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Proving that commutator subgroup of the group of Mobius Transformations is infinite
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I'm not going to define the group of Mobius Transformations here. I'll be referring to the group of Mobius Transformations as M and I'll refer to the commutator subgroup of M as M'.
I understand that you can use an element of M' to generate an infinite cyclic subgroup, and this element is (1, - 1, 0, 1). Since (1, - 1, 0, 1)^n=(1, -n, 0, 1) must also be in the commutator subgroup, and n can go till infinity, M' is infinite as M' must contain the infinite cyclic subgroup generated by (1, - 1, 0, 1).
Now I'm not satisfied with this explanation. (1, - 1, 0, 1) seems to come out of nowhere. Even if it is in M', there's nothing that really says that I have to put it as this arbitrary element that will generate the infinite cyclic subgroup.
I was wondering if there is an alternative to this proof that is more intuitive. Initally when I approached this question I was thinking of a mapping from M to M' and proving that it is a bijective mapping. Does it work?
I'd like to know if there are any more intuitive proofs of this theorem. Using an arbitrary element like (1, - 1, 0, 1) isn't something I can come up with without prior knowledge.
group-theory
asked Jul 28 at 8:44
Yip Jung Hon
18011
add a comment |Â
up vote
0
down vote
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I'm not going to define the group of Mobius Transformations here. I'll be referring to the group of Mobius Transformations as M and I'll refer to the commutator subgroup of M as M'.
I understand that you can use an element of M' to generate an infinite cyclic subgroup, and this element is (1, - 1, 0, 1). Since (1, - 1, 0, 1)^n=(1, -n, 0, 1) must also be in the commutator subgroup, and n can go till infinity, M' is infinite as M' must contain the infinite cyclic subgroup generated by (1, - 1, 0, 1).
Now I'm not satisfied with this explanation. (1, - 1, 0, 1) seems to come out of nowhere. Even if it is in M', there's nothing that really says that I have to put it as this arbitrary element that will generate the infinite cyclic subgroup.
I was wondering if there is an alternative to this proof that is more intuitive. Initally when I approached this question I was thinking of a mapping from M to M' and proving that it is a bijective mapping. Does it work?
I'd like to know if there are any more intuitive proofs of this theorem. Using an arbitrary element like (1, - 1, 0, 1) isn't something I can come up with without prior knowledge.
group-theory
asked Jul 28 at 8:44
Yip Jung Hon
18011
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 28 at 9:05
So far as "why pick this element" - it suffices to show that some element in the commutator subgroup has infinite order, so exhibiting an example of such completes the proof. That is an entirely standard method. [Note a group can have infinite order without having any elements of infinite order] As to how this element is found, that is another question.
– Mark Bennet
Jul 28 at 9:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm not going to define the group of Mobius Transformations here. I'll be referring to the group of Mobius Transformations as M and I'll refer to the commutator subgroup of M as M'.
I understand that you can use an element of M' to generate an infinite cyclic subgroup, and this element is (1, - 1, 0, 1). Since (1, - 1, 0, 1)^n=(1, -n, 0, 1) must also be in the commutator subgroup, and n can go till infinity, M' is infinite as M' must contain the infinite cyclic subgroup generated by (1, - 1, 0, 1).
Now I'm not satisfied with this explanation. (1, - 1, 0, 1) seems to come out of nowhere. Even if it is in M', there's nothing that really says that I have to put it as this arbitrary element that will generate the infinite cyclic subgroup.
I was wondering if there is an alternative to this proof that is more intuitive. Initally when I approached this question I was thinking of a mapping from M to M' and proving that it is a bijective mapping. Does it work?
I'd like to know if there are any more intuitive proofs of this theorem. Using an arbitrary element like (1, - 1, 0, 1) isn't something I can come up with without prior knowledge.
group-theory
asked Jul 28 at 8:44
Yip Jung Hon
18011
I'm not going to define the group of Mobius Transformations here. I'll be referring to the group of Mobius Transformations as M and I'll refer to the commutator subgroup of M as M'.
I understand that you can use an element of M' to generate an infinite cyclic subgroup, and this element is (1, - 1, 0, 1). Since (1, - 1, 0, 1)^n=(1, -n, 0, 1) must also be in the commutator subgroup, and n can go till infinity, M' is infinite as M' must contain the infinite cyclic subgroup generated by (1, - 1, 0, 1).
Now I'm not satisfied with this explanation. (1, - 1, 0, 1) seems to come out of nowhere. Even if it is in M', there's nothing that really says that I have to put it as this arbitrary element that will generate the infinite cyclic subgroup.
I was wondering if there is an alternative to this proof that is more intuitive. Initally when I approached this question I was thinking of a mapping from M to M' and proving that it is a bijective mapping. Does it work?
I'd like to know if there are any more intuitive proofs of this theorem. Using an arbitrary element like (1, - 1, 0, 1) isn't something I can come up with without prior knowledge.
group-theory
asked Jul 28 at 8:44
Yip Jung Hon
18011
asked Jul 28 at 8:44
Yip Jung Hon
18011
asked Jul 28 at 8:44
Yip Jung Hon
18011
asked Jul 28 at 8:44
Yip Jung Hon
18011
18011
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 28 at 9:05
So far as "why pick this element" - it suffices to show that some element in the commutator subgroup has infinite order, so exhibiting an example of such completes the proof. That is an entirely standard method. [Note a group can have infinite order without having any elements of infinite order] As to how this element is found, that is another question.
– Mark Bennet
Jul 28 at 9:27
add a comment |Â
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 28 at 9:05
So far as "why pick this element" - it suffices to show that some element in the commutator subgroup has infinite order, so exhibiting an example of such completes the proof. That is an entirely standard method. [Note a group can have infinite order without having any elements of infinite order] As to how this element is found, that is another question.
– Mark Bennet
Jul 28 at 9:27
1
1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 28 at 9:05
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 28 at 9:05
So far as "why pick this element" - it suffices to show that some element in the commutator subgroup has infinite order, so exhibiting an example of such completes the proof. That is an entirely standard method. [Note a group can have infinite order without having any elements of infinite order] As to how this element is found, that is another question.
– Mark Bennet
Jul 28 at 9:27
So far as "why pick this element" - it suffices to show that some element in the commutator subgroup has infinite order, so exhibiting an example of such completes the proof. That is an entirely standard method. [Note a group can have infinite order without having any elements of infinite order] As to how this element is found, that is another question.
– Mark Bennet
Jul 28 at 9:27
add a comment |Â
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1
Please see math.meta.stackexchange.com/questions/5020
– Lord Shark the Unknown
Jul 28 at 9:05
So far as "why pick this element" - it suffices to show that some element in the commutator subgroup has infinite order, so exhibiting an example of such completes the proof. That is an entirely standard method. [Note a group can have infinite order without having any elements of infinite order] As to how this element is found, that is another question.
– Mark Bennet
Jul 28 at 9:27