Mixing
Is it possible to find all finite groups which has subgroup of order $n$?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Is it possible to find all finite group which has a subgroup of certain order $n$ (here say for example $2$) ? I am thinking about this problem for few days, but not getting how to approach.
I am thinking that the theorem -
Every finite group is isomorphic to a subgroup of $S_n$.
can give something about it. Can anyone answer please? I have searched for it, but it is not posted here and not available in internet.
abstract-algebra group-theory
asked Jul 28 at 5:17
Ken Ono
404
 |Â
show 1 more comment
up vote
2
down vote
favorite
Is it possible to find all finite group which has a subgroup of certain order $n$ (here say for example $2$) ? I am thinking about this problem for few days, but not getting how to approach.
I am thinking that the theorem -
Every finite group is isomorphic to a subgroup of $S_n$.
can give something about it. Can anyone answer please? I have searched for it, but it is not posted here and not available in internet.
abstract-algebra group-theory
asked Jul 28 at 5:17
Ken Ono
404
8
If I'm not mistaken, the trivial group of order 1 is a subgroup of all groups. Thus, a special case of your problem is equivalent to the classification of all finite groups, which is much harder than the classification of all finite simple groups, which is a problem that took decades of work by hundreds of group theorists. Good luck!
– Display name
Jul 28 at 5:26
1
@Displayname Ok, then for some specific $n$ is it possible to find all groups, which has subgroup of order $n$ ? as, I have mentioned suppose - $n=2$.
– Ken Ono
Jul 28 at 5:34
1
Also, every finite group whose order is divisible by a prime number $p$ has a subgroup of order $p$ (Cauchy's theorem). Thus, another special case of your problem is equivalent to the classification of all finite groups that are not $p'$-groups.
– Orat
Jul 28 at 5:42
Well, to take the help of Calay's theorem, can we find all subgroups of $S_n$ that has subgroup of order $2$ ? If we can find, then it can be stated - "The groups which has subgroup of order $2$ are $cdots$, upto isomorphism"
– Ken Ono
Jul 28 at 6:26
1
@KenOno By Cauchy's Theorem, every group of even order has a subgroup of order $2$. By Cayley's theorem, every group is isomorphic to a subgroup of $S_n$ for some $n$. So finding all such subgroups is equivalent to finding all groups of even order. But, given any finite group $G$, the group $G times mathbbZ/2mathbbZ$ is a group of even order, so again, this is equivalent to finding all finite groups, a problem very far from being solved as it stands now..
– Henrique Augusto Souza
Jul 28 at 6:35
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Is it possible to find all finite group which has a subgroup of certain order $n$ (here say for example $2$) ? I am thinking about this problem for few days, but not getting how to approach.
I am thinking that the theorem -
Every finite group is isomorphic to a subgroup of $S_n$.
can give something about it. Can anyone answer please? I have searched for it, but it is not posted here and not available in internet.
abstract-algebra group-theory
asked Jul 28 at 5:17
Ken Ono
404
Is it possible to find all finite group which has a subgroup of certain order $n$ (here say for example $2$) ? I am thinking about this problem for few days, but not getting how to approach.
I am thinking that the theorem -
Every finite group is isomorphic to a subgroup of $S_n$.
can give something about it. Can anyone answer please? I have searched for it, but it is not posted here and not available in internet.
abstract-algebra group-theory
asked Jul 28 at 5:17
Ken Ono
404
asked Jul 28 at 5:17
Ken Ono
404
asked Jul 28 at 5:17
Ken Ono
404
asked Jul 28 at 5:17
Ken Ono
404
404
8
If I'm not mistaken, the trivial group of order 1 is a subgroup of all groups. Thus, a special case of your problem is equivalent to the classification of all finite groups, which is much harder than the classification of all finite simple groups, which is a problem that took decades of work by hundreds of group theorists. Good luck!
– Display name
Jul 28 at 5:26
1
@Displayname Ok, then for some specific $n$ is it possible to find all groups, which has subgroup of order $n$ ? as, I have mentioned suppose - $n=2$.
– Ken Ono
Jul 28 at 5:34
1
Also, every finite group whose order is divisible by a prime number $p$ has a subgroup of order $p$ (Cauchy's theorem). Thus, another special case of your problem is equivalent to the classification of all finite groups that are not $p'$-groups.
– Orat
Jul 28 at 5:42
Well, to take the help of Calay's theorem, can we find all subgroups of $S_n$ that has subgroup of order $2$ ? If we can find, then it can be stated - "The groups which has subgroup of order $2$ are $cdots$, upto isomorphism"
– Ken Ono
Jul 28 at 6:26
1
@KenOno By Cauchy's Theorem, every group of even order has a subgroup of order $2$. By Cayley's theorem, every group is isomorphic to a subgroup of $S_n$ for some $n$. So finding all such subgroups is equivalent to finding all groups of even order. But, given any finite group $G$, the group $G times mathbbZ/2mathbbZ$ is a group of even order, so again, this is equivalent to finding all finite groups, a problem very far from being solved as it stands now..
– Henrique Augusto Souza
Jul 28 at 6:35
 |Â
show 1 more comment
8
If I'm not mistaken, the trivial group of order 1 is a subgroup of all groups. Thus, a special case of your problem is equivalent to the classification of all finite groups, which is much harder than the classification of all finite simple groups, which is a problem that took decades of work by hundreds of group theorists. Good luck!
– Display name
Jul 28 at 5:26
1
@Displayname Ok, then for some specific $n$ is it possible to find all groups, which has subgroup of order $n$ ? as, I have mentioned suppose - $n=2$.
– Ken Ono
Jul 28 at 5:34
1
Also, every finite group whose order is divisible by a prime number $p$ has a subgroup of order $p$ (Cauchy's theorem). Thus, another special case of your problem is equivalent to the classification of all finite groups that are not $p'$-groups.
– Orat
Jul 28 at 5:42
Well, to take the help of Calay's theorem, can we find all subgroups of $S_n$ that has subgroup of order $2$ ? If we can find, then it can be stated - "The groups which has subgroup of order $2$ are $cdots$, upto isomorphism"
– Ken Ono
Jul 28 at 6:26
1
@KenOno By Cauchy's Theorem, every group of even order has a subgroup of order $2$. By Cayley's theorem, every group is isomorphic to a subgroup of $S_n$ for some $n$. So finding all such subgroups is equivalent to finding all groups of even order. But, given any finite group $G$, the group $G times mathbbZ/2mathbbZ$ is a group of even order, so again, this is equivalent to finding all finite groups, a problem very far from being solved as it stands now..
– Henrique Augusto Souza
Jul 28 at 6:35
8
8
If I'm not mistaken, the trivial group of order 1 is a subgroup of all groups. Thus, a special case of your problem is equivalent to the classification of all finite groups, which is much harder than the classification of all finite simple groups, which is a problem that took decades of work by hundreds of group theorists. Good luck!
– Display name
Jul 28 at 5:26
If I'm not mistaken, the trivial group of order 1 is a subgroup of all groups. Thus, a special case of your problem is equivalent to the classification of all finite groups, which is much harder than the classification of all finite simple groups, which is a problem that took decades of work by hundreds of group theorists. Good luck!
– Display name
Jul 28 at 5:26
1
1
@Displayname Ok, then for some specific $n$ is it possible to find all groups, which has subgroup of order $n$ ? as, I have mentioned suppose - $n=2$.
– Ken Ono
Jul 28 at 5:34
@Displayname Ok, then for some specific $n$ is it possible to find all groups, which has subgroup of order $n$ ? as, I have mentioned suppose - $n=2$.
– Ken Ono
Jul 28 at 5:34
1
1
Also, every finite group whose order is divisible by a prime number $p$ has a subgroup of order $p$ (Cauchy's theorem). Thus, another special case of your problem is equivalent to the classification of all finite groups that are not $p'$-groups.
– Orat
Jul 28 at 5:42
Also, every finite group whose order is divisible by a prime number $p$ has a subgroup of order $p$ (Cauchy's theorem). Thus, another special case of your problem is equivalent to the classification of all finite groups that are not $p'$-groups.
– Orat
Jul 28 at 5:42
Well, to take the help of Calay's theorem, can we find all subgroups of $S_n$ that has subgroup of order $2$ ? If we can find, then it can be stated - "The groups which has subgroup of order $2$ are $cdots$, upto isomorphism"
– Ken Ono
Jul 28 at 6:26
Well, to take the help of Calay's theorem, can we find all subgroups of $S_n$ that has subgroup of order $2$ ? If we can find, then it can be stated - "The groups which has subgroup of order $2$ are $cdots$, upto isomorphism"
– Ken Ono
Jul 28 at 6:26
1
1
@KenOno By Cauchy's Theorem, every group of even order has a subgroup of order $2$. By Cayley's theorem, every group is isomorphic to a subgroup of $S_n$ for some $n$. So finding all such subgroups is equivalent to finding all groups of even order. But, given any finite group $G$, the group $G times mathbbZ/2mathbbZ$ is a group of even order, so again, this is equivalent to finding all finite groups, a problem very far from being solved as it stands now..
– Henrique Augusto Souza
Jul 28 at 6:35
@KenOno By Cauchy's Theorem, every group of even order has a subgroup of order $2$. By Cayley's theorem, every group is isomorphic to a subgroup of $S_n$ for some $n$. So finding all such subgroups is equivalent to finding all groups of even order. But, given any finite group $G$, the group $G times mathbbZ/2mathbbZ$ is a group of even order, so again, this is equivalent to finding all finite groups, a problem very far from being solved as it stands now..
– Henrique Augusto Souza
Jul 28 at 6:35
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
2
down vote
First, the result regarding $S_n$ that you are referring to is known as "Cayley's theorem".
In fact, a necessary condition is $n$ divides $|G|$, the order of the group $G$.
Indeed, if there exists a subgroup of $G$ that has order $n$, then $n$ will divide $|G|$ by Lagrange's theorem.
Under a certain condition, a converse holds.
Suppose that $n$ divides $|G|$, and pick Sylow groups $P_1, ldots, P_r$, where we assume that $n = p_1^k_1 cdots p_r^k_r$ and $P_j$ is a $p_j$-Sylow group.
Now if $P$ is any $p$-group, then it has a normal subgroup of order $p^m$, whenever $p^m$ divides $|P|$. For $m=1$ this is Cauchy's theorem, and for general $m$ use Cauchy's theorem and divide out the corresponding cyclic group and then use induction.
Thus, pick for each $j$ a subgroup $Q_j le P_j$ of order $p_j^k_j$. Then the $Q_j$ are pairwise disjoint for order reasons, so that they define a subgroup
$$
H := langle Q_1, ldots, Q_r rangle
$$
that is a direct product of $Q_1, ldots, Q_r$ (if each $Q_j$ is a normal subgroup of $H$) and hence has order $n$.
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
1
That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator.
– AlgebraicsAnonymous
Jul 28 at 6:19
Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1.
– AlgebraicsAnonymous
Jul 28 at 6:22
add a comment |Â
up vote
2
down vote
As noted in the comments by Display name and Orat, it is (as of today) improbable for general or even for a given $n$ to find all groups containing a subgroup of order $n$ in the sense of classification - to formulate a list of groups such as that any group is isomorphic to one in the list.
On the other hand, sometimes necessary and/or sufficient conditions can be given. One simple necessary condition is that $n$ divides the order of $G$, and one simple sufficient condition is that $G$ has order $n$.
For specific $n$, better conditions can be given. If $n = p$ is a prime, then Cauchy's theorem states that a necessary and sufficient condition is that $p$ divides the order of $G$. This is strengthened by Sylow's theorems: a necessary and sufficient condition for a finite group $G$ to have a subgroup of order $p^a$ is that $p^a$ divides the order of $G$. In terms of classification, even $p$-groups are hard to tell, let alone all groups who contain them.
When $n$ is no longer a prime, things get complicated. If we factor $n = p_1^a_1cdots p_k^a_k$, then for a group $G$ whose order is a multiple of $n$ we can easily find subgroups $H_i$ with order $p_i^a_i$ for every $1 leq i leq k$. But we cannot guarantee in general that their product is again a subgroup (altough it will have the correct order as a set and every subgroup of order $n$ can be written in such way). I don't have a construction out of my pocket, but I believe that arbitrarily bad behaved examples can be found.
If your group $G$ is nilpotent, than each of the $H_i$ in the above construction are normal in $G$ and you can guarantee that the product $H = H_1cdots H_k$ is again a subgroup of $G$ of order $n$. Classify all nilpotent finite groups - that's also a hard problem.
According to GroupWiki, your property lies somewhere between supersolvable (which is weaker than nilpotency) and solvable, as the alternating group on four letters $A_4$ is a solvable group that does not satisty this property. It also states that any group satisfying it must be solvable. Note that I'm talking about an even stronger property: a group $G$ having subgroups of every order possible (at least one subgroup for each divisor of it's order). If we talk about a fixed $n$, then we go back to the original problem: if $H$ has order $n$ and $G$ is any finite group, $Gtimes H$ is a group containing a subgroup of order $n$, and their classification involves classifying all finite groups.
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
Can you say something about the proof for the nilpotent case?
– AlgebraicsAnonymous
Jul 28 at 6:26
1
@AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/…
– Henrique Augusto Souza
Jul 28 at 6:28
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
First, the result regarding $S_n$ that you are referring to is known as "Cayley's theorem".
In fact, a necessary condition is $n$ divides $|G|$, the order of the group $G$.
Indeed, if there exists a subgroup of $G$ that has order $n$, then $n$ will divide $|G|$ by Lagrange's theorem.
Under a certain condition, a converse holds.
Suppose that $n$ divides $|G|$, and pick Sylow groups $P_1, ldots, P_r$, where we assume that $n = p_1^k_1 cdots p_r^k_r$ and $P_j$ is a $p_j$-Sylow group.
Now if $P$ is any $p$-group, then it has a normal subgroup of order $p^m$, whenever $p^m$ divides $|P|$. For $m=1$ this is Cauchy's theorem, and for general $m$ use Cauchy's theorem and divide out the corresponding cyclic group and then use induction.
Thus, pick for each $j$ a subgroup $Q_j le P_j$ of order $p_j^k_j$. Then the $Q_j$ are pairwise disjoint for order reasons, so that they define a subgroup
$$
H := langle Q_1, ldots, Q_r rangle
$$
that is a direct product of $Q_1, ldots, Q_r$ (if each $Q_j$ is a normal subgroup of $H$) and hence has order $n$.
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
1
That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator.
– AlgebraicsAnonymous
Jul 28 at 6:19
Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1.
– AlgebraicsAnonymous
Jul 28 at 6:22
add a comment |Â
up vote
2
down vote
First, the result regarding $S_n$ that you are referring to is known as "Cayley's theorem".
In fact, a necessary condition is $n$ divides $|G|$, the order of the group $G$.
Indeed, if there exists a subgroup of $G$ that has order $n$, then $n$ will divide $|G|$ by Lagrange's theorem.
Under a certain condition, a converse holds.
Suppose that $n$ divides $|G|$, and pick Sylow groups $P_1, ldots, P_r$, where we assume that $n = p_1^k_1 cdots p_r^k_r$ and $P_j$ is a $p_j$-Sylow group.
Now if $P$ is any $p$-group, then it has a normal subgroup of order $p^m$, whenever $p^m$ divides $|P|$. For $m=1$ this is Cauchy's theorem, and for general $m$ use Cauchy's theorem and divide out the corresponding cyclic group and then use induction.
Thus, pick for each $j$ a subgroup $Q_j le P_j$ of order $p_j^k_j$. Then the $Q_j$ are pairwise disjoint for order reasons, so that they define a subgroup
$$
H := langle Q_1, ldots, Q_r rangle
$$
that is a direct product of $Q_1, ldots, Q_r$ (if each $Q_j$ is a normal subgroup of $H$) and hence has order $n$.
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
1
That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator.
– AlgebraicsAnonymous
Jul 28 at 6:19
Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1.
– AlgebraicsAnonymous
Jul 28 at 6:22
add a comment |Â
up vote
2
down vote
up vote
2
down vote
First, the result regarding $S_n$ that you are referring to is known as "Cayley's theorem".
In fact, a necessary condition is $n$ divides $|G|$, the order of the group $G$.
Indeed, if there exists a subgroup of $G$ that has order $n$, then $n$ will divide $|G|$ by Lagrange's theorem.
Under a certain condition, a converse holds.
Suppose that $n$ divides $|G|$, and pick Sylow groups $P_1, ldots, P_r$, where we assume that $n = p_1^k_1 cdots p_r^k_r$ and $P_j$ is a $p_j$-Sylow group.
Now if $P$ is any $p$-group, then it has a normal subgroup of order $p^m$, whenever $p^m$ divides $|P|$. For $m=1$ this is Cauchy's theorem, and for general $m$ use Cauchy's theorem and divide out the corresponding cyclic group and then use induction.
Thus, pick for each $j$ a subgroup $Q_j le P_j$ of order $p_j^k_j$. Then the $Q_j$ are pairwise disjoint for order reasons, so that they define a subgroup
$$
H := langle Q_1, ldots, Q_r rangle
$$
that is a direct product of $Q_1, ldots, Q_r$ (if each $Q_j$ is a normal subgroup of $H$) and hence has order $n$.
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
First, the result regarding $S_n$ that you are referring to is known as "Cayley's theorem".
In fact, a necessary condition is $n$ divides $|G|$, the order of the group $G$.
Indeed, if there exists a subgroup of $G$ that has order $n$, then $n$ will divide $|G|$ by Lagrange's theorem.
Under a certain condition, a converse holds.
Suppose that $n$ divides $|G|$, and pick Sylow groups $P_1, ldots, P_r$, where we assume that $n = p_1^k_1 cdots p_r^k_r$ and $P_j$ is a $p_j$-Sylow group.
Now if $P$ is any $p$-group, then it has a normal subgroup of order $p^m$, whenever $p^m$ divides $|P|$. For $m=1$ this is Cauchy's theorem, and for general $m$ use Cauchy's theorem and divide out the corresponding cyclic group and then use induction.
Thus, pick for each $j$ a subgroup $Q_j le P_j$ of order $p_j^k_j$. Then the $Q_j$ are pairwise disjoint for order reasons, so that they define a subgroup
$$
H := langle Q_1, ldots, Q_r rangle
$$
that is a direct product of $Q_1, ldots, Q_r$ (if each $Q_j$ is a normal subgroup of $H$) and hence has order $n$.
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
answered Jul 28 at 6:03
AlgebraicsAnonymous
66611
66611
1
That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator.
– AlgebraicsAnonymous
Jul 28 at 6:19
Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1.
– AlgebraicsAnonymous
Jul 28 at 6:22
add a comment |Â
1
That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator.
– AlgebraicsAnonymous
Jul 28 at 6:19
Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1.
– AlgebraicsAnonymous
Jul 28 at 6:22
1
1
That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator.
– AlgebraicsAnonymous
Jul 28 at 6:19
That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator.
– AlgebraicsAnonymous
Jul 28 at 6:19
Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1.
– AlgebraicsAnonymous
Jul 28 at 6:22
Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1.
– AlgebraicsAnonymous
Jul 28 at 6:22
add a comment |Â
up vote
2
down vote
As noted in the comments by Display name and Orat, it is (as of today) improbable for general or even for a given $n$ to find all groups containing a subgroup of order $n$ in the sense of classification - to formulate a list of groups such as that any group is isomorphic to one in the list.
On the other hand, sometimes necessary and/or sufficient conditions can be given. One simple necessary condition is that $n$ divides the order of $G$, and one simple sufficient condition is that $G$ has order $n$.
For specific $n$, better conditions can be given. If $n = p$ is a prime, then Cauchy's theorem states that a necessary and sufficient condition is that $p$ divides the order of $G$. This is strengthened by Sylow's theorems: a necessary and sufficient condition for a finite group $G$ to have a subgroup of order $p^a$ is that $p^a$ divides the order of $G$. In terms of classification, even $p$-groups are hard to tell, let alone all groups who contain them.
When $n$ is no longer a prime, things get complicated. If we factor $n = p_1^a_1cdots p_k^a_k$, then for a group $G$ whose order is a multiple of $n$ we can easily find subgroups $H_i$ with order $p_i^a_i$ for every $1 leq i leq k$. But we cannot guarantee in general that their product is again a subgroup (altough it will have the correct order as a set and every subgroup of order $n$ can be written in such way). I don't have a construction out of my pocket, but I believe that arbitrarily bad behaved examples can be found.
If your group $G$ is nilpotent, than each of the $H_i$ in the above construction are normal in $G$ and you can guarantee that the product $H = H_1cdots H_k$ is again a subgroup of $G$ of order $n$. Classify all nilpotent finite groups - that's also a hard problem.
According to GroupWiki, your property lies somewhere between supersolvable (which is weaker than nilpotency) and solvable, as the alternating group on four letters $A_4$ is a solvable group that does not satisty this property. It also states that any group satisfying it must be solvable. Note that I'm talking about an even stronger property: a group $G$ having subgroups of every order possible (at least one subgroup for each divisor of it's order). If we talk about a fixed $n$, then we go back to the original problem: if $H$ has order $n$ and $G$ is any finite group, $Gtimes H$ is a group containing a subgroup of order $n$, and their classification involves classifying all finite groups.
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
Can you say something about the proof for the nilpotent case?
– AlgebraicsAnonymous
Jul 28 at 6:26
1
@AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/…
– Henrique Augusto Souza
Jul 28 at 6:28
add a comment |Â
up vote
2
down vote
As noted in the comments by Display name and Orat, it is (as of today) improbable for general or even for a given $n$ to find all groups containing a subgroup of order $n$ in the sense of classification - to formulate a list of groups such as that any group is isomorphic to one in the list.
On the other hand, sometimes necessary and/or sufficient conditions can be given. One simple necessary condition is that $n$ divides the order of $G$, and one simple sufficient condition is that $G$ has order $n$.
For specific $n$, better conditions can be given. If $n = p$ is a prime, then Cauchy's theorem states that a necessary and sufficient condition is that $p$ divides the order of $G$. This is strengthened by Sylow's theorems: a necessary and sufficient condition for a finite group $G$ to have a subgroup of order $p^a$ is that $p^a$ divides the order of $G$. In terms of classification, even $p$-groups are hard to tell, let alone all groups who contain them.
When $n$ is no longer a prime, things get complicated. If we factor $n = p_1^a_1cdots p_k^a_k$, then for a group $G$ whose order is a multiple of $n$ we can easily find subgroups $H_i$ with order $p_i^a_i$ for every $1 leq i leq k$. But we cannot guarantee in general that their product is again a subgroup (altough it will have the correct order as a set and every subgroup of order $n$ can be written in such way). I don't have a construction out of my pocket, but I believe that arbitrarily bad behaved examples can be found.
If your group $G$ is nilpotent, than each of the $H_i$ in the above construction are normal in $G$ and you can guarantee that the product $H = H_1cdots H_k$ is again a subgroup of $G$ of order $n$. Classify all nilpotent finite groups - that's also a hard problem.
According to GroupWiki, your property lies somewhere between supersolvable (which is weaker than nilpotency) and solvable, as the alternating group on four letters $A_4$ is a solvable group that does not satisty this property. It also states that any group satisfying it must be solvable. Note that I'm talking about an even stronger property: a group $G$ having subgroups of every order possible (at least one subgroup for each divisor of it's order). If we talk about a fixed $n$, then we go back to the original problem: if $H$ has order $n$ and $G$ is any finite group, $Gtimes H$ is a group containing a subgroup of order $n$, and their classification involves classifying all finite groups.
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
Can you say something about the proof for the nilpotent case?
– AlgebraicsAnonymous
Jul 28 at 6:26
1
@AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/…
– Henrique Augusto Souza
Jul 28 at 6:28
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As noted in the comments by Display name and Orat, it is (as of today) improbable for general or even for a given $n$ to find all groups containing a subgroup of order $n$ in the sense of classification - to formulate a list of groups such as that any group is isomorphic to one in the list.
On the other hand, sometimes necessary and/or sufficient conditions can be given. One simple necessary condition is that $n$ divides the order of $G$, and one simple sufficient condition is that $G$ has order $n$.
For specific $n$, better conditions can be given. If $n = p$ is a prime, then Cauchy's theorem states that a necessary and sufficient condition is that $p$ divides the order of $G$. This is strengthened by Sylow's theorems: a necessary and sufficient condition for a finite group $G$ to have a subgroup of order $p^a$ is that $p^a$ divides the order of $G$. In terms of classification, even $p$-groups are hard to tell, let alone all groups who contain them.
When $n$ is no longer a prime, things get complicated. If we factor $n = p_1^a_1cdots p_k^a_k$, then for a group $G$ whose order is a multiple of $n$ we can easily find subgroups $H_i$ with order $p_i^a_i$ for every $1 leq i leq k$. But we cannot guarantee in general that their product is again a subgroup (altough it will have the correct order as a set and every subgroup of order $n$ can be written in such way). I don't have a construction out of my pocket, but I believe that arbitrarily bad behaved examples can be found.
If your group $G$ is nilpotent, than each of the $H_i$ in the above construction are normal in $G$ and you can guarantee that the product $H = H_1cdots H_k$ is again a subgroup of $G$ of order $n$. Classify all nilpotent finite groups - that's also a hard problem.
According to GroupWiki, your property lies somewhere between supersolvable (which is weaker than nilpotency) and solvable, as the alternating group on four letters $A_4$ is a solvable group that does not satisty this property. It also states that any group satisfying it must be solvable. Note that I'm talking about an even stronger property: a group $G$ having subgroups of every order possible (at least one subgroup for each divisor of it's order). If we talk about a fixed $n$, then we go back to the original problem: if $H$ has order $n$ and $G$ is any finite group, $Gtimes H$ is a group containing a subgroup of order $n$, and their classification involves classifying all finite groups.
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
As noted in the comments by Display name and Orat, it is (as of today) improbable for general or even for a given $n$ to find all groups containing a subgroup of order $n$ in the sense of classification - to formulate a list of groups such as that any group is isomorphic to one in the list.
On the other hand, sometimes necessary and/or sufficient conditions can be given. One simple necessary condition is that $n$ divides the order of $G$, and one simple sufficient condition is that $G$ has order $n$.
For specific $n$, better conditions can be given. If $n = p$ is a prime, then Cauchy's theorem states that a necessary and sufficient condition is that $p$ divides the order of $G$. This is strengthened by Sylow's theorems: a necessary and sufficient condition for a finite group $G$ to have a subgroup of order $p^a$ is that $p^a$ divides the order of $G$. In terms of classification, even $p$-groups are hard to tell, let alone all groups who contain them.
When $n$ is no longer a prime, things get complicated. If we factor $n = p_1^a_1cdots p_k^a_k$, then for a group $G$ whose order is a multiple of $n$ we can easily find subgroups $H_i$ with order $p_i^a_i$ for every $1 leq i leq k$. But we cannot guarantee in general that their product is again a subgroup (altough it will have the correct order as a set and every subgroup of order $n$ can be written in such way). I don't have a construction out of my pocket, but I believe that arbitrarily bad behaved examples can be found.
If your group $G$ is nilpotent, than each of the $H_i$ in the above construction are normal in $G$ and you can guarantee that the product $H = H_1cdots H_k$ is again a subgroup of $G$ of order $n$. Classify all nilpotent finite groups - that's also a hard problem.
According to GroupWiki, your property lies somewhere between supersolvable (which is weaker than nilpotency) and solvable, as the alternating group on four letters $A_4$ is a solvable group that does not satisty this property. It also states that any group satisfying it must be solvable. Note that I'm talking about an even stronger property: a group $G$ having subgroups of every order possible (at least one subgroup for each divisor of it's order). If we talk about a fixed $n$, then we go back to the original problem: if $H$ has order $n$ and $G$ is any finite group, $Gtimes H$ is a group containing a subgroup of order $n$, and their classification involves classifying all finite groups.
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
edited Jul 28 at 6:42
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
answered Jul 28 at 6:10
Henrique Augusto Souza
1,226314
1,226314
Can you say something about the proof for the nilpotent case?
– AlgebraicsAnonymous
Jul 28 at 6:26
1
@AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/…
– Henrique Augusto Souza
Jul 28 at 6:28
add a comment |Â
Can you say something about the proof for the nilpotent case?
– AlgebraicsAnonymous
Jul 28 at 6:26
1
@AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/…
– Henrique Augusto Souza
Jul 28 at 6:28
Can you say something about the proof for the nilpotent case?
– AlgebraicsAnonymous
Jul 28 at 6:26
Can you say something about the proof for the nilpotent case?
– AlgebraicsAnonymous
Jul 28 at 6:26
1
1
@AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/…
– Henrique Augusto Souza
Jul 28 at 6:28
@AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/…
– Henrique Augusto Souza
Jul 28 at 6:28
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865002%2fis-it-possible-to-find-all-finite-groups-which-has-subgroup-of-order-n%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
8
If I'm not mistaken, the trivial group of order 1 is a subgroup of all groups. Thus, a special case of your problem is equivalent to the classification of all finite groups, which is much harder than the classification of all finite simple groups, which is a problem that took decades of work by hundreds of group theorists. Good luck!
– Display name
Jul 28 at 5:26
1
@Displayname Ok, then for some specific $n$ is it possible to find all groups, which has subgroup of order $n$ ? as, I have mentioned suppose - $n=2$.
– Ken Ono
Jul 28 at 5:34
1
Also, every finite group whose order is divisible by a prime number $p$ has a subgroup of order $p$ (Cauchy's theorem). Thus, another special case of your problem is equivalent to the classification of all finite groups that are not $p'$-groups.
– Orat
Jul 28 at 5:42
Well, to take the help of Calay's theorem, can we find all subgroups of $S_n$ that has subgroup of order $2$ ? If we can find, then it can be stated - "The groups which has subgroup of order $2$ are $cdots$, upto isomorphism"
– Ken Ono
Jul 28 at 6:26
1
@KenOno By Cauchy's Theorem, every group of even order has a subgroup of order $2$. By Cayley's theorem, every group is isomorphic to a subgroup of $S_n$ for some $n$. So finding all such subgroups is equivalent to finding all groups of even order. But, given any finite group $G$, the group $G times mathbbZ/2mathbbZ$ is a group of even order, so again, this is equivalent to finding all finite groups, a problem very far from being solved as it stands now..
– Henrique Augusto Souza
Jul 28 at 6:35