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?


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













15












$begingroup$



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 :)










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Do you know short exact sequences from other contexts like vector spaces or modules?
    $endgroup$
    – lush
    Apr 4 at 12:08






  • 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
    Apr 4 at 12:11






  • 5




    $begingroup$
    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
    Apr 4 at 12:42
















15












$begingroup$



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 :)










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Do you know short exact sequences from other contexts like vector spaces or modules?
    $endgroup$
    – lush
    Apr 4 at 12:08






  • 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
    Apr 4 at 12:11






  • 5




    $begingroup$
    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
    Apr 4 at 12:42














15












15








15


2



$begingroup$



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 :)










share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 4 at 11:43









ShaunShaun

10.4k113686




10.4k113686







  • 1




    $begingroup$
    Do you know short exact sequences from other contexts like vector spaces or modules?
    $endgroup$
    – lush
    Apr 4 at 12:08






  • 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
    Apr 4 at 12:11






  • 5




    $begingroup$
    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
    Apr 4 at 12:42













  • 1




    $begingroup$
    Do you know short exact sequences from other contexts like vector spaces or modules?
    $endgroup$
    – lush
    Apr 4 at 12:08






  • 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
    Apr 4 at 12:11






  • 5




    $begingroup$
    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
    Apr 4 at 12:42








1




1




$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
$endgroup$
– lush
Apr 4 at 12:08




$begingroup$
Do you know short exact sequences from other contexts like vector spaces or modules?
$endgroup$
– lush
Apr 4 at 12:08




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
Apr 4 at 12:11




$begingroup$
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
Apr 4 at 12:11




5




5




$begingroup$
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
Apr 4 at 12:42





$begingroup$
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
Apr 4 at 12:42











2 Answers
2






active

oldest

votes


















19












$begingroup$

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






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    Apr 4 at 17:56


















15












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Is $G$ here the same as $A$ in the OP?
    $endgroup$
    – Mitch
    Apr 4 at 16:27










  • $begingroup$
    Very nice and enlightening answer!
    $endgroup$
    – lush
    Apr 4 at 17:25











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2 Answers
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2 Answers
2






active

oldest

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active

oldest

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active

oldest

votes









19












$begingroup$

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






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    Apr 4 at 17:56















19












$begingroup$

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






share|cite|improve this answer











$endgroup$












  • $begingroup$
    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
    Apr 4 at 17:56













19












19








19





$begingroup$

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






share|cite|improve this answer











$endgroup$



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







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edited Apr 4 at 13:29

























answered Apr 4 at 13:17









user1729user1729

17.7k64294




17.7k64294











  • $begingroup$
    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
    Apr 4 at 17:56
















  • $begingroup$
    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
    Apr 4 at 17:56















$begingroup$
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
Apr 4 at 17:56




$begingroup$
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
Apr 4 at 17:56











15












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Is $G$ here the same as $A$ in the OP?
    $endgroup$
    – Mitch
    Apr 4 at 16:27










  • $begingroup$
    Very nice and enlightening answer!
    $endgroup$
    – lush
    Apr 4 at 17:25















15












$begingroup$

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.






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Is $G$ here the same as $A$ in the OP?
    $endgroup$
    – Mitch
    Apr 4 at 16:27










  • $begingroup$
    Very nice and enlightening answer!
    $endgroup$
    – lush
    Apr 4 at 17:25













15












15








15





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 4 at 13:37









hunterhunter

15.6k32643




15.6k32643







  • 1




    $begingroup$
    Is $G$ here the same as $A$ in the OP?
    $endgroup$
    – Mitch
    Apr 4 at 16:27










  • $begingroup$
    Very nice and enlightening answer!
    $endgroup$
    – lush
    Apr 4 at 17:25












  • 1




    $begingroup$
    Is $G$ here the same as $A$ in the OP?
    $endgroup$
    – Mitch
    Apr 4 at 16:27










  • $begingroup$
    Very nice and enlightening answer!
    $endgroup$
    – lush
    Apr 4 at 17:25







1




1




$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
Apr 4 at 16:27




$begingroup$
Is $G$ here the same as $A$ in the OP?
$endgroup$
– Mitch
Apr 4 at 16:27












$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
Apr 4 at 17:25




$begingroup$
Very nice and enlightening answer!
$endgroup$
– lush
Apr 4 at 17:25

















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