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Can we show that every subgroup of a free group is free using Kurosh's theorem?
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Question: I have seen classic proofs about " every subgroup of a free group is free". My question now is Can we use Kurosh's theorem to prove the statement?
Kurosh's theorem may be found in the book: "Introduction to group theory, Oleg Bogopolski, page 92." Thanks for any feedback!
geometric-group-theory
asked Jul 28 at 23:18
Djinola
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Question: I have seen classic proofs about " every subgroup of a free group is free". My question now is Can we use Kurosh's theorem to prove the statement?
Kurosh's theorem may be found in the book: "Introduction to group theory, Oleg Bogopolski, page 92." Thanks for any feedback!
geometric-group-theory
asked Jul 28 at 23:18
Djinola
697
1
Yes, it's a straightforward corollary.
– Qiaochu Yuan
Jul 28 at 23:23
@ Qiaochu Yuan, Kurosh's roughly says that every subgroup G of free product H(of free groups Hi) is a free product of some free group F and the groups G intersection with some conjugates of Hi. When you say that it is a straightforward corollary, do you mean it is in the case where H is just the free group H? Or if not, can you be more explicit?
– Djinola
Jul 28 at 23:33
1
Write a free group as the free product of copies of $mathbbZ$.
– Qiaochu Yuan
Jul 28 at 23:37
Can you give a little bit more detail up to the point where I can apply Kurosh's theorem? It looks like any subgroup is gonna be again a free product of copies of Z so far, right?
– Djinola
Jul 29 at 0:03
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question: I have seen classic proofs about " every subgroup of a free group is free". My question now is Can we use Kurosh's theorem to prove the statement?
Kurosh's theorem may be found in the book: "Introduction to group theory, Oleg Bogopolski, page 92." Thanks for any feedback!
geometric-group-theory
asked Jul 28 at 23:18
Djinola
697
Question: I have seen classic proofs about " every subgroup of a free group is free". My question now is Can we use Kurosh's theorem to prove the statement?
Kurosh's theorem may be found in the book: "Introduction to group theory, Oleg Bogopolski, page 92." Thanks for any feedback!
geometric-group-theory
asked Jul 28 at 23:18
Djinola
697
asked Jul 28 at 23:18
Djinola
697
asked Jul 28 at 23:18
Djinola
697
asked Jul 28 at 23:18
Djinola
697
697
1
Yes, it's a straightforward corollary.
– Qiaochu Yuan
Jul 28 at 23:23
@ Qiaochu Yuan, Kurosh's roughly says that every subgroup G of free product H(of free groups Hi) is a free product of some free group F and the groups G intersection with some conjugates of Hi. When you say that it is a straightforward corollary, do you mean it is in the case where H is just the free group H? Or if not, can you be more explicit?
– Djinola
Jul 28 at 23:33
1
Write a free group as the free product of copies of $mathbbZ$.
– Qiaochu Yuan
Jul 28 at 23:37
Can you give a little bit more detail up to the point where I can apply Kurosh's theorem? It looks like any subgroup is gonna be again a free product of copies of Z so far, right?
– Djinola
Jul 29 at 0:03
add a comment |Â
1
Yes, it's a straightforward corollary.
– Qiaochu Yuan
Jul 28 at 23:23
@ Qiaochu Yuan, Kurosh's roughly says that every subgroup G of free product H(of free groups Hi) is a free product of some free group F and the groups G intersection with some conjugates of Hi. When you say that it is a straightforward corollary, do you mean it is in the case where H is just the free group H? Or if not, can you be more explicit?
– Djinola
Jul 28 at 23:33
1
Write a free group as the free product of copies of $mathbbZ$.
– Qiaochu Yuan
Jul 28 at 23:37
Can you give a little bit more detail up to the point where I can apply Kurosh's theorem? It looks like any subgroup is gonna be again a free product of copies of Z so far, right?
– Djinola
Jul 29 at 0:03
1
1
Yes, it's a straightforward corollary.
– Qiaochu Yuan
Jul 28 at 23:23
Yes, it's a straightforward corollary.
– Qiaochu Yuan
Jul 28 at 23:23
@ Qiaochu Yuan, Kurosh's roughly says that every subgroup G of free product H(of free groups Hi) is a free product of some free group F and the groups G intersection with some conjugates of Hi. When you say that it is a straightforward corollary, do you mean it is in the case where H is just the free group H? Or if not, can you be more explicit?
– Djinola
Jul 28 at 23:33
@ Qiaochu Yuan, Kurosh's roughly says that every subgroup G of free product H(of free groups Hi) is a free product of some free group F and the groups G intersection with some conjugates of Hi. When you say that it is a straightforward corollary, do you mean it is in the case where H is just the free group H? Or if not, can you be more explicit?
– Djinola
Jul 28 at 23:33
1
1
Write a free group as the free product of copies of $mathbbZ$.
– Qiaochu Yuan
Jul 28 at 23:37
Write a free group as the free product of copies of $mathbbZ$.
– Qiaochu Yuan
Jul 28 at 23:37
Can you give a little bit more detail up to the point where I can apply Kurosh's theorem? It looks like any subgroup is gonna be again a free product of copies of Z so far, right?
– Djinola
Jul 29 at 0:03
Can you give a little bit more detail up to the point where I can apply Kurosh's theorem? It looks like any subgroup is gonna be again a free product of copies of Z so far, right?
– Djinola
Jul 29 at 0:03
add a comment |Â
1 Answer
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The answer to your question is "yes", as @QiaochuYuan suggests in the comments. Let me give a detailed answer. But, after that, I will express some reservations about this as a method of proof.
Consider a free group $F$. As QiaochuYuan suggests, write $F$ as a free product $F = bigstar_iin I , Z_i$ where each group $Z_i$ is infinite cyclic. Given a subgroup $A < F$, Kurosh's Theorem let's you conclude that the group $A$ is a free product of the form $A = (bigstar_jin J , A_j) * B$ with the following properties:
- The free factors $A_j$ ($j in J$) are all of the nontrivial subgroups of $A$ that can be written in the form $A cap g Z_i g^-1$ for $i in I$ and $g in F$. It follows that each $A_i$ is a nontrivial subgroup of an infinite cyclic group, and so is itself infinite cyclic.
- The subgroup $B$ is a free group.
Thus $A$ is a free product of infinite cyclic groups and one free group, from which one can conclude that $A$ is a free group.
So, what are my reservations? Roughly speaking, in the topological approach to studying free groups, one deduces that $B$ is a free group in pretty much the same way as one proves that a subgroup of a free group is free. Namely, in both cases, one arrives at the conclusion that the group in question is free by constructing a free action of that group on a simplicial tree; then one proves that any group that acts freely on a simplicial tree is a free group. For $B$ itself, the tree that one uses is the Bass-Serre tree of the free product decomposition $F = bigstar_i in I , Z_i$. For a subgroup of a free group, the tree that one uses is the Cayley tree of the free group with respect to a free basis.
In a nutshell: the proof of the Kurosh subgroup theorem can be regarded as an extension of the proof that subgroups of free groups are free, an extension which requires the additional machinery of Bass-Serre theory to carry out in full. So it is technologically less expensive to just prove that subgroups of free groups are free than to deduce that from the Kurosh subgroup theorem.
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The answer to your question is "yes", as @QiaochuYuan suggests in the comments. Let me give a detailed answer. But, after that, I will express some reservations about this as a method of proof.
Consider a free group $F$. As QiaochuYuan suggests, write $F$ as a free product $F = bigstar_iin I , Z_i$ where each group $Z_i$ is infinite cyclic. Given a subgroup $A < F$, Kurosh's Theorem let's you conclude that the group $A$ is a free product of the form $A = (bigstar_jin J , A_j) * B$ with the following properties:
- The free factors $A_j$ ($j in J$) are all of the nontrivial subgroups of $A$ that can be written in the form $A cap g Z_i g^-1$ for $i in I$ and $g in F$. It follows that each $A_i$ is a nontrivial subgroup of an infinite cyclic group, and so is itself infinite cyclic.
- The subgroup $B$ is a free group.
Thus $A$ is a free product of infinite cyclic groups and one free group, from which one can conclude that $A$ is a free group.
So, what are my reservations? Roughly speaking, in the topological approach to studying free groups, one deduces that $B$ is a free group in pretty much the same way as one proves that a subgroup of a free group is free. Namely, in both cases, one arrives at the conclusion that the group in question is free by constructing a free action of that group on a simplicial tree; then one proves that any group that acts freely on a simplicial tree is a free group. For $B$ itself, the tree that one uses is the Bass-Serre tree of the free product decomposition $F = bigstar_i in I , Z_i$. For a subgroup of a free group, the tree that one uses is the Cayley tree of the free group with respect to a free basis.
In a nutshell: the proof of the Kurosh subgroup theorem can be regarded as an extension of the proof that subgroups of free groups are free, an extension which requires the additional machinery of Bass-Serre theory to carry out in full. So it is technologically less expensive to just prove that subgroups of free groups are free than to deduce that from the Kurosh subgroup theorem.
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
add a comment |Â
up vote
2
down vote
accepted
The answer to your question is "yes", as @QiaochuYuan suggests in the comments. Let me give a detailed answer. But, after that, I will express some reservations about this as a method of proof.
Consider a free group $F$. As QiaochuYuan suggests, write $F$ as a free product $F = bigstar_iin I , Z_i$ where each group $Z_i$ is infinite cyclic. Given a subgroup $A < F$, Kurosh's Theorem let's you conclude that the group $A$ is a free product of the form $A = (bigstar_jin J , A_j) * B$ with the following properties:
- The free factors $A_j$ ($j in J$) are all of the nontrivial subgroups of $A$ that can be written in the form $A cap g Z_i g^-1$ for $i in I$ and $g in F$. It follows that each $A_i$ is a nontrivial subgroup of an infinite cyclic group, and so is itself infinite cyclic.
- The subgroup $B$ is a free group.
Thus $A$ is a free product of infinite cyclic groups and one free group, from which one can conclude that $A$ is a free group.
So, what are my reservations? Roughly speaking, in the topological approach to studying free groups, one deduces that $B$ is a free group in pretty much the same way as one proves that a subgroup of a free group is free. Namely, in both cases, one arrives at the conclusion that the group in question is free by constructing a free action of that group on a simplicial tree; then one proves that any group that acts freely on a simplicial tree is a free group. For $B$ itself, the tree that one uses is the Bass-Serre tree of the free product decomposition $F = bigstar_i in I , Z_i$. For a subgroup of a free group, the tree that one uses is the Cayley tree of the free group with respect to a free basis.
In a nutshell: the proof of the Kurosh subgroup theorem can be regarded as an extension of the proof that subgroups of free groups are free, an extension which requires the additional machinery of Bass-Serre theory to carry out in full. So it is technologically less expensive to just prove that subgroups of free groups are free than to deduce that from the Kurosh subgroup theorem.
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The answer to your question is "yes", as @QiaochuYuan suggests in the comments. Let me give a detailed answer. But, after that, I will express some reservations about this as a method of proof.
Consider a free group $F$. As QiaochuYuan suggests, write $F$ as a free product $F = bigstar_iin I , Z_i$ where each group $Z_i$ is infinite cyclic. Given a subgroup $A < F$, Kurosh's Theorem let's you conclude that the group $A$ is a free product of the form $A = (bigstar_jin J , A_j) * B$ with the following properties:
- The free factors $A_j$ ($j in J$) are all of the nontrivial subgroups of $A$ that can be written in the form $A cap g Z_i g^-1$ for $i in I$ and $g in F$. It follows that each $A_i$ is a nontrivial subgroup of an infinite cyclic group, and so is itself infinite cyclic.
- The subgroup $B$ is a free group.
Thus $A$ is a free product of infinite cyclic groups and one free group, from which one can conclude that $A$ is a free group.
So, what are my reservations? Roughly speaking, in the topological approach to studying free groups, one deduces that $B$ is a free group in pretty much the same way as one proves that a subgroup of a free group is free. Namely, in both cases, one arrives at the conclusion that the group in question is free by constructing a free action of that group on a simplicial tree; then one proves that any group that acts freely on a simplicial tree is a free group. For $B$ itself, the tree that one uses is the Bass-Serre tree of the free product decomposition $F = bigstar_i in I , Z_i$. For a subgroup of a free group, the tree that one uses is the Cayley tree of the free group with respect to a free basis.
In a nutshell: the proof of the Kurosh subgroup theorem can be regarded as an extension of the proof that subgroups of free groups are free, an extension which requires the additional machinery of Bass-Serre theory to carry out in full. So it is technologically less expensive to just prove that subgroups of free groups are free than to deduce that from the Kurosh subgroup theorem.
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
The answer to your question is "yes", as @QiaochuYuan suggests in the comments. Let me give a detailed answer. But, after that, I will express some reservations about this as a method of proof.
Consider a free group $F$. As QiaochuYuan suggests, write $F$ as a free product $F = bigstar_iin I , Z_i$ where each group $Z_i$ is infinite cyclic. Given a subgroup $A < F$, Kurosh's Theorem let's you conclude that the group $A$ is a free product of the form $A = (bigstar_jin J , A_j) * B$ with the following properties:
- The free factors $A_j$ ($j in J$) are all of the nontrivial subgroups of $A$ that can be written in the form $A cap g Z_i g^-1$ for $i in I$ and $g in F$. It follows that each $A_i$ is a nontrivial subgroup of an infinite cyclic group, and so is itself infinite cyclic.
- The subgroup $B$ is a free group.
Thus $A$ is a free product of infinite cyclic groups and one free group, from which one can conclude that $A$ is a free group.
So, what are my reservations? Roughly speaking, in the topological approach to studying free groups, one deduces that $B$ is a free group in pretty much the same way as one proves that a subgroup of a free group is free. Namely, in both cases, one arrives at the conclusion that the group in question is free by constructing a free action of that group on a simplicial tree; then one proves that any group that acts freely on a simplicial tree is a free group. For $B$ itself, the tree that one uses is the Bass-Serre tree of the free product decomposition $F = bigstar_i in I , Z_i$. For a subgroup of a free group, the tree that one uses is the Cayley tree of the free group with respect to a free basis.
In a nutshell: the proof of the Kurosh subgroup theorem can be regarded as an extension of the proof that subgroups of free groups are free, an extension which requires the additional machinery of Bass-Serre theory to carry out in full. So it is technologically less expensive to just prove that subgroups of free groups are free than to deduce that from the Kurosh subgroup theorem.
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
edited Jul 30 at 12:20
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
answered Jul 30 at 2:19
Lee Mosher
45.4k33478
45.4k33478
add a comment |Â
add a comment |Â
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1
Yes, it's a straightforward corollary.
– Qiaochu Yuan
Jul 28 at 23:23
@ Qiaochu Yuan, Kurosh's roughly says that every subgroup G of free product H(of free groups Hi) is a free product of some free group F and the groups G intersection with some conjugates of Hi. When you say that it is a straightforward corollary, do you mean it is in the case where H is just the free group H? Or if not, can you be more explicit?
– Djinola
Jul 28 at 23:33
1
Write a free group as the free product of copies of $mathbbZ$.
– Qiaochu Yuan
Jul 28 at 23:37
Can you give a little bit more detail up to the point where I can apply Kurosh's theorem? It looks like any subgroup is gonna be again a free product of copies of Z so far, right?
– Djinola
Jul 29 at 0:03