Mixing
A group in which every proper subgroup is contained in a maximal subgroup
Clash Royale CLAN TAG#URR8PPP
up vote
11
down vote
favorite
2
Let $G$ be a group in which every proper subgroup is contained in a maximal subgroup of $G$.
Can we conclude that $G$ is finitely generated? (@Max commented that converse of this statement is true.)
abstract-algebra group-theory
edited Jul 15 at 10:13
Derek Holt
49.8k53366
asked Jul 15 at 5:47
mesel
10.1k21644
 |Â
show 4 more comments
up vote
11
down vote
favorite
2
Let $G$ be a group in which every proper subgroup is contained in a maximal subgroup of $G$.
Can we conclude that $G$ is finitely generated? (@Max commented that converse of this statement is true.)
abstract-algebra group-theory
edited Jul 15 at 10:13
Derek Holt
49.8k53366
asked Jul 15 at 5:47
mesel
10.1k21644
You must mean every proper subgroup is contained in a maximal subgroup. How about a direct product of countably many groups of prime order $p$? The quotient group with any subgroup has the same property and has maximal subgroups.
– Derek Holt
Jul 15 at 7:27
The converse is true : in a finitely generated group, every proper subgroup is contained in a maximal subgroup (hint : given a finite generating family $F$, a subgroup of $G$ is $G$ iff it contains $F$)
– Max
Jul 15 at 9:38
@Max: So I completely misremember :) I did not really understand your hint. But I thought the following. Let $H<G$ and let $mathfrak L$ be the set of all proper subgroups of $G$ that contains $H$. Assume that $mathfrak L$ does not have a maximal element. Then there exists a totoally ordered set $mathbb S$ such that $bigcup_Hin SHnotin mathfrak L$, and hence $bigcup_Hin SH=G$. Since $G$ is finitly generated, $Gin mathbb S$. This is a contradiction as $G$ is not proper in $G$.
– mesel
Jul 15 at 9:51
How do you deduce $Gin S$ ?
– Max
Jul 15 at 10:21
@Max: $G$ is finitely generated, that is $langle x_1,x_2,...,x_krangle=G$. As $bigcup_Hin mathbb SH=G$, set $x_iin H_i$. Then $G=H_j$ for some $j$. Thus, $Gin mathbb S$.
– mesel
Jul 15 at 12:55
 |Â
show 4 more comments
up vote
11
down vote
favorite
2
up vote
11
down vote
favorite
2
2
Let $G$ be a group in which every proper subgroup is contained in a maximal subgroup of $G$.
Can we conclude that $G$ is finitely generated? (@Max commented that converse of this statement is true.)
abstract-algebra group-theory
edited Jul 15 at 10:13
Derek Holt
49.8k53366
asked Jul 15 at 5:47
mesel
10.1k21644
Let $G$ be a group in which every proper subgroup is contained in a maximal subgroup of $G$.
Can we conclude that $G$ is finitely generated? (@Max commented that converse of this statement is true.)
abstract-algebra group-theory
edited Jul 15 at 10:13
Derek Holt
49.8k53366
asked Jul 15 at 5:47
mesel
10.1k21644
edited Jul 15 at 10:13
Derek Holt
49.8k53366
edited Jul 15 at 10:13
Derek Holt
49.8k53366
edited Jul 15 at 10:13
Derek Holt
49.8k53366
49.8k53366
asked Jul 15 at 5:47
mesel
10.1k21644
asked Jul 15 at 5:47
mesel
10.1k21644
asked Jul 15 at 5:47
mesel
10.1k21644
10.1k21644
You must mean every proper subgroup is contained in a maximal subgroup. How about a direct product of countably many groups of prime order $p$? The quotient group with any subgroup has the same property and has maximal subgroups.
– Derek Holt
Jul 15 at 7:27
The converse is true : in a finitely generated group, every proper subgroup is contained in a maximal subgroup (hint : given a finite generating family $F$, a subgroup of $G$ is $G$ iff it contains $F$)
– Max
Jul 15 at 9:38
@Max: So I completely misremember :) I did not really understand your hint. But I thought the following. Let $H<G$ and let $mathfrak L$ be the set of all proper subgroups of $G$ that contains $H$. Assume that $mathfrak L$ does not have a maximal element. Then there exists a totoally ordered set $mathbb S$ such that $bigcup_Hin SHnotin mathfrak L$, and hence $bigcup_Hin SH=G$. Since $G$ is finitly generated, $Gin mathbb S$. This is a contradiction as $G$ is not proper in $G$.
– mesel
Jul 15 at 9:51
How do you deduce $Gin S$ ?
– Max
Jul 15 at 10:21
@Max: $G$ is finitely generated, that is $langle x_1,x_2,...,x_krangle=G$. As $bigcup_Hin mathbb SH=G$, set $x_iin H_i$. Then $G=H_j$ for some $j$. Thus, $Gin mathbb S$.
– mesel
Jul 15 at 12:55
 |Â
show 4 more comments
You must mean every proper subgroup is contained in a maximal subgroup. How about a direct product of countably many groups of prime order $p$? The quotient group with any subgroup has the same property and has maximal subgroups.
– Derek Holt
Jul 15 at 7:27
The converse is true : in a finitely generated group, every proper subgroup is contained in a maximal subgroup (hint : given a finite generating family $F$, a subgroup of $G$ is $G$ iff it contains $F$)
– Max
Jul 15 at 9:38
@Max: So I completely misremember :) I did not really understand your hint. But I thought the following. Let $H<G$ and let $mathfrak L$ be the set of all proper subgroups of $G$ that contains $H$. Assume that $mathfrak L$ does not have a maximal element. Then there exists a totoally ordered set $mathbb S$ such that $bigcup_Hin SHnotin mathfrak L$, and hence $bigcup_Hin SH=G$. Since $G$ is finitly generated, $Gin mathbb S$. This is a contradiction as $G$ is not proper in $G$.
– mesel
Jul 15 at 9:51
How do you deduce $Gin S$ ?
– Max
Jul 15 at 10:21
@Max: $G$ is finitely generated, that is $langle x_1,x_2,...,x_krangle=G$. As $bigcup_Hin mathbb SH=G$, set $x_iin H_i$. Then $G=H_j$ for some $j$. Thus, $Gin mathbb S$.
– mesel
Jul 15 at 12:55
You must mean every proper subgroup is contained in a maximal subgroup. How about a direct product of countably many groups of prime order $p$? The quotient group with any subgroup has the same property and has maximal subgroups.
– Derek Holt
Jul 15 at 7:27
You must mean every proper subgroup is contained in a maximal subgroup. How about a direct product of countably many groups of prime order $p$? The quotient group with any subgroup has the same property and has maximal subgroups.
– Derek Holt
Jul 15 at 7:27
The converse is true : in a finitely generated group, every proper subgroup is contained in a maximal subgroup (hint : given a finite generating family $F$, a subgroup of $G$ is $G$ iff it contains $F$)
– Max
Jul 15 at 9:38
The converse is true : in a finitely generated group, every proper subgroup is contained in a maximal subgroup (hint : given a finite generating family $F$, a subgroup of $G$ is $G$ iff it contains $F$)
– Max
Jul 15 at 9:38
@Max: So I completely misremember :) I did not really understand your hint. But I thought the following. Let $H<G$ and let $mathfrak L$ be the set of all proper subgroups of $G$ that contains $H$. Assume that $mathfrak L$ does not have a maximal element. Then there exists a totoally ordered set $mathbb S$ such that $bigcup_Hin SHnotin mathfrak L$, and hence $bigcup_Hin SH=G$. Since $G$ is finitly generated, $Gin mathbb S$. This is a contradiction as $G$ is not proper in $G$.
– mesel
Jul 15 at 9:51
@Max: So I completely misremember :) I did not really understand your hint. But I thought the following. Let $H<G$ and let $mathfrak L$ be the set of all proper subgroups of $G$ that contains $H$. Assume that $mathfrak L$ does not have a maximal element. Then there exists a totoally ordered set $mathbb S$ such that $bigcup_Hin SHnotin mathfrak L$, and hence $bigcup_Hin SH=G$. Since $G$ is finitly generated, $Gin mathbb S$. This is a contradiction as $G$ is not proper in $G$.
– mesel
Jul 15 at 9:51
How do you deduce $Gin S$ ?
– Max
Jul 15 at 10:21
How do you deduce $Gin S$ ?
– Max
Jul 15 at 10:21
@Max: $G$ is finitely generated, that is $langle x_1,x_2,...,x_krangle=G$. As $bigcup_Hin mathbb SH=G$, set $x_iin H_i$. Then $G=H_j$ for some $j$. Thus, $Gin mathbb S$.
– mesel
Jul 15 at 12:55
@Max: $G$ is finitely generated, that is $langle x_1,x_2,...,x_krangle=G$. As $bigcup_Hin mathbb SH=G$, set $x_iin H_i$. Then $G=H_j$ for some $j$. Thus, $Gin mathbb S$.
– mesel
Jul 15 at 12:55
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
10
down vote
accepted
No. For instance, let $G$ be a vector space over $mathbbF_p$ for some prime $p$. Then every proper subgroup (i.e., subspace) is contained in a maximal subgroup (i.e., codimension 1 subspace) but $G$ is not finitely generated unless it is finite dimensional.
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
No. For instance, let $G$ be a vector space over $mathbbF_p$ for some prime $p$. Then every proper subgroup (i.e., subspace) is contained in a maximal subgroup (i.e., codimension 1 subspace) but $G$ is not finitely generated unless it is finite dimensional.
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
add a comment |Â
up vote
10
down vote
accepted
No. For instance, let $G$ be a vector space over $mathbbF_p$ for some prime $p$. Then every proper subgroup (i.e., subspace) is contained in a maximal subgroup (i.e., codimension 1 subspace) but $G$ is not finitely generated unless it is finite dimensional.
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
add a comment |Â
up vote
10
down vote
accepted
up vote
10
down vote
accepted
No. For instance, let $G$ be a vector space over $mathbbF_p$ for some prime $p$. Then every proper subgroup (i.e., subspace) is contained in a maximal subgroup (i.e., codimension 1 subspace) but $G$ is not finitely generated unless it is finite dimensional.
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
No. For instance, let $G$ be a vector space over $mathbbF_p$ for some prime $p$. Then every proper subgroup (i.e., subspace) is contained in a maximal subgroup (i.e., codimension 1 subspace) but $G$ is not finitely generated unless it is finite dimensional.
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
edited Jul 15 at 7:44
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
answered Jul 15 at 7:26
Eric Wofsey
163k12189300
163k12189300
add a comment |Â
add a comment |Â
Â
draft saved
draft discarded
Â
draft saved
draft discarded
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%2f2852218%2fa-group-in-which-every-proper-subgroup-is-contained-in-a-maximal-subgroup%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
You must mean every proper subgroup is contained in a maximal subgroup. How about a direct product of countably many groups of prime order $p$? The quotient group with any subgroup has the same property and has maximal subgroups.
– Derek Holt
Jul 15 at 7:27
The converse is true : in a finitely generated group, every proper subgroup is contained in a maximal subgroup (hint : given a finite generating family $F$, a subgroup of $G$ is $G$ iff it contains $F$)
– Max
Jul 15 at 9:38
@Max: So I completely misremember :) I did not really understand your hint. But I thought the following. Let $H<G$ and let $mathfrak L$ be the set of all proper subgroups of $G$ that contains $H$. Assume that $mathfrak L$ does not have a maximal element. Then there exists a totoally ordered set $mathbb S$ such that $bigcup_Hin SHnotin mathfrak L$, and hence $bigcup_Hin SH=G$. Since $G$ is finitly generated, $Gin mathbb S$. This is a contradiction as $G$ is not proper in $G$.
– mesel
Jul 15 at 9:51
How do you deduce $Gin S$ ?
– Max
Jul 15 at 10:21
@Max: $G$ is finitely generated, that is $langle x_1,x_2,...,x_krangle=G$. As $bigcup_Hin mathbb SH=G$, set $x_iin H_i$. Then $G=H_j$ for some $j$. Thus, $Gin mathbb S$.
– mesel
Jul 15 at 12:55