyclic groups in a real world problem?".
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 19:20
1
$begingroup$
Knowing something exists doesn't necessarily help you find it. Even if we know the order of a group, and we know that it has to contain an element of order $5$, we still have to go through and compute the order of elements by hand to find which one that is. Sometimes we can use other facts about the group, but order of elements is an elusive thing
$endgroup$
– leibnewtz
Oct 9 '17 at 2:42
1
$begingroup$
Although I do think OP's question is kind of ambiguous. I'm not sure if they want to find cyclic subgroups up to isomorphism, or if they want to explicitly write out the subgroups. In the former case, theorems like Cauchy's theorem would give you at least a partial answer
$endgroup$
– leibnewtz
Oct 9 '17 at 2:44
|
show 3 more comments
$begingroup$
Is this kind of result are you looking for?
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 1:29
$begingroup$
The results you've posted are mostly (partial) converses to Lagrange's theorem. The OP's question has a much simpler answer: The cyclic subgroups of a group $G$ are those generated by one of its elements.
$endgroup$
– leibnewtz
Oct 8 '17 at 18:50
$begingroup$
Yes, it's the definition and it's easy to calculate all cyclic subgroup in $U(16)$, but it's not always the case (and it's really boring). I interpreted the question as "how can we find cyclic groups in a real world problem?".
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 19:20
1
$begingroup$
Knowing something exists doesn't necessarily help you find it. Even if we know the order of a group, and we know that it has to contain an element of order $5$, we still have to go through and compute the order of elements by hand to find which one that is. Sometimes we can use other facts about the group, but order of elements is an elusive thing
$endgroup$
– leibnewtz
Oct 9 '17 at 2:42
1
$begingroup$
Although I do think OP's question is kind of ambiguous. I'm not sure if they want to find cyclic subgroups up to isomorphism, or if they want to explicitly write out the subgroups. In the former case, theorems like Cauchy's theorem would give you at least a partial answer
$endgroup$
– leibnewtz
Oct 9 '17 at 2:44
$begingroup$
Is this kind of result are you looking for?
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 1:29
$begingroup$
Is this kind of result are you looking for?
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 1:29
$begingroup$
The results you've posted are mostly (partial) converses to Lagrange's theorem. The OP's question has a much simpler answer: The cyclic subgroups of a group $G$ are those generated by one of its elements.
$endgroup$
– leibnewtz
Oct 8 '17 at 18:50
$begingroup$
The results you've posted are mostly (partial) converses to Lagrange's theorem. The OP's question has a much simpler answer: The cyclic subgroups of a group $G$ are those generated by one of its elements.
$endgroup$
– leibnewtz
Oct 8 '17 at 18:50
$begingroup$
Yes, it's the definition and it's easy to calculate all cyclic subgroup in $U(16)$, but it's not always the case (and it's really boring). I interpreted the question as "how can we find cyclic groups in a real world problem?".
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 19:20
$begingroup$
Yes, it's the definition and it's easy to calculate all cyclic subgroup in $U(16)$, but it's not always the case (and it's really boring). I interpreted the question as "how can we find cyclic groups in a real world problem?".
$endgroup$
– Thadeu Henrique Costa
Oct 8 '17 at 19:20
1
1
$begingroup$
Knowing something exists doesn't necessarily help you find it. Even if we know the order of a group, and we know that it has to contain an element of order $5$, we still have to go through and compute the order of elements by hand to find which one that is. Sometimes we can use other facts about the group, but order of elements is an elusive thing
$endgroup$
– leibnewtz
Oct 9 '17 at 2:42
$begingroup$
Knowing something exists doesn't necessarily help you find it. Even if we know the order of a group, and we know that it has to contain an element of order $5$, we still have to go through and compute the order of elements by hand to find which one that is. Sometimes we can use other facts about the group, but order of elements is an elusive thing
$endgroup$
– leibnewtz
Oct 9 '17 at 2:42
1
1
$begingroup$
Although I do think OP's question is kind of ambiguous. I'm not sure if they want to find cyclic subgroups up to isomorphism, or if they want to explicitly write out the subgroups. In the former case, theorems like Cauchy's theorem would give you at least a partial answer
$endgroup$
– leibnewtz
Oct 9 '17 at 2:44
$begingroup$
Although I do think OP's question is kind of ambiguous. I'm not sure if they want to find cyclic subgroups up to isomorphism, or if they want to explicitly write out the subgroups. In the former case, theorems like Cauchy's theorem would give you at least a partial answer
$endgroup$
– leibnewtz
Oct 9 '17 at 2:44
|
show 3 more comments
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$begingroup$
A group is cyclic if and only if it is generated by one element. What are the subgroups of a group generated by one element?
$endgroup$
– leibnewtz
Oct 8 '17 at 18:45