If I have the presentation of a group, how can I find the commutator subgroup of it?
$begingroup$
I have the group given by the presentation $G= langle a,bmid a^2,b^2rangle$
How can I in general find $G',G/G',G''$ ?
thanks for any hints.
abstract-algebra group-theory group-presentation
edited Dec 19 '18 at 12:33
Shaun
9,294113684
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
$endgroup$
add a comment |
$begingroup$
I have the group given by the presentation $G= langle a,bmid a^2,b^2rangle$
How can I in general find $G',G/G',G''$ ?
thanks for any hints.
abstract-algebra group-theory group-presentation
edited Dec 19 '18 at 12:33
Shaun
9,294113684
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
$endgroup$
$begingroup$
have a look at this post for an example: math.stackexchange.com/questions/416617/…
$endgroup$
– citedcorpse
Jun 11 '13 at 17:38
$begingroup$
also, i think finding a presentation even for just the commutator subgroup itself is not possible in general
$endgroup$
– citedcorpse
Jun 11 '13 at 17:40
add a comment |
$begingroup$
I have the group given by the presentation $G= langle a,bmid a^2,b^2rangle$
How can I in general find $G',G/G',G''$ ?
thanks for any hints.
abstract-algebra group-theory group-presentation
edited Dec 19 '18 at 12:33
Shaun
9,294113684
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
$endgroup$
I have the group given by the presentation $G= langle a,bmid a^2,b^2rangle$
How can I in general find $G',G/G',G''$ ?
thanks for any hints.
abstract-algebra group-theory group-presentation
abstract-algebra group-theory group-presentation
edited Dec 19 '18 at 12:33
Shaun
9,294113684
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
edited Dec 19 '18 at 12:33
Shaun
9,294113684
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
edited Dec 19 '18 at 12:33
Shaun
9,294113684
edited Dec 19 '18 at 12:33
Shaun
9,294113684
edited Dec 19 '18 at 12:33
Shaun
9,294113684
9,294113684
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
asked Jun 11 '13 at 17:37
kiranovalobaskiranovalobas
419213
419213
$begingroup$
have a look at this post for an example: math.stackexchange.com/questions/416617/…
$endgroup$
– citedcorpse
Jun 11 '13 at 17:38
$begingroup$
also, i think finding a presentation even for just the commutator subgroup itself is not possible in general
$endgroup$
– citedcorpse
Jun 11 '13 at 17:40
add a comment |
$begingroup$
have a look at this post for an example: math.stackexchange.com/questions/416617/…
$endgroup$
– citedcorpse
Jun 11 '13 at 17:38
$begingroup$
also, i think finding a presentation even for just the commutator subgroup itself is not possible in general
$endgroup$
– citedcorpse
Jun 11 '13 at 17:40
$begingroup$
have a look at this post for an example: math.stackexchange.com/questions/416617/…
$endgroup$
– citedcorpse
Jun 11 '13 at 17:38
$begingroup$
have a look at this post for an example: math.stackexchange.com/questions/416617/…
$endgroup$
– citedcorpse
Jun 11 '13 at 17:38
$begingroup$
also, i think finding a presentation even for just the commutator subgroup itself is not possible in general
$endgroup$
– citedcorpse
Jun 11 '13 at 17:40
$begingroup$
also, i think finding a presentation even for just the commutator subgroup itself is not possible in general
$endgroup$
– citedcorpse
Jun 11 '13 at 17:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In general, $G/G^{prime}$ is the group attained by making the generators pairwise commute, and is called the abelianisation of $G$ (*). As we only have two generators, here $G/G^{prime}$ is the group $langle a, b; a^2, b^2, [a, b]rangle$. I will leave you to work out what this is isomorphic to.
To find the derived subgroup $G^{prime}$, you can (in general) use something called "Reidemeister-Schreier" (if you want to know more, I once wrote out an example - note that this will only spit out a finite presentation if $G$ is given by a finite presentation and $G/G^{prime}$ is finite). However, you needn't do anything quite so fancy here! Indeed, $$G^{prime}=langle (ab)^2rangle.$$ Why? Well, $(ab)^2=abab=a^{-1}b^{-1}ab=[a, b]$, while $langle (ab)^2rangle$ is normal in $G$ (why?). Can you see why this is sufficient?
Now, $G^{prime}$ is cyclic. What does this mean for $G^{primeprime}$?
(*) A constructive example of the abelianisation would be something like $langle a, b, c; aba^{-1}c^{-1}, a^2, b^3rangle$, then your abelinisation is $langle a, b, c; aba^{-1}c^{-1}, a^2, c^3, [a, b], [a, c], [b,c]rangle$...but...as $a$ and $b$ now commute the first relation means that $b=c$, so your group becomes $langle a, b; a^2, b^3, [a, b]rangle$. Note that $c^3$ has become $b^3$. Can you see why this is?
edited Apr 13 '17 at 12:20
Community♦
1
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
$endgroup$
1
$begingroup$
There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example
$endgroup$
– Jack Schmidt
Jun 12 '13 at 14:19
$begingroup$
@JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.)
$endgroup$
– user1729
Jun 12 '13 at 15:29
$begingroup$
(And it should also be mentioned that if $G/G^{prime}$ is infinite then $G^{prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.)
$endgroup$
– user1729
Jun 12 '13 at 15:30
$begingroup$
Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation.
$endgroup$
– Jack Schmidt
Jun 12 '13 at 15:34
add a comment |
$begingroup$
Another way to find $G'$ is to notice that $G$ is the free product $mathbb{Z}_2 ast mathbb{Z}_2$. Then $G'$ is the kernel of the canonical projection $mathbb{Z}_2 ast mathbb{Z}_2 to mathbb{Z}_2 times mathbb{Z}_2$, and
Lemma: Let $H$ and $K$ be two groups. The kernel of $H ast K to H times K$ is free over the set ${[h,k] mid h in H backslash {1}, k in K backslash {1}}$.
For example, you can see this answer.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
$endgroup$
2
$begingroup$
This then implies that unless $H$ and $K$ are both cyclic, $G^{primeprime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group.
$endgroup$
– user1729
Jun 12 '13 at 10:06
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
In general, $G/G^{prime}$ is the group attained by making the generators pairwise commute, and is called the abelianisation of $G$ (*). As we only have two generators, here $G/G^{prime}$ is the group $langle a, b; a^2, b^2, [a, b]rangle$. I will leave you to work out what this is isomorphic to.
To find the derived subgroup $G^{prime}$, you can (in general) use something called "Reidemeister-Schreier" (if you want to know more, I once wrote out an example - note that this will only spit out a finite presentation if $G$ is given by a finite presentation and $G/G^{prime}$ is finite). However, you needn't do anything quite so fancy here! Indeed, $$G^{prime}=langle (ab)^2rangle.$$ Why? Well, $(ab)^2=abab=a^{-1}b^{-1}ab=[a, b]$, while $langle (ab)^2rangle$ is normal in $G$ (why?). Can you see why this is sufficient?
Now, $G^{prime}$ is cyclic. What does this mean for $G^{primeprime}$?
(*) A constructive example of the abelianisation would be something like $langle a, b, c; aba^{-1}c^{-1}, a^2, b^3rangle$, then your abelinisation is $langle a, b, c; aba^{-1}c^{-1}, a^2, c^3, [a, b], [a, c], [b,c]rangle$...but...as $a$ and $b$ now commute the first relation means that $b=c$, so your group becomes $langle a, b; a^2, b^3, [a, b]rangle$. Note that $c^3$ has become $b^3$. Can you see why this is?
edited Apr 13 '17 at 12:20
Community♦
1
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
$endgroup$
1
$begingroup$
There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example
$endgroup$
– Jack Schmidt
Jun 12 '13 at 14:19
$begingroup$
@JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.)
$endgroup$
– user1729
Jun 12 '13 at 15:29
$begingroup$
(And it should also be mentioned that if $G/G^{prime}$ is infinite then $G^{prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.)
$endgroup$
– user1729
Jun 12 '13 at 15:30
$begingroup$
Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation.
$endgroup$
– Jack Schmidt
Jun 12 '13 at 15:34
add a comment |
$begingroup$
In general, $G/G^{prime}$ is the group attained by making the generators pairwise commute, and is called the abelianisation of $G$ (*). As we only have two generators, here $G/G^{prime}$ is the group $langle a, b; a^2, b^2, [a, b]rangle$. I will leave you to work out what this is isomorphic to.
To find the derived subgroup $G^{prime}$, you can (in general) use something called "Reidemeister-Schreier" (if you want to know more, I once wrote out an example - note that this will only spit out a finite presentation if $G$ is given by a finite presentation and $G/G^{prime}$ is finite). However, you needn't do anything quite so fancy here! Indeed, $$G^{prime}=langle (ab)^2rangle.$$ Why? Well, $(ab)^2=abab=a^{-1}b^{-1}ab=[a, b]$, while $langle (ab)^2rangle$ is normal in $G$ (why?). Can you see why this is sufficient?
Now, $G^{prime}$ is cyclic. What does this mean for $G^{primeprime}$?
(*) A constructive example of the abelianisation would be something like $langle a, b, c; aba^{-1}c^{-1}, a^2, b^3rangle$, then your abelinisation is $langle a, b, c; aba^{-1}c^{-1}, a^2, c^3, [a, b], [a, c], [b,c]rangle$...but...as $a$ and $b$ now commute the first relation means that $b=c$, so your group becomes $langle a, b; a^2, b^3, [a, b]rangle$. Note that $c^3$ has become $b^3$. Can you see why this is?
edited Apr 13 '17 at 12:20
Community♦
1
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
$endgroup$
1
$begingroup$
There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example
$endgroup$
– Jack Schmidt
Jun 12 '13 at 14:19
$begingroup$
@JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.)
$endgroup$
– user1729
Jun 12 '13 at 15:29
$begingroup$
(And it should also be mentioned that if $G/G^{prime}$ is infinite then $G^{prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.)
$endgroup$
– user1729
Jun 12 '13 at 15:30
$begingroup$
Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation.
$endgroup$
– Jack Schmidt
Jun 12 '13 at 15:34
add a comment |
$begingroup$
In general, $G/G^{prime}$ is the group attained by making the generators pairwise commute, and is called the abelianisation of $G$ (*). As we only have two generators, here $G/G^{prime}$ is the group $langle a, b; a^2, b^2, [a, b]rangle$. I will leave you to work out what this is isomorphic to.
To find the derived subgroup $G^{prime}$, you can (in general) use something called "Reidemeister-Schreier" (if you want to know more, I once wrote out an example - note that this will only spit out a finite presentation if $G$ is given by a finite presentation and $G/G^{prime}$ is finite). However, you needn't do anything quite so fancy here! Indeed, $$G^{prime}=langle (ab)^2rangle.$$ Why? Well, $(ab)^2=abab=a^{-1}b^{-1}ab=[a, b]$, while $langle (ab)^2rangle$ is normal in $G$ (why?). Can you see why this is sufficient?
Now, $G^{prime}$ is cyclic. What does this mean for $G^{primeprime}$?
(*) A constructive example of the abelianisation would be something like $langle a, b, c; aba^{-1}c^{-1}, a^2, b^3rangle$, then your abelinisation is $langle a, b, c; aba^{-1}c^{-1}, a^2, c^3, [a, b], [a, c], [b,c]rangle$...but...as $a$ and $b$ now commute the first relation means that $b=c$, so your group becomes $langle a, b; a^2, b^3, [a, b]rangle$. Note that $c^3$ has become $b^3$. Can you see why this is?
edited Apr 13 '17 at 12:20
Community♦
1
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
$endgroup$
In general, $G/G^{prime}$ is the group attained by making the generators pairwise commute, and is called the abelianisation of $G$ (*). As we only have two generators, here $G/G^{prime}$ is the group $langle a, b; a^2, b^2, [a, b]rangle$. I will leave you to work out what this is isomorphic to.
To find the derived subgroup $G^{prime}$, you can (in general) use something called "Reidemeister-Schreier" (if you want to know more, I once wrote out an example - note that this will only spit out a finite presentation if $G$ is given by a finite presentation and $G/G^{prime}$ is finite). However, you needn't do anything quite so fancy here! Indeed, $$G^{prime}=langle (ab)^2rangle.$$ Why? Well, $(ab)^2=abab=a^{-1}b^{-1}ab=[a, b]$, while $langle (ab)^2rangle$ is normal in $G$ (why?). Can you see why this is sufficient?
Now, $G^{prime}$ is cyclic. What does this mean for $G^{primeprime}$?
(*) A constructive example of the abelianisation would be something like $langle a, b, c; aba^{-1}c^{-1}, a^2, b^3rangle$, then your abelinisation is $langle a, b, c; aba^{-1}c^{-1}, a^2, c^3, [a, b], [a, c], [b,c]rangle$...but...as $a$ and $b$ now commute the first relation means that $b=c$, so your group becomes $langle a, b; a^2, b^3, [a, b]rangle$. Note that $c^3$ has become $b^3$. Can you see why this is?
edited Apr 13 '17 at 12:20
Community♦
1
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
answered Jun 11 '13 at 18:32
user1729user1729
17.2k64193
17.2k64193
1
$begingroup$
There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example
$endgroup$
– Jack Schmidt
Jun 12 '13 at 14:19
$begingroup$
@JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.)
$endgroup$
– user1729
Jun 12 '13 at 15:29
$begingroup$
(And it should also be mentioned that if $G/G^{prime}$ is infinite then $G^{prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.)
$endgroup$
– user1729
Jun 12 '13 at 15:30
$begingroup$
Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation.
$endgroup$
– Jack Schmidt
Jun 12 '13 at 15:34
add a comment |
1
$begingroup$
There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example
$endgroup$
– Jack Schmidt
Jun 12 '13 at 14:19
$begingroup$
@JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.)
$endgroup$
– user1729
Jun 12 '13 at 15:29
$begingroup$
(And it should also be mentioned that if $G/G^{prime}$ is infinite then $G^{prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.)
$endgroup$
– user1729
Jun 12 '13 at 15:30
$begingroup$
Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation.
$endgroup$
– Jack Schmidt
Jun 12 '13 at 15:34
1
1
$begingroup$
There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example
$endgroup$
– Jack Schmidt
Jun 12 '13 at 14:19
$begingroup$
There is one potential restriction on the method in this answer: $G/G'$ should be finite for Reidemeister-Schreier to give a finite presentation. (+1) for easy methods and second example
$endgroup$
– Jack Schmidt
Jun 12 '13 at 14:19
$begingroup$
@JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.)
$endgroup$
– user1729
Jun 12 '13 at 15:29
$begingroup$
@JackSchmidt Good point - I have added this in. However, because finite presentability has been mention, anyone who is reading this should note that one can still use Reidemeister-Schreier even if $G/G^{prime}$ is infinite. You just need to bring your thinking-cap with you! (Although this is impractical unless $G/G^{prime}$ is infinite cyclic, or has a similar, very nice, Cayley graph.)
$endgroup$
– user1729
Jun 12 '13 at 15:29
$begingroup$
(And it should also be mentioned that if $G/G^{prime}$ is infinite then $G^{prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.)
$endgroup$
– user1729
Jun 12 '13 at 15:30
$begingroup$
(And it should also be mentioned that if $G/G^{prime}$ is infinite then $G^{prime}$ is not necessarily finitely presented, or even finitely generated! See Seirios' answer.)
$endgroup$
– user1729
Jun 12 '13 at 15:30
$begingroup$
Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation.
$endgroup$
– Jack Schmidt
Jun 12 '13 at 15:34
$begingroup$
Thanks! Yes, I had to remove my gloom and doom statements from my comment. A computer cannot do much of anything when $G/G'$ (or in my actual examples $G'/G''$) is infinite, but humans are pretty clever and can often recognize free groups at least, even if R-S is not obviously (to a computer) giving the trivial presentation.
$endgroup$
– Jack Schmidt
Jun 12 '13 at 15:34
add a comment |
$begingroup$
Another way to find $G'$ is to notice that $G$ is the free product $mathbb{Z}_2 ast mathbb{Z}_2$. Then $G'$ is the kernel of the canonical projection $mathbb{Z}_2 ast mathbb{Z}_2 to mathbb{Z}_2 times mathbb{Z}_2$, and
Lemma: Let $H$ and $K$ be two groups. The kernel of $H ast K to H times K$ is free over the set ${[h,k] mid h in H backslash {1}, k in K backslash {1}}$.
For example, you can see this answer.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
$endgroup$
2
$begingroup$
This then implies that unless $H$ and $K$ are both cyclic, $G^{primeprime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group.
$endgroup$
– user1729
Jun 12 '13 at 10:06
add a comment |
$begingroup$
Another way to find $G'$ is to notice that $G$ is the free product $mathbb{Z}_2 ast mathbb{Z}_2$. Then $G'$ is the kernel of the canonical projection $mathbb{Z}_2 ast mathbb{Z}_2 to mathbb{Z}_2 times mathbb{Z}_2$, and
Lemma: Let $H$ and $K$ be two groups. The kernel of $H ast K to H times K$ is free over the set ${[h,k] mid h in H backslash {1}, k in K backslash {1}}$.
For example, you can see this answer.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
$endgroup$
2
$begingroup$
This then implies that unless $H$ and $K$ are both cyclic, $G^{primeprime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group.
$endgroup$
– user1729
Jun 12 '13 at 10:06
add a comment |
$begingroup$
Another way to find $G'$ is to notice that $G$ is the free product $mathbb{Z}_2 ast mathbb{Z}_2$. Then $G'$ is the kernel of the canonical projection $mathbb{Z}_2 ast mathbb{Z}_2 to mathbb{Z}_2 times mathbb{Z}_2$, and
Lemma: Let $H$ and $K$ be two groups. The kernel of $H ast K to H times K$ is free over the set ${[h,k] mid h in H backslash {1}, k in K backslash {1}}$.
For example, you can see this answer.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
$endgroup$
Another way to find $G'$ is to notice that $G$ is the free product $mathbb{Z}_2 ast mathbb{Z}_2$. Then $G'$ is the kernel of the canonical projection $mathbb{Z}_2 ast mathbb{Z}_2 to mathbb{Z}_2 times mathbb{Z}_2$, and
Lemma: Let $H$ and $K$ be two groups. The kernel of $H ast K to H times K$ is free over the set ${[h,k] mid h in H backslash {1}, k in K backslash {1}}$.
For example, you can see this answer.
edited Apr 13 '17 at 12:19
Community♦
1
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
answered Jun 12 '13 at 9:38
SeiriosSeirios
24.2k34799
24.2k34799
2
$begingroup$
This then implies that unless $H$ and $K$ are both cyclic, $G^{primeprime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group.
$endgroup$
– user1729
Jun 12 '13 at 10:06
add a comment |
2
$begingroup$
This then implies that unless $H$ and $K$ are both cyclic, $G^{primeprime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group.
$endgroup$
– user1729
Jun 12 '13 at 10:06
2
2
$begingroup$
This then implies that unless $H$ and $K$ are both cyclic, $G^{primeprime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group.
$endgroup$
– user1729
Jun 12 '13 at 10:06
$begingroup$
This then implies that unless $H$ and $K$ are both cyclic, $G^{primeprime}$ is free on countably many generators, as it is the commutator subgroup of a non-abelian free group.
$endgroup$
– user1729
Jun 12 '13 at 10:06
add a comment |
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$begingroup$
have a look at this post for an example: math.stackexchange.com/questions/416617/…
$endgroup$
– citedcorpse
Jun 11 '13 at 17:38
$begingroup$
also, i think finding a presentation even for just the commutator subgroup itself is not possible in general
$endgroup$
– citedcorpse
Jun 11 '13 at 17:40