Mixing
Torsion-free abelian groups of finite rank and a subgroup of finite index (Fuchs' problem) - self study
$begingroup$
I'm trying to solve the following exercise (Fuchs, "Infinite Abelian Groups", Vol. $2$, pp. $153$, Ex.$5$):
Let $A$ be a torsion-free abelian group of finite rank.
If $phi$ is an isomorphism of $A$ into itself, then $phi A$ is of finite index in $A$.
[Hint: if $F$ is a free subgroup of $A$ of the same rank as $A$, then $Fcap phi F$ is of finite index in $F$; apply Ex. $4$ to $phi A / (Fcap phi F)le A/(Fcap phi F)$, noting that $A/Fcong phi A/phi F$.]
In order to get some intuition, I've considered the finitely generated free group $F=oplus_{i=1}^nlangle x_iranglecong oplus_{i=1}^n mathbb{Z}$.
If $phi$ is a monomorphism of A into intself, then $Fcap phi Fcongoplus_{i=1}^m mathbb{Z}$ for some $mle n$ since every subgroup of a free group is again free.
But now $F/(Fcapphi F)congmathbb{Z}^n/mathbb{Z}^mcong mathbb{Z}^{n-m}$, which doesn't have finite index if $m < n$.
What am I missing?
Thanks in advance for your help.
abstract-algebra group-theory abelian-groups infinite-groups
edited Feb 17 at 18:49
user26857
39.5k124283
asked Jan 29 at 19:50
LBJFSLBJFS
337111
$endgroup$
|
show 2 more comments
$begingroup$
I'm trying to solve the following exercise (Fuchs, "Infinite Abelian Groups", Vol. $2$, pp. $153$, Ex.$5$):
Let $A$ be a torsion-free abelian group of finite rank.
If $phi$ is an isomorphism of $A$ into itself, then $phi A$ is of finite index in $A$.
[Hint: if $F$ is a free subgroup of $A$ of the same rank as $A$, then $Fcap phi F$ is of finite index in $F$; apply Ex. $4$ to $phi A / (Fcap phi F)le A/(Fcap phi F)$, noting that $A/Fcong phi A/phi F$.]
In order to get some intuition, I've considered the finitely generated free group $F=oplus_{i=1}^nlangle x_iranglecong oplus_{i=1}^n mathbb{Z}$.
If $phi$ is a monomorphism of A into intself, then $Fcap phi Fcongoplus_{i=1}^m mathbb{Z}$ for some $mle n$ since every subgroup of a free group is again free.
But now $F/(Fcapphi F)congmathbb{Z}^n/mathbb{Z}^mcong mathbb{Z}^{n-m}$, which doesn't have finite index if $m < n$.
What am I missing?
Thanks in advance for your help.
abstract-algebra group-theory abelian-groups infinite-groups
edited Feb 17 at 18:49
user26857
39.5k124283
asked Jan 29 at 19:50
LBJFSLBJFS
337111
$endgroup$
2
$begingroup$
First, note that $Fcapphi F$ is not equal to a sum of copies of $mathbb{Z}$, it is isomorphic to such a sum (the map $mathbb{Z}tomathbb{Z}$ sending $1$ to $2$ has image isomorphic to $mathbb{Z}$, but not equal to $mathbb{Z}$). Second: what you are noting is that the intersection cannot by free of rank less than $n$.
$endgroup$
– Arturo Magidin
Jan 29 at 20:39
1
$begingroup$
Tangential to your question, but I would warn you that referring to an "isomorphism of $A$ into itself" is very old-fashioned and is not standard terminology anymore (it is now nearly universal that isomorphisms must be surjective).
$endgroup$
– Eric Wofsey
Jan 29 at 20:54
$begingroup$
@ArturoMagidin I think I've got your suggestion, but I still don't see how to prove the statement :-(
$endgroup$
– LBJFS
Jan 29 at 21:00
$begingroup$
@EricWofsey Thanks :-)
$endgroup$
– LBJFS
Jan 29 at 21:01
$begingroup$
If $Fcapphi F$ were of infinite index, then you would be able to take a basis for $F$, and add an element $x$ with $langle xranglecapphi F$ trivial, which would give you a subgroup of $F$ of rank strictly larger than that of $F$...
$endgroup$
– Arturo Magidin
Jan 29 at 21:18
|
show 2 more comments
$begingroup$
I'm trying to solve the following exercise (Fuchs, "Infinite Abelian Groups", Vol. $2$, pp. $153$, Ex.$5$):
Let $A$ be a torsion-free abelian group of finite rank.
If $phi$ is an isomorphism of $A$ into itself, then $phi A$ is of finite index in $A$.
[Hint: if $F$ is a free subgroup of $A$ of the same rank as $A$, then $Fcap phi F$ is of finite index in $F$; apply Ex. $4$ to $phi A / (Fcap phi F)le A/(Fcap phi F)$, noting that $A/Fcong phi A/phi F$.]
In order to get some intuition, I've considered the finitely generated free group $F=oplus_{i=1}^nlangle x_iranglecong oplus_{i=1}^n mathbb{Z}$.
If $phi$ is a monomorphism of A into intself, then $Fcap phi Fcongoplus_{i=1}^m mathbb{Z}$ for some $mle n$ since every subgroup of a free group is again free.
But now $F/(Fcapphi F)congmathbb{Z}^n/mathbb{Z}^mcong mathbb{Z}^{n-m}$, which doesn't have finite index if $m < n$.
What am I missing?
Thanks in advance for your help.
abstract-algebra group-theory abelian-groups infinite-groups
edited Feb 17 at 18:49
user26857
39.5k124283
asked Jan 29 at 19:50
LBJFSLBJFS
337111
$endgroup$
I'm trying to solve the following exercise (Fuchs, "Infinite Abelian Groups", Vol. $2$, pp. $153$, Ex.$5$):
Let $A$ be a torsion-free abelian group of finite rank.
If $phi$ is an isomorphism of $A$ into itself, then $phi A$ is of finite index in $A$.
[Hint: if $F$ is a free subgroup of $A$ of the same rank as $A$, then $Fcap phi F$ is of finite index in $F$; apply Ex. $4$ to $phi A / (Fcap phi F)le A/(Fcap phi F)$, noting that $A/Fcong phi A/phi F$.]
In order to get some intuition, I've considered the finitely generated free group $F=oplus_{i=1}^nlangle x_iranglecong oplus_{i=1}^n mathbb{Z}$.
If $phi$ is a monomorphism of A into intself, then $Fcap phi Fcongoplus_{i=1}^m mathbb{Z}$ for some $mle n$ since every subgroup of a free group is again free.
But now $F/(Fcapphi F)congmathbb{Z}^n/mathbb{Z}^mcong mathbb{Z}^{n-m}$, which doesn't have finite index if $m < n$.
What am I missing?
Thanks in advance for your help.
abstract-algebra group-theory abelian-groups infinite-groups
abstract-algebra group-theory abelian-groups infinite-groups
edited Feb 17 at 18:49
user26857
39.5k124283
asked Jan 29 at 19:50
LBJFSLBJFS
337111
edited Feb 17 at 18:49
user26857
39.5k124283
asked Jan 29 at 19:50
LBJFSLBJFS
337111
edited Feb 17 at 18:49
user26857
39.5k124283
edited Feb 17 at 18:49
user26857
39.5k124283
edited Feb 17 at 18:49
user26857
39.5k124283
39.5k124283
asked Jan 29 at 19:50
LBJFSLBJFS
337111
asked Jan 29 at 19:50
LBJFSLBJFS
337111
asked Jan 29 at 19:50
LBJFSLBJFS
337111
337111
2
$begingroup$
First, note that $Fcapphi F$ is not equal to a sum of copies of $mathbb{Z}$, it is isomorphic to such a sum (the map $mathbb{Z}tomathbb{Z}$ sending $1$ to $2$ has image isomorphic to $mathbb{Z}$, but not equal to $mathbb{Z}$). Second: what you are noting is that the intersection cannot by free of rank less than $n$.
$endgroup$
– Arturo Magidin
Jan 29 at 20:39
1
$begingroup$
Tangential to your question, but I would warn you that referring to an "isomorphism of $A$ into itself" is very old-fashioned and is not standard terminology anymore (it is now nearly universal that isomorphisms must be surjective).
$endgroup$
– Eric Wofsey
Jan 29 at 20:54
$begingroup$
@ArturoMagidin I think I've got your suggestion, but I still don't see how to prove the statement :-(
$endgroup$
– LBJFS
Jan 29 at 21:00
$begingroup$
@EricWofsey Thanks :-)
$endgroup$
– LBJFS
Jan 29 at 21:01
$begingroup$
If $Fcapphi F$ were of infinite index, then you would be able to take a basis for $F$, and add an element $x$ with $langle xranglecapphi F$ trivial, which would give you a subgroup of $F$ of rank strictly larger than that of $F$...
$endgroup$
– Arturo Magidin
Jan 29 at 21:18
|
show 2 more comments
2
$begingroup$
First, note that $Fcapphi F$ is not equal to a sum of copies of $mathbb{Z}$, it is isomorphic to such a sum (the map $mathbb{Z}tomathbb{Z}$ sending $1$ to $2$ has image isomorphic to $mathbb{Z}$, but not equal to $mathbb{Z}$). Second: what you are noting is that the intersection cannot by free of rank less than $n$.
$endgroup$
– Arturo Magidin
Jan 29 at 20:39
1
$begingroup$
Tangential to your question, but I would warn you that referring to an "isomorphism of $A$ into itself" is very old-fashioned and is not standard terminology anymore (it is now nearly universal that isomorphisms must be surjective).
$endgroup$
– Eric Wofsey
Jan 29 at 20:54
$begingroup$
@ArturoMagidin I think I've got your suggestion, but I still don't see how to prove the statement :-(
$endgroup$
– LBJFS
Jan 29 at 21:00
$begingroup$
@EricWofsey Thanks :-)
$endgroup$
– LBJFS
Jan 29 at 21:01
$begingroup$
If $Fcapphi F$ were of infinite index, then you would be able to take a basis for $F$, and add an element $x$ with $langle xranglecapphi F$ trivial, which would give you a subgroup of $F$ of rank strictly larger than that of $F$...
$endgroup$
– Arturo Magidin
Jan 29 at 21:18
2
2
$begingroup$
First, note that $Fcapphi F$ is not equal to a sum of copies of $mathbb{Z}$, it is isomorphic to such a sum (the map $mathbb{Z}tomathbb{Z}$ sending $1$ to $2$ has image isomorphic to $mathbb{Z}$, but not equal to $mathbb{Z}$). Second: what you are noting is that the intersection cannot by free of rank less than $n$.
$endgroup$
– Arturo Magidin
Jan 29 at 20:39
$begingroup$
First, note that $Fcapphi F$ is not equal to a sum of copies of $mathbb{Z}$, it is isomorphic to such a sum (the map $mathbb{Z}tomathbb{Z}$ sending $1$ to $2$ has image isomorphic to $mathbb{Z}$, but not equal to $mathbb{Z}$). Second: what you are noting is that the intersection cannot by free of rank less than $n$.
$endgroup$
– Arturo Magidin
Jan 29 at 20:39
1
1
$begingroup$
Tangential to your question, but I would warn you that referring to an "isomorphism of $A$ into itself" is very old-fashioned and is not standard terminology anymore (it is now nearly universal that isomorphisms must be surjective).
$endgroup$
– Eric Wofsey
Jan 29 at 20:54
$begingroup$
Tangential to your question, but I would warn you that referring to an "isomorphism of $A$ into itself" is very old-fashioned and is not standard terminology anymore (it is now nearly universal that isomorphisms must be surjective).
$endgroup$
– Eric Wofsey
Jan 29 at 20:54
$begingroup$
@ArturoMagidin I think I've got your suggestion, but I still don't see how to prove the statement :-(
$endgroup$
– LBJFS
Jan 29 at 21:00
$begingroup$
@ArturoMagidin I think I've got your suggestion, but I still don't see how to prove the statement :-(
$endgroup$
– LBJFS
Jan 29 at 21:00
$begingroup$
@EricWofsey Thanks :-)
$endgroup$
– LBJFS
Jan 29 at 21:01
$begingroup$
@EricWofsey Thanks :-)
$endgroup$
– LBJFS
Jan 29 at 21:01
$begingroup$
If $Fcapphi F$ were of infinite index, then you would be able to take a basis for $F$, and add an element $x$ with $langle xranglecapphi F$ trivial, which would give you a subgroup of $F$ of rank strictly larger than that of $F$...
$endgroup$
– Arturo Magidin
Jan 29 at 21:18
$begingroup$
If $Fcapphi F$ were of infinite index, then you would be able to take a basis for $F$, and add an element $x$ with $langle xranglecapphi F$ trivial, which would give you a subgroup of $F$ of rank strictly larger than that of $F$...
$endgroup$
– Arturo Magidin
Jan 29 at 21:18
|
show 2 more comments
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$begingroup$
First, note that $Fcapphi F$ is not equal to a sum of copies of $mathbb{Z}$, it is isomorphic to such a sum (the map $mathbb{Z}tomathbb{Z}$ sending $1$ to $2$ has image isomorphic to $mathbb{Z}$, but not equal to $mathbb{Z}$). Second: what you are noting is that the intersection cannot by free of rank less than $n$.
$endgroup$
– Arturo Magidin
Jan 29 at 20:39
1
$begingroup$
Tangential to your question, but I would warn you that referring to an "isomorphism of $A$ into itself" is very old-fashioned and is not standard terminology anymore (it is now nearly universal that isomorphisms must be surjective).
$endgroup$
– Eric Wofsey
Jan 29 at 20:54
$begingroup$
@ArturoMagidin I think I've got your suggestion, but I still don't see how to prove the statement :-(
$endgroup$
– LBJFS
Jan 29 at 21:00
$begingroup$
@EricWofsey Thanks :-)
$endgroup$
– LBJFS
Jan 29 at 21:01
$begingroup$
If $Fcapphi F$ were of infinite index, then you would be able to take a basis for $F$, and add an element $x$ with $langle xranglecapphi F$ trivial, which would give you a subgroup of $F$ of rank strictly larger than that of $F$...
$endgroup$
– Arturo Magidin
Jan 29 at 21:18