Mixing
Subgroups of free groups question
$begingroup$
I am just reading Allufi chapter 0. I have a specific question in regards to a comment that the book made.
"By Proposition 6.9, every nontrivial subgroup of $mathbb{Z}$ is in fact iso-morphic to $mathbb{Z}$. Putting this a little strangely, it says that every subgroup of the free group on one generator is free."
I am not sure I understand this. Is it saying that subgroup of free groups on one generator is isomorphic to a subgroup of free groups of one generator ? I am guessing I am right in my understanding, because the second comment it is saying beware that free groups on two generators contain subgroups isomorphic to free group of arbitrary generators.
abstract-algebra category-theory free-groups
asked Jan 29 at 21:17
NewbieNewbie
454312
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add a comment |
$begingroup$
I am just reading Allufi chapter 0. I have a specific question in regards to a comment that the book made.
"By Proposition 6.9, every nontrivial subgroup of $mathbb{Z}$ is in fact iso-morphic to $mathbb{Z}$. Putting this a little strangely, it says that every subgroup of the free group on one generator is free."
I am not sure I understand this. Is it saying that subgroup of free groups on one generator is isomorphic to a subgroup of free groups of one generator ? I am guessing I am right in my understanding, because the second comment it is saying beware that free groups on two generators contain subgroups isomorphic to free group of arbitrary generators.
abstract-algebra category-theory free-groups
asked Jan 29 at 21:17
NewbieNewbie
454312
$endgroup$
$begingroup$
As a minor note, the restatement is a slightly different claim. First, the restatement only states that the subgroups are free and not that they are isomorphic to $mathbb Z$ (which would also be false given the next point). Second, the restatement doesn't restrict to nontrivial subgroups, but that's fine because the trivial (sub)group is also a free group (on zero generators). That said, once you do establish that all subgroups of $mathbb Z$ are free, it is very easy to prove that they must be free groups generated by at most one generator.
$endgroup$
– Derek Elkins
Jan 29 at 21:47
add a comment |
$begingroup$
I am just reading Allufi chapter 0. I have a specific question in regards to a comment that the book made.
"By Proposition 6.9, every nontrivial subgroup of $mathbb{Z}$ is in fact iso-morphic to $mathbb{Z}$. Putting this a little strangely, it says that every subgroup of the free group on one generator is free."
I am not sure I understand this. Is it saying that subgroup of free groups on one generator is isomorphic to a subgroup of free groups of one generator ? I am guessing I am right in my understanding, because the second comment it is saying beware that free groups on two generators contain subgroups isomorphic to free group of arbitrary generators.
abstract-algebra category-theory free-groups
asked Jan 29 at 21:17
NewbieNewbie
454312
$endgroup$
I am just reading Allufi chapter 0. I have a specific question in regards to a comment that the book made.
"By Proposition 6.9, every nontrivial subgroup of $mathbb{Z}$ is in fact iso-morphic to $mathbb{Z}$. Putting this a little strangely, it says that every subgroup of the free group on one generator is free."
I am not sure I understand this. Is it saying that subgroup of free groups on one generator is isomorphic to a subgroup of free groups of one generator ? I am guessing I am right in my understanding, because the second comment it is saying beware that free groups on two generators contain subgroups isomorphic to free group of arbitrary generators.
abstract-algebra category-theory free-groups
abstract-algebra category-theory free-groups
asked Jan 29 at 21:17
NewbieNewbie
454312
asked Jan 29 at 21:17
NewbieNewbie
454312
asked Jan 29 at 21:17
NewbieNewbie
454312
asked Jan 29 at 21:17
NewbieNewbie
454312
asked Jan 29 at 21:17
NewbieNewbie
454312
454312
$begingroup$
As a minor note, the restatement is a slightly different claim. First, the restatement only states that the subgroups are free and not that they are isomorphic to $mathbb Z$ (which would also be false given the next point). Second, the restatement doesn't restrict to nontrivial subgroups, but that's fine because the trivial (sub)group is also a free group (on zero generators). That said, once you do establish that all subgroups of $mathbb Z$ are free, it is very easy to prove that they must be free groups generated by at most one generator.
$endgroup$
– Derek Elkins
Jan 29 at 21:47
add a comment |
$begingroup$
As a minor note, the restatement is a slightly different claim. First, the restatement only states that the subgroups are free and not that they are isomorphic to $mathbb Z$ (which would also be false given the next point). Second, the restatement doesn't restrict to nontrivial subgroups, but that's fine because the trivial (sub)group is also a free group (on zero generators). That said, once you do establish that all subgroups of $mathbb Z$ are free, it is very easy to prove that they must be free groups generated by at most one generator.
$endgroup$
– Derek Elkins
Jan 29 at 21:47
$begingroup$
As a minor note, the restatement is a slightly different claim. First, the restatement only states that the subgroups are free and not that they are isomorphic to $mathbb Z$ (which would also be false given the next point). Second, the restatement doesn't restrict to nontrivial subgroups, but that's fine because the trivial (sub)group is also a free group (on zero generators). That said, once you do establish that all subgroups of $mathbb Z$ are free, it is very easy to prove that they must be free groups generated by at most one generator.
$endgroup$
– Derek Elkins
Jan 29 at 21:47
$begingroup$
As a minor note, the restatement is a slightly different claim. First, the restatement only states that the subgroups are free and not that they are isomorphic to $mathbb Z$ (which would also be false given the next point). Second, the restatement doesn't restrict to nontrivial subgroups, but that's fine because the trivial (sub)group is also a free group (on zero generators). That said, once you do establish that all subgroups of $mathbb Z$ are free, it is very easy to prove that they must be free groups generated by at most one generator.
$endgroup$
– Derek Elkins
Jan 29 at 21:47
add a comment |
1 Answer
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Yes, you understood it right. The only free group of one generator is $mathbb{Z}$ up to isomorphism. So its non trivial subgroups are isomorphic to it. With free groups of more than one generator it becomes much more complicated.
answered Jan 29 at 21:20
MarkMark
10.4k1622
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$begingroup$
Yes, you understood it right. The only free group of one generator is $mathbb{Z}$ up to isomorphism. So its non trivial subgroups are isomorphic to it. With free groups of more than one generator it becomes much more complicated.
answered Jan 29 at 21:20
MarkMark
10.4k1622
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add a comment |
$begingroup$
Yes, you understood it right. The only free group of one generator is $mathbb{Z}$ up to isomorphism. So its non trivial subgroups are isomorphic to it. With free groups of more than one generator it becomes much more complicated.
answered Jan 29 at 21:20
MarkMark
10.4k1622
$endgroup$
add a comment |
$begingroup$
Yes, you understood it right. The only free group of one generator is $mathbb{Z}$ up to isomorphism. So its non trivial subgroups are isomorphic to it. With free groups of more than one generator it becomes much more complicated.
answered Jan 29 at 21:20
MarkMark
10.4k1622
$endgroup$
Yes, you understood it right. The only free group of one generator is $mathbb{Z}$ up to isomorphism. So its non trivial subgroups are isomorphic to it. With free groups of more than one generator it becomes much more complicated.
answered Jan 29 at 21:20
MarkMark
10.4k1622
answered Jan 29 at 21:20
MarkMark
10.4k1622
answered Jan 29 at 21:20
MarkMark
10.4k1622
answered Jan 29 at 21:20
MarkMark
10.4k1622
10.4k1622
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$begingroup$
As a minor note, the restatement is a slightly different claim. First, the restatement only states that the subgroups are free and not that they are isomorphic to $mathbb Z$ (which would also be false given the next point). Second, the restatement doesn't restrict to nontrivial subgroups, but that's fine because the trivial (sub)group is also a free group (on zero generators). That said, once you do establish that all subgroups of $mathbb Z$ are free, it is very easy to prove that they must be free groups generated by at most one generator.
$endgroup$
– Derek Elkins
Jan 29 at 21:47