Mixing
Projective objects in the category of finite groups
$begingroup$
It's clear $1$ is a projective object in the category of finite groups. Are there any others?
Note that the dual problem, for injective objects, is comparatively easy (using Cayley's theorem).
category-theory finite-groups
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
$endgroup$
add a comment |
$begingroup$
It's clear $1$ is a projective object in the category of finite groups. Are there any others?
Note that the dual problem, for injective objects, is comparatively easy (using Cayley's theorem).
category-theory finite-groups
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
$endgroup$
$begingroup$
I don't know the answer yet, I'm thinking about it; I found that such a $P$ was non abelian, and then improved the result to prove that $P$ was perfect.
$endgroup$
– Max
Jan 31 at 15:25
$begingroup$
@Max Indeed, Suppose $P$ has a nontrivial abelian quotient. Then its ha a non-trivial cyclic quotient $Bbb Z/nBbb Z$. But that surejction $Pto Bbb Z/nBbb Z$ cannot be lifted to $Bbb Z/nmBbb Z$ for $mgg 1$: some element of $P$ would have to be mapped to a generator.
$endgroup$
– Hagen von Eitzen
Feb 3 at 16:09
$begingroup$
@HagenvonEitzen : yup; I was trying some other things like using the Cayley embedding, but I don't know any nonsplit surjection $Kto mathfrak{S}_n$; or I tried to see if there was any reason why we could find a finite quotient of the free group on $P$ generators that was larger than $P$ but no luck there.
$endgroup$
– Max
Feb 3 at 17:13
add a comment |
$begingroup$
It's clear $1$ is a projective object in the category of finite groups. Are there any others?
Note that the dual problem, for injective objects, is comparatively easy (using Cayley's theorem).
category-theory finite-groups
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
$endgroup$
It's clear $1$ is a projective object in the category of finite groups. Are there any others?
Note that the dual problem, for injective objects, is comparatively easy (using Cayley's theorem).
category-theory finite-groups
category-theory finite-groups
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
asked Jan 31 at 13:02
Mr. ChipMr. Chip
3,2701129
3,2701129
$begingroup$
I don't know the answer yet, I'm thinking about it; I found that such a $P$ was non abelian, and then improved the result to prove that $P$ was perfect.
$endgroup$
– Max
Jan 31 at 15:25
$begingroup$
@Max Indeed, Suppose $P$ has a nontrivial abelian quotient. Then its ha a non-trivial cyclic quotient $Bbb Z/nBbb Z$. But that surejction $Pto Bbb Z/nBbb Z$ cannot be lifted to $Bbb Z/nmBbb Z$ for $mgg 1$: some element of $P$ would have to be mapped to a generator.
$endgroup$
– Hagen von Eitzen
Feb 3 at 16:09
$begingroup$
@HagenvonEitzen : yup; I was trying some other things like using the Cayley embedding, but I don't know any nonsplit surjection $Kto mathfrak{S}_n$; or I tried to see if there was any reason why we could find a finite quotient of the free group on $P$ generators that was larger than $P$ but no luck there.
$endgroup$
– Max
Feb 3 at 17:13
add a comment |
$begingroup$
I don't know the answer yet, I'm thinking about it; I found that such a $P$ was non abelian, and then improved the result to prove that $P$ was perfect.
$endgroup$
– Max
Jan 31 at 15:25
$begingroup$
@Max Indeed, Suppose $P$ has a nontrivial abelian quotient. Then its ha a non-trivial cyclic quotient $Bbb Z/nBbb Z$. But that surejction $Pto Bbb Z/nBbb Z$ cannot be lifted to $Bbb Z/nmBbb Z$ for $mgg 1$: some element of $P$ would have to be mapped to a generator.
$endgroup$
– Hagen von Eitzen
Feb 3 at 16:09
$begingroup$
@HagenvonEitzen : yup; I was trying some other things like using the Cayley embedding, but I don't know any nonsplit surjection $Kto mathfrak{S}_n$; or I tried to see if there was any reason why we could find a finite quotient of the free group on $P$ generators that was larger than $P$ but no luck there.
$endgroup$
– Max
Feb 3 at 17:13
$begingroup$
I don't know the answer yet, I'm thinking about it; I found that such a $P$ was non abelian, and then improved the result to prove that $P$ was perfect.
$endgroup$
– Max
Jan 31 at 15:25
$begingroup$
I don't know the answer yet, I'm thinking about it; I found that such a $P$ was non abelian, and then improved the result to prove that $P$ was perfect.
$endgroup$
– Max
Jan 31 at 15:25
$begingroup$
@Max Indeed, Suppose $P$ has a nontrivial abelian quotient. Then its ha a non-trivial cyclic quotient $Bbb Z/nBbb Z$. But that surejction $Pto Bbb Z/nBbb Z$ cannot be lifted to $Bbb Z/nmBbb Z$ for $mgg 1$: some element of $P$ would have to be mapped to a generator.
$endgroup$
– Hagen von Eitzen
Feb 3 at 16:09
$begingroup$
@Max Indeed, Suppose $P$ has a nontrivial abelian quotient. Then its ha a non-trivial cyclic quotient $Bbb Z/nBbb Z$. But that surejction $Pto Bbb Z/nBbb Z$ cannot be lifted to $Bbb Z/nmBbb Z$ for $mgg 1$: some element of $P$ would have to be mapped to a generator.
$endgroup$
– Hagen von Eitzen
Feb 3 at 16:09
$begingroup$
@HagenvonEitzen : yup; I was trying some other things like using the Cayley embedding, but I don't know any nonsplit surjection $Kto mathfrak{S}_n$; or I tried to see if there was any reason why we could find a finite quotient of the free group on $P$ generators that was larger than $P$ but no luck there.
$endgroup$
– Max
Feb 3 at 17:13
$begingroup$
@HagenvonEitzen : yup; I was trying some other things like using the Cayley embedding, but I don't know any nonsplit surjection $Kto mathfrak{S}_n$; or I tried to see if there was any reason why we could find a finite quotient of the free group on $P$ generators that was larger than $P$ but no luck there.
$endgroup$
– Max
Feb 3 at 17:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are no others. Here's a proof using well-known facts about the cohomology of finite groups.
Let $G$ be a nontrivial finite group, and let $p$ be a prime dividing the order of $G$. Then there is a finite dimensional (and hence finite) $mathbb{F}_pG$-module $M$ with $H^2(G,M)neq0$. A nonzero cohomology class represents a non-split extension $1to Mtotilde{G}to Gto1$.
If the existence of $M$ is not familiar, but you believe that $H^k(G,mathbb{F}_p)neq0$ for some $k>0$, then you can produce $M$ from the trivial $mathbb{F}_pG$-module $mathbb{F}_p$ by dimension shifting. Or for a specific $M$, take an exact sequence
$$0to Mto P_1to P_0tomathbb{F}_pto0,$$
where $P_1to P_0tomathbb{F}_pto0$ is a projective presentation of the trivial $mathbb{F}_pG$-module.
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
1
$begingroup$
group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$?
$endgroup$
– hunter
Feb 4 at 11:32
1
$begingroup$
@hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $Gtotilde{G}$ such that the composition $Gtotilde{G}to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $tilde{G}to G$.
$endgroup$
– Jeremy Rickard
Feb 4 at 11:37
$begingroup$
Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1to Mto Ato Gto 1$ such that the induced action is the given one
$endgroup$
– Max
Feb 4 at 15:44
$begingroup$
@hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup
$endgroup$
– Max
Feb 4 at 15:51
$begingroup$
@JeremyRickard yes now that I think about it this is just fine.
$endgroup$
– hunter
Feb 4 at 15:51
|
show 5 more comments
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1 Answer
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$begingroup$
There are no others. Here's a proof using well-known facts about the cohomology of finite groups.
Let $G$ be a nontrivial finite group, and let $p$ be a prime dividing the order of $G$. Then there is a finite dimensional (and hence finite) $mathbb{F}_pG$-module $M$ with $H^2(G,M)neq0$. A nonzero cohomology class represents a non-split extension $1to Mtotilde{G}to Gto1$.
If the existence of $M$ is not familiar, but you believe that $H^k(G,mathbb{F}_p)neq0$ for some $k>0$, then you can produce $M$ from the trivial $mathbb{F}_pG$-module $mathbb{F}_p$ by dimension shifting. Or for a specific $M$, take an exact sequence
$$0to Mto P_1to P_0tomathbb{F}_pto0,$$
where $P_1to P_0tomathbb{F}_pto0$ is a projective presentation of the trivial $mathbb{F}_pG$-module.
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
1
$begingroup$
group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$?
$endgroup$
– hunter
Feb 4 at 11:32
1
$begingroup$
@hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $Gtotilde{G}$ such that the composition $Gtotilde{G}to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $tilde{G}to G$.
$endgroup$
– Jeremy Rickard
Feb 4 at 11:37
$begingroup$
Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1to Mto Ato Gto 1$ such that the induced action is the given one
$endgroup$
– Max
Feb 4 at 15:44
$begingroup$
@hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup
$endgroup$
– Max
Feb 4 at 15:51
$begingroup$
@JeremyRickard yes now that I think about it this is just fine.
$endgroup$
– hunter
Feb 4 at 15:51
|
show 5 more comments
$begingroup$
There are no others. Here's a proof using well-known facts about the cohomology of finite groups.
Let $G$ be a nontrivial finite group, and let $p$ be a prime dividing the order of $G$. Then there is a finite dimensional (and hence finite) $mathbb{F}_pG$-module $M$ with $H^2(G,M)neq0$. A nonzero cohomology class represents a non-split extension $1to Mtotilde{G}to Gto1$.
If the existence of $M$ is not familiar, but you believe that $H^k(G,mathbb{F}_p)neq0$ for some $k>0$, then you can produce $M$ from the trivial $mathbb{F}_pG$-module $mathbb{F}_p$ by dimension shifting. Or for a specific $M$, take an exact sequence
$$0to Mto P_1to P_0tomathbb{F}_pto0,$$
where $P_1to P_0tomathbb{F}_pto0$ is a projective presentation of the trivial $mathbb{F}_pG$-module.
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
1
$begingroup$
group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$?
$endgroup$
– hunter
Feb 4 at 11:32
1
$begingroup$
@hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $Gtotilde{G}$ such that the composition $Gtotilde{G}to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $tilde{G}to G$.
$endgroup$
– Jeremy Rickard
Feb 4 at 11:37
$begingroup$
Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1to Mto Ato Gto 1$ such that the induced action is the given one
$endgroup$
– Max
Feb 4 at 15:44
$begingroup$
@hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup
$endgroup$
– Max
Feb 4 at 15:51
$begingroup$
@JeremyRickard yes now that I think about it this is just fine.
$endgroup$
– hunter
Feb 4 at 15:51
|
show 5 more comments
$begingroup$
There are no others. Here's a proof using well-known facts about the cohomology of finite groups.
Let $G$ be a nontrivial finite group, and let $p$ be a prime dividing the order of $G$. Then there is a finite dimensional (and hence finite) $mathbb{F}_pG$-module $M$ with $H^2(G,M)neq0$. A nonzero cohomology class represents a non-split extension $1to Mtotilde{G}to Gto1$.
If the existence of $M$ is not familiar, but you believe that $H^k(G,mathbb{F}_p)neq0$ for some $k>0$, then you can produce $M$ from the trivial $mathbb{F}_pG$-module $mathbb{F}_p$ by dimension shifting. Or for a specific $M$, take an exact sequence
$$0to Mto P_1to P_0tomathbb{F}_pto0,$$
where $P_1to P_0tomathbb{F}_pto0$ is a projective presentation of the trivial $mathbb{F}_pG$-module.
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
$endgroup$
There are no others. Here's a proof using well-known facts about the cohomology of finite groups.
Let $G$ be a nontrivial finite group, and let $p$ be a prime dividing the order of $G$. Then there is a finite dimensional (and hence finite) $mathbb{F}_pG$-module $M$ with $H^2(G,M)neq0$. A nonzero cohomology class represents a non-split extension $1to Mtotilde{G}to Gto1$.
If the existence of $M$ is not familiar, but you believe that $H^k(G,mathbb{F}_p)neq0$ for some $k>0$, then you can produce $M$ from the trivial $mathbb{F}_pG$-module $mathbb{F}_p$ by dimension shifting. Or for a specific $M$, take an exact sequence
$$0to Mto P_1to P_0tomathbb{F}_pto0,$$
where $P_1to P_0tomathbb{F}_pto0$ is a projective presentation of the trivial $mathbb{F}_pG$-module.
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
answered Feb 4 at 11:20
Jeremy RickardJeremy Rickard
16.9k11746
16.9k11746
1
$begingroup$
group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$?
$endgroup$
– hunter
Feb 4 at 11:32
1
$begingroup$
@hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $Gtotilde{G}$ such that the composition $Gtotilde{G}to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $tilde{G}to G$.
$endgroup$
– Jeremy Rickard
Feb 4 at 11:37
$begingroup$
Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1to Mto Ato Gto 1$ such that the induced action is the given one
$endgroup$
– Max
Feb 4 at 15:44
$begingroup$
@hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup
$endgroup$
– Max
Feb 4 at 15:51
$begingroup$
@JeremyRickard yes now that I think about it this is just fine.
$endgroup$
– hunter
Feb 4 at 15:51
|
show 5 more comments
1
$begingroup$
group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$?
$endgroup$
– hunter
Feb 4 at 11:32
1
$begingroup$
@hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $Gtotilde{G}$ such that the composition $Gtotilde{G}to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $tilde{G}to G$.
$endgroup$
– Jeremy Rickard
Feb 4 at 11:37
$begingroup$
Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1to Mto Ato Gto 1$ such that the induced action is the given one
$endgroup$
– Max
Feb 4 at 15:44
$begingroup$
@hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup
$endgroup$
– Max
Feb 4 at 15:51
$begingroup$
@JeremyRickard yes now that I think about it this is just fine.
$endgroup$
– hunter
Feb 4 at 15:51
1
1
$begingroup$
group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$?
$endgroup$
– hunter
Feb 4 at 11:32
$begingroup$
group cohomology question -- What if there is a splitting of the extension as abstract groups which is incompatible with the given action of $G$ on $M$?
$endgroup$
– hunter
Feb 4 at 11:32
1
1
$begingroup$
@hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $Gtotilde{G}$ such that the composition $Gtotilde{G}to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $tilde{G}to G$.
$endgroup$
– Jeremy Rickard
Feb 4 at 11:37
$begingroup$
@hunter I'm not sure what you mean by a splitting being compatible with the action on $M$. A splitting is simply a group homomorphism $Gtotilde{G}$ such that the composition $Gtotilde{G}to G$ is the identity map. In any case, the action of $G$ on $M$ is determined by the map $tilde{G}to G$.
$endgroup$
– Jeremy Rickard
Feb 4 at 11:37
$begingroup$
Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1to Mto Ato Gto 1$ such that the induced action is the given one
$endgroup$
– Max
Feb 4 at 15:44
$begingroup$
Yes, but @hunter is right in that $H^2(G,M)$ assumes a given action on $M$, and it classifies extensions $1to Mto Ato Gto 1$ such that the induced action is the given one
$endgroup$
– Max
Feb 4 at 15:44
$begingroup$
@hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup
$endgroup$
– Max
Feb 4 at 15:51
$begingroup$
@hunter : ah actually I just realized it doesn't make sense to ask that : a nontrivial cohomology class in $H^2(G,M)$ provides an extension $tilde{G}$ that induces the action on $M$; and that isn't split. Sorry Jeremy for the mixup
$endgroup$
– Max
Feb 4 at 15:51
$begingroup$
@JeremyRickard yes now that I think about it this is just fine.
$endgroup$
– hunter
Feb 4 at 15:51
$begingroup$
@JeremyRickard yes now that I think about it this is just fine.
$endgroup$
– hunter
Feb 4 at 15:51
|
show 5 more comments
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$begingroup$
I don't know the answer yet, I'm thinking about it; I found that such a $P$ was non abelian, and then improved the result to prove that $P$ was perfect.
$endgroup$
– Max
Jan 31 at 15:25
$begingroup$
@Max Indeed, Suppose $P$ has a nontrivial abelian quotient. Then its ha a non-trivial cyclic quotient $Bbb Z/nBbb Z$. But that surejction $Pto Bbb Z/nBbb Z$ cannot be lifted to $Bbb Z/nmBbb Z$ for $mgg 1$: some element of $P$ would have to be mapped to a generator.
$endgroup$
– Hagen von Eitzen
Feb 3 at 16:09
$begingroup$
@HagenvonEitzen : yup; I was trying some other things like using the Cayley embedding, but I don't know any nonsplit surjection $Kto mathfrak{S}_n$; or I tried to see if there was any reason why we could find a finite quotient of the free group on $P$ generators that was larger than $P$ but no luck there.
$endgroup$
– Max
Feb 3 at 17:13