What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?Group presentation for semidirect productsAre split exact sequences exact in the opposite direction?Can we produce a long exact sequence in cohomology from more than just short exact sequences?What is a short exact sequence?Properties of Short Exact SequencesProposition 3.2.7 of Bland on short exact sequencesProposition 3.2.7 of Bland on short exact sequences, part 2Why is a direct summand of a pretriangle a pretriangle?Why are semi-simple abelian categories pre-triangulated?Do long exact Mayer Vietoris sequences decompose into short exact sequences $0to H_n(Acap B)to H_n(A)oplus H_n(B)to H_n(X)to 0$?Condition for pullback to split
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What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
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What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
Group presentation for semidirect productsAre split exact sequences exact in the opposite direction?Can we produce a long exact sequence in cohomology from more than just short exact sequences?What is a short exact sequence?Properties of Short Exact SequencesProposition 3.2.7 of Bland on short exact sequencesProposition 3.2.7 of Bland on short exact sequences, part 2Why is a direct summand of a pretriangle a pretriangle?Why are semi-simple abelian categories pre-triangulated?Do long exact Mayer Vietoris sequences decompose into short exact sequences $0to H_n(Acap B)to H_n(A)oplus H_n(B)to H_n(X)to 0$?Condition for pullback to split
$begingroup$
What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
I'm sorry that the definitions below are a bit haphazard but they're how I learnt about them, chronologically.
In Johnson's "Presentation$colorreds$ of Groups," page 100, there is the following . . .
Definition 1: A diagram in a category $mathfrakC$, which consists of objects $A_nmid ninBbb Z$ and morphisms $$partial_n: A_nto A_n+1, nin Bbb Z,tag6$$ is called a sequence in $mathfrakC$. Such a sequence is called exact if $$operatornameImpartial_n=ker partial_n+1,,text for all nin Bbb Z$$ [. . .] A short exact sequence in the category $mathfrakC_Bbb R$ of right $Bbb R$-modules is an exact sequence of the form $(6)$ with all but three consecutive terms equal to zero. [. . .]
Also, ibid., page 101, is this:
It is fairly obvious that a sequence
$$0longrightarrow AstackrelthetalongrightarrowBstackrelphilongrightarrowClongrightarrow 0$$
is a short exact sequence if and only if the following conditions hold:
$theta$ is one-to-one,
$phi$ is onto,
$thetaphi=0$,
$ker phileoperatornameImtheta$.
I'm reading Baumslag's "Topics in Combinatorial Group Theory". Section III.2 on semidirect products starts with
Let $$1longrightarrow AstackrelalphalongrightarrowEstackrelbetalongrightarrowQlongrightarrow 1$$ be a short exact sequence of groups. We term $E$ an extension of $A$ by $Q$.
Thoughts:
I'm aware that semidirect products can be seen as short exact sequences but this is not something I understand yet. My view of semidirect products is as if they are defined by a particular presentation and my go-to examples are the dihedral groups.
Please help :)
group-theory soft-question category-theory intuition exact-sequence
$endgroup$
add a comment |
$begingroup$
What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
I'm sorry that the definitions below are a bit haphazard but they're how I learnt about them, chronologically.
In Johnson's "Presentation$colorreds$ of Groups," page 100, there is the following . . .
Definition 1: A diagram in a category $mathfrakC$, which consists of objects $A_nmid ninBbb Z$ and morphisms $$partial_n: A_nto A_n+1, nin Bbb Z,tag6$$ is called a sequence in $mathfrakC$. Such a sequence is called exact if $$operatornameImpartial_n=ker partial_n+1,,text for all nin Bbb Z$$ [. . .] A short exact sequence in the category $mathfrakC_Bbb R$ of right $Bbb R$-modules is an exact sequence of the form $(6)$ with all but three consecutive terms equal to zero. [. . .]
Also, ibid., page 101, is this:
It is fairly obvious that a sequence
$$0longrightarrow AstackrelthetalongrightarrowBstackrelphilongrightarrowClongrightarrow 0$$
is a short exact sequence if and only if the following conditions hold:
$theta$ is one-to-one,
$phi$ is onto,
$thetaphi=0$,
$ker phileoperatornameImtheta$.
I'm reading Baumslag's "Topics in Combinatorial Group Theory". Section III.2 on semidirect products starts with
Let $$1longrightarrow AstackrelalphalongrightarrowEstackrelbetalongrightarrowQlongrightarrow 1$$ be a short exact sequence of groups. We term $E$ an extension of $A$ by $Q$.
Thoughts:
I'm aware that semidirect products can be seen as short exact sequences but this is not something I understand yet. My view of semidirect products is as if they are defined by a particular presentation and my go-to examples are the dihedral groups.
Please help :)
group-theory soft-question category-theory intuition exact-sequence
$endgroup$
1
$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
$endgroup$
– lush
14 hours ago
1
$begingroup$
I'm afraid not, @lush. I'm aware of extensions in the context of fields and how they can be seen as vector spaces though, like in Galois Theory or perhaps in Algebraic Number Theory.
$endgroup$
– Shaun
14 hours ago
4
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
$endgroup$
– user1729
13 hours ago
add a comment |
$begingroup$
What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
I'm sorry that the definitions below are a bit haphazard but they're how I learnt about them, chronologically.
In Johnson's "Presentation$colorreds$ of Groups," page 100, there is the following . . .
Definition 1: A diagram in a category $mathfrakC$, which consists of objects $A_nmid ninBbb Z$ and morphisms $$partial_n: A_nto A_n+1, nin Bbb Z,tag6$$ is called a sequence in $mathfrakC$. Such a sequence is called exact if $$operatornameImpartial_n=ker partial_n+1,,text for all nin Bbb Z$$ [. . .] A short exact sequence in the category $mathfrakC_Bbb R$ of right $Bbb R$-modules is an exact sequence of the form $(6)$ with all but three consecutive terms equal to zero. [. . .]
Also, ibid., page 101, is this:
It is fairly obvious that a sequence
$$0longrightarrow AstackrelthetalongrightarrowBstackrelphilongrightarrowClongrightarrow 0$$
is a short exact sequence if and only if the following conditions hold:
$theta$ is one-to-one,
$phi$ is onto,
$thetaphi=0$,
$ker phileoperatornameImtheta$.
I'm reading Baumslag's "Topics in Combinatorial Group Theory". Section III.2 on semidirect products starts with
Let $$1longrightarrow AstackrelalphalongrightarrowEstackrelbetalongrightarrowQlongrightarrow 1$$ be a short exact sequence of groups. We term $E$ an extension of $A$ by $Q$.
Thoughts:
I'm aware that semidirect products can be seen as short exact sequences but this is not something I understand yet. My view of semidirect products is as if they are defined by a particular presentation and my go-to examples are the dihedral groups.
Please help :)
group-theory soft-question category-theory intuition exact-sequence
$endgroup$
What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
I'm sorry that the definitions below are a bit haphazard but they're how I learnt about them, chronologically.
In Johnson's "Presentation$colorreds$ of Groups," page 100, there is the following . . .
Definition 1: A diagram in a category $mathfrakC$, which consists of objects $A_nmid ninBbb Z$ and morphisms $$partial_n: A_nto A_n+1, nin Bbb Z,tag6$$ is called a sequence in $mathfrakC$. Such a sequence is called exact if $$operatornameImpartial_n=ker partial_n+1,,text for all nin Bbb Z$$ [. . .] A short exact sequence in the category $mathfrakC_Bbb R$ of right $Bbb R$-modules is an exact sequence of the form $(6)$ with all but three consecutive terms equal to zero. [. . .]
Also, ibid., page 101, is this:
It is fairly obvious that a sequence
$$0longrightarrow AstackrelthetalongrightarrowBstackrelphilongrightarrowClongrightarrow 0$$
is a short exact sequence if and only if the following conditions hold:
$theta$ is one-to-one,
$phi$ is onto,
$thetaphi=0$,
$ker phileoperatornameImtheta$.
I'm reading Baumslag's "Topics in Combinatorial Group Theory". Section III.2 on semidirect products starts with
Let $$1longrightarrow AstackrelalphalongrightarrowEstackrelbetalongrightarrowQlongrightarrow 1$$ be a short exact sequence of groups. We term $E$ an extension of $A$ by $Q$.
Thoughts:
I'm aware that semidirect products can be seen as short exact sequences but this is not something I understand yet. My view of semidirect products is as if they are defined by a particular presentation and my go-to examples are the dihedral groups.
Please help :)
group-theory soft-question category-theory intuition exact-sequence
group-theory soft-question category-theory intuition exact-sequence
asked 14 hours ago
ShaunShaun
10.3k113686
10.3k113686
1
$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
$endgroup$
– lush
14 hours ago
1
$begingroup$
I'm afraid not, @lush. I'm aware of extensions in the context of fields and how they can be seen as vector spaces though, like in Galois Theory or perhaps in Algebraic Number Theory.
$endgroup$
– Shaun
14 hours ago
4
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
$endgroup$
– user1729
13 hours ago
add a comment |
1
$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
$endgroup$
– lush
14 hours ago
1
$begingroup$
I'm afraid not, @lush. I'm aware of extensions in the context of fields and how they can be seen as vector spaces though, like in Galois Theory or perhaps in Algebraic Number Theory.
$endgroup$
– Shaun
14 hours ago
4
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
$endgroup$
– user1729
13 hours ago
1
1
$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
$endgroup$
– lush
14 hours ago
$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
$endgroup$
– lush
14 hours ago
1
1
$begingroup$
I'm afraid not, @lush. I'm aware of extensions in the context of fields and how they can be seen as vector spaces though, like in Galois Theory or perhaps in Algebraic Number Theory.
$endgroup$
– Shaun
14 hours ago
$begingroup$
I'm afraid not, @lush. I'm aware of extensions in the context of fields and how they can be seen as vector spaces though, like in Galois Theory or perhaps in Algebraic Number Theory.
$endgroup$
– Shaun
14 hours ago
4
4
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
$endgroup$
– user1729
13 hours ago
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
$endgroup$
– user1729
13 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
Because you care about presentations: if $A$ has presentation $langle mathbfxmidmathbfrrangle$ and $Q$ has presentation $langle mathbfymidmathbfsrangle$ then the group $E$ given by the above short exact sequence has presentation of the form:
$$
langle mathbfx, ymid SW_S^-1 (Sinmathbfs), mathbfr, mathbftrangle
$$
where $W_Sin F(mathbfx)$ for all $Sinmathbfs$, and $mathbft$ consists of words of the form $y^-epsilonxy^epsilonX^-1$ with $xinmathbfx$, $yinmathbfy$ and $Xin F(mathbfx)$. The intuition here is that relators in $mathbft$ ensure normality of $A$, and so removing all the $x$-terms makes sense. When they are removed you get the presentation $langle mathbfymidmathbfsrangle$, because of the relators $SW_S^-1$. I will leave you to work out where the maps $alpha$ and $beta$ fold in to this description.
The above presentation justifies the term extension of $A$ by $Q$: we've started with a presentation for $A$, and then added in the presentation for $Q$ in a specific way to obtain a presentation for $E$.
For a worked example of the above (with some genuinely amazing applications, both in the paper and in subsequent research), look at the paper Rips, E. (1982), Subgroups of small Cancellation Groups. Bull. Lond. Math. Soc. 14: 45-47. doi:10.1112/blms/14.1.45
It is an interesting question when a presentation of the above form does actually define a group extension. This was studied in the paper Pride, S., Harlander, J. & Baik, Y. (1998). The geometry of group extensions. J. Group Theory, 1(4), pp. 395-416. doi:10.1515/jgth.1998.028
$endgroup$
$begingroup$
It's probably worth saying something about how this interacts with the principle that we generally only care about things "up to isomorphism".
$endgroup$
– Hurkyl
8 hours ago
add a comment |
$begingroup$
Let's say you want to classify all the finite groups (up to isomorphism). You know that there are simple groups, the groups which have no non-trivial normal subgroups. You can think of these as the "atoms." Now you want to classify all the finite groups, not just the simple ones. You might hope that an arbitrary finite group is a product of the simple ones, but unfortunately it's not that... simple.
For example, the dihedral group of order 6 is "built out of" $mathbbZ/2mathbbZ$ and $mathbbZ/3mathbbZ$ in the sense that it has a normal subgroup isomorphic to $mathbbZ/3mathbbZ$ and the quotient is isomorphic to $mathbbZ/2mathbbZ$, but it's definitely not their product. So to finish the classification problem, you need some set of rules for forming "molecules" from the atoms.
The "molecule" rule is equivalent to this: given two groups $G$ and $Q$, classify all groups $E$ such $G$ is a normal subgroup of $E$ with the quotient isomorphic to $Q$. This is exactly the extension problem. If we have an effective way to calculate this for every pair of groups, and we also know all the simple groups, then we have solved the classificaiton problem.
$endgroup$
1
$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
10 hours ago
$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
9 hours ago
add a comment |
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$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
Because you care about presentations: if $A$ has presentation $langle mathbfxmidmathbfrrangle$ and $Q$ has presentation $langle mathbfymidmathbfsrangle$ then the group $E$ given by the above short exact sequence has presentation of the form:
$$
langle mathbfx, ymid SW_S^-1 (Sinmathbfs), mathbfr, mathbftrangle
$$
where $W_Sin F(mathbfx)$ for all $Sinmathbfs$, and $mathbft$ consists of words of the form $y^-epsilonxy^epsilonX^-1$ with $xinmathbfx$, $yinmathbfy$ and $Xin F(mathbfx)$. The intuition here is that relators in $mathbft$ ensure normality of $A$, and so removing all the $x$-terms makes sense. When they are removed you get the presentation $langle mathbfymidmathbfsrangle$, because of the relators $SW_S^-1$. I will leave you to work out where the maps $alpha$ and $beta$ fold in to this description.
The above presentation justifies the term extension of $A$ by $Q$: we've started with a presentation for $A$, and then added in the presentation for $Q$ in a specific way to obtain a presentation for $E$.
For a worked example of the above (with some genuinely amazing applications, both in the paper and in subsequent research), look at the paper Rips, E. (1982), Subgroups of small Cancellation Groups. Bull. Lond. Math. Soc. 14: 45-47. doi:10.1112/blms/14.1.45
It is an interesting question when a presentation of the above form does actually define a group extension. This was studied in the paper Pride, S., Harlander, J. & Baik, Y. (1998). The geometry of group extensions. J. Group Theory, 1(4), pp. 395-416. doi:10.1515/jgth.1998.028
$endgroup$
$begingroup$
It's probably worth saying something about how this interacts with the principle that we generally only care about things "up to isomorphism".
$endgroup$
– Hurkyl
8 hours ago
add a comment |
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
Because you care about presentations: if $A$ has presentation $langle mathbfxmidmathbfrrangle$ and $Q$ has presentation $langle mathbfymidmathbfsrangle$ then the group $E$ given by the above short exact sequence has presentation of the form:
$$
langle mathbfx, ymid SW_S^-1 (Sinmathbfs), mathbfr, mathbftrangle
$$
where $W_Sin F(mathbfx)$ for all $Sinmathbfs$, and $mathbft$ consists of words of the form $y^-epsilonxy^epsilonX^-1$ with $xinmathbfx$, $yinmathbfy$ and $Xin F(mathbfx)$. The intuition here is that relators in $mathbft$ ensure normality of $A$, and so removing all the $x$-terms makes sense. When they are removed you get the presentation $langle mathbfymidmathbfsrangle$, because of the relators $SW_S^-1$. I will leave you to work out where the maps $alpha$ and $beta$ fold in to this description.
The above presentation justifies the term extension of $A$ by $Q$: we've started with a presentation for $A$, and then added in the presentation for $Q$ in a specific way to obtain a presentation for $E$.
For a worked example of the above (with some genuinely amazing applications, both in the paper and in subsequent research), look at the paper Rips, E. (1982), Subgroups of small Cancellation Groups. Bull. Lond. Math. Soc. 14: 45-47. doi:10.1112/blms/14.1.45
It is an interesting question when a presentation of the above form does actually define a group extension. This was studied in the paper Pride, S., Harlander, J. & Baik, Y. (1998). The geometry of group extensions. J. Group Theory, 1(4), pp. 395-416. doi:10.1515/jgth.1998.028
$endgroup$
$begingroup$
It's probably worth saying something about how this interacts with the principle that we generally only care about things "up to isomorphism".
$endgroup$
– Hurkyl
8 hours ago
add a comment |
$begingroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
Because you care about presentations: if $A$ has presentation $langle mathbfxmidmathbfrrangle$ and $Q$ has presentation $langle mathbfymidmathbfsrangle$ then the group $E$ given by the above short exact sequence has presentation of the form:
$$
langle mathbfx, ymid SW_S^-1 (Sinmathbfs), mathbfr, mathbftrangle
$$
where $W_Sin F(mathbfx)$ for all $Sinmathbfs$, and $mathbft$ consists of words of the form $y^-epsilonxy^epsilonX^-1$ with $xinmathbfx$, $yinmathbfy$ and $Xin F(mathbfx)$. The intuition here is that relators in $mathbft$ ensure normality of $A$, and so removing all the $x$-terms makes sense. When they are removed you get the presentation $langle mathbfymidmathbfsrangle$, because of the relators $SW_S^-1$. I will leave you to work out where the maps $alpha$ and $beta$ fold in to this description.
The above presentation justifies the term extension of $A$ by $Q$: we've started with a presentation for $A$, and then added in the presentation for $Q$ in a specific way to obtain a presentation for $E$.
For a worked example of the above (with some genuinely amazing applications, both in the paper and in subsequent research), look at the paper Rips, E. (1982), Subgroups of small Cancellation Groups. Bull. Lond. Math. Soc. 14: 45-47. doi:10.1112/blms/14.1.45
It is an interesting question when a presentation of the above form does actually define a group extension. This was studied in the paper Pride, S., Harlander, J. & Baik, Y. (1998). The geometry of group extensions. J. Group Theory, 1(4), pp. 395-416. doi:10.1515/jgth.1998.028
$endgroup$
A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
Because you care about presentations: if $A$ has presentation $langle mathbfxmidmathbfrrangle$ and $Q$ has presentation $langle mathbfymidmathbfsrangle$ then the group $E$ given by the above short exact sequence has presentation of the form:
$$
langle mathbfx, ymid SW_S^-1 (Sinmathbfs), mathbfr, mathbftrangle
$$
where $W_Sin F(mathbfx)$ for all $Sinmathbfs$, and $mathbft$ consists of words of the form $y^-epsilonxy^epsilonX^-1$ with $xinmathbfx$, $yinmathbfy$ and $Xin F(mathbfx)$. The intuition here is that relators in $mathbft$ ensure normality of $A$, and so removing all the $x$-terms makes sense. When they are removed you get the presentation $langle mathbfymidmathbfsrangle$, because of the relators $SW_S^-1$. I will leave you to work out where the maps $alpha$ and $beta$ fold in to this description.
The above presentation justifies the term extension of $A$ by $Q$: we've started with a presentation for $A$, and then added in the presentation for $Q$ in a specific way to obtain a presentation for $E$.
For a worked example of the above (with some genuinely amazing applications, both in the paper and in subsequent research), look at the paper Rips, E. (1982), Subgroups of small Cancellation Groups. Bull. Lond. Math. Soc. 14: 45-47. doi:10.1112/blms/14.1.45
It is an interesting question when a presentation of the above form does actually define a group extension. This was studied in the paper Pride, S., Harlander, J. & Baik, Y. (1998). The geometry of group extensions. J. Group Theory, 1(4), pp. 395-416. doi:10.1515/jgth.1998.028
edited 13 hours ago
answered 13 hours ago
user1729user1729
17.6k64294
17.6k64294
$begingroup$
It's probably worth saying something about how this interacts with the principle that we generally only care about things "up to isomorphism".
$endgroup$
– Hurkyl
8 hours ago
add a comment |
$begingroup$
It's probably worth saying something about how this interacts with the principle that we generally only care about things "up to isomorphism".
$endgroup$
– Hurkyl
8 hours ago
$begingroup$
It's probably worth saying something about how this interacts with the principle that we generally only care about things "up to isomorphism".
$endgroup$
– Hurkyl
8 hours ago
$begingroup$
It's probably worth saying something about how this interacts with the principle that we generally only care about things "up to isomorphism".
$endgroup$
– Hurkyl
8 hours ago
add a comment |
$begingroup$
Let's say you want to classify all the finite groups (up to isomorphism). You know that there are simple groups, the groups which have no non-trivial normal subgroups. You can think of these as the "atoms." Now you want to classify all the finite groups, not just the simple ones. You might hope that an arbitrary finite group is a product of the simple ones, but unfortunately it's not that... simple.
For example, the dihedral group of order 6 is "built out of" $mathbbZ/2mathbbZ$ and $mathbbZ/3mathbbZ$ in the sense that it has a normal subgroup isomorphic to $mathbbZ/3mathbbZ$ and the quotient is isomorphic to $mathbbZ/2mathbbZ$, but it's definitely not their product. So to finish the classification problem, you need some set of rules for forming "molecules" from the atoms.
The "molecule" rule is equivalent to this: given two groups $G$ and $Q$, classify all groups $E$ such $G$ is a normal subgroup of $E$ with the quotient isomorphic to $Q$. This is exactly the extension problem. If we have an effective way to calculate this for every pair of groups, and we also know all the simple groups, then we have solved the classificaiton problem.
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1
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Is $G$ here the same as $A$ in the OP?
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– Mitch
10 hours ago
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Very nice and enlightening answer!
$endgroup$
– lush
9 hours ago
add a comment |
$begingroup$
Let's say you want to classify all the finite groups (up to isomorphism). You know that there are simple groups, the groups which have no non-trivial normal subgroups. You can think of these as the "atoms." Now you want to classify all the finite groups, not just the simple ones. You might hope that an arbitrary finite group is a product of the simple ones, but unfortunately it's not that... simple.
For example, the dihedral group of order 6 is "built out of" $mathbbZ/2mathbbZ$ and $mathbbZ/3mathbbZ$ in the sense that it has a normal subgroup isomorphic to $mathbbZ/3mathbbZ$ and the quotient is isomorphic to $mathbbZ/2mathbbZ$, but it's definitely not their product. So to finish the classification problem, you need some set of rules for forming "molecules" from the atoms.
The "molecule" rule is equivalent to this: given two groups $G$ and $Q$, classify all groups $E$ such $G$ is a normal subgroup of $E$ with the quotient isomorphic to $Q$. This is exactly the extension problem. If we have an effective way to calculate this for every pair of groups, and we also know all the simple groups, then we have solved the classificaiton problem.
$endgroup$
1
$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
10 hours ago
$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
9 hours ago
add a comment |
$begingroup$
Let's say you want to classify all the finite groups (up to isomorphism). You know that there are simple groups, the groups which have no non-trivial normal subgroups. You can think of these as the "atoms." Now you want to classify all the finite groups, not just the simple ones. You might hope that an arbitrary finite group is a product of the simple ones, but unfortunately it's not that... simple.
For example, the dihedral group of order 6 is "built out of" $mathbbZ/2mathbbZ$ and $mathbbZ/3mathbbZ$ in the sense that it has a normal subgroup isomorphic to $mathbbZ/3mathbbZ$ and the quotient is isomorphic to $mathbbZ/2mathbbZ$, but it's definitely not their product. So to finish the classification problem, you need some set of rules for forming "molecules" from the atoms.
The "molecule" rule is equivalent to this: given two groups $G$ and $Q$, classify all groups $E$ such $G$ is a normal subgroup of $E$ with the quotient isomorphic to $Q$. This is exactly the extension problem. If we have an effective way to calculate this for every pair of groups, and we also know all the simple groups, then we have solved the classificaiton problem.
$endgroup$
Let's say you want to classify all the finite groups (up to isomorphism). You know that there are simple groups, the groups which have no non-trivial normal subgroups. You can think of these as the "atoms." Now you want to classify all the finite groups, not just the simple ones. You might hope that an arbitrary finite group is a product of the simple ones, but unfortunately it's not that... simple.
For example, the dihedral group of order 6 is "built out of" $mathbbZ/2mathbbZ$ and $mathbbZ/3mathbbZ$ in the sense that it has a normal subgroup isomorphic to $mathbbZ/3mathbbZ$ and the quotient is isomorphic to $mathbbZ/2mathbbZ$, but it's definitely not their product. So to finish the classification problem, you need some set of rules for forming "molecules" from the atoms.
The "molecule" rule is equivalent to this: given two groups $G$ and $Q$, classify all groups $E$ such $G$ is a normal subgroup of $E$ with the quotient isomorphic to $Q$. This is exactly the extension problem. If we have an effective way to calculate this for every pair of groups, and we also know all the simple groups, then we have solved the classificaiton problem.
answered 12 hours ago
hunterhunter
15.6k32640
15.6k32640
1
$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
10 hours ago
$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
9 hours ago
add a comment |
1
$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
10 hours ago
$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
9 hours ago
1
1
$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
10 hours ago
$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
10 hours ago
$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
9 hours ago
$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
9 hours ago
add a comment |
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$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
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– lush
14 hours ago
1
$begingroup$
I'm afraid not, @lush. I'm aware of extensions in the context of fields and how they can be seen as vector spaces though, like in Galois Theory or perhaps in Algebraic Number Theory.
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– Shaun
14 hours ago
4
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A short exact sequence $1rightarrow Arightarrow Erightarrow Qrightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/Acong Q$". [The sequence also gives the isomorphism $beta: E/Arightarrow Q$, while $alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
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– user1729
13 hours ago