Mixing
Is there a simple abelian group $G$ with infinite order?
$begingroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem in this book:
On p.80
Problem 8:
Show that a simple abelian group $G neq {e}$ is a cyclic group whose order is prime.
Did the author intend the following problem?
If this problem were the following, I could solve it:
Problem 8':
Show that a simple finite abelian group $G neq {e}$ is a cyclic group whose order is prime.
Proof:
Because $G neq {e}$, there is an element $g in G$ which is not equal to $e$.
$H := {g^i | i in mathbb{Z}}$ is a subgroup of $G$ and $G$ is abelian.
So $H$ is a normal subgroup of $G$.
And $G$ is simple.
And $H ni g ne e$.
So $H = G$.
Let $n := #H = #G$.
Then, $n$ is prime.
If $n$ is not prime, then we can write $n = d d'$, $1 < d < n$, $1 < d' < n$.
Then $H' := {(g^d)^i | i in mathbb{Z}}$ is a subgroup of $G$ whose order is $d'$.
So, $H'$ is neither ${e}$ nor $G$.
Becasue $G$ is abelian, $H'$ is a normal subgroup of $G$.
And $G$ is simple.
This is a contradiction.
abstract-algebra group-theory simple-groups
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
$endgroup$
add a comment |
$begingroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem in this book:
On p.80
Problem 8:
Show that a simple abelian group $G neq {e}$ is a cyclic group whose order is prime.
Did the author intend the following problem?
If this problem were the following, I could solve it:
Problem 8':
Show that a simple finite abelian group $G neq {e}$ is a cyclic group whose order is prime.
Proof:
Because $G neq {e}$, there is an element $g in G$ which is not equal to $e$.
$H := {g^i | i in mathbb{Z}}$ is a subgroup of $G$ and $G$ is abelian.
So $H$ is a normal subgroup of $G$.
And $G$ is simple.
And $H ni g ne e$.
So $H = G$.
Let $n := #H = #G$.
Then, $n$ is prime.
If $n$ is not prime, then we can write $n = d d'$, $1 < d < n$, $1 < d' < n$.
Then $H' := {(g^d)^i | i in mathbb{Z}}$ is a subgroup of $G$ whose order is $d'$.
So, $H'$ is neither ${e}$ nor $G$.
Becasue $G$ is abelian, $H'$ is a normal subgroup of $G$.
And $G$ is simple.
This is a contradiction.
abstract-algebra group-theory simple-groups
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
$endgroup$
4
$begingroup$
No, the problem as stated in the book, is correct. You are supposed to prove that any simple abelian group is finite. It is not difficult, so you should try and do it yourself.
$endgroup$
– Derek Holt
Feb 1 at 9:20
$begingroup$
Thank you very very much, for your answer. Derek Holt. I will try.
$endgroup$
– tchappy ha
Feb 1 at 9:21
add a comment |
$begingroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem in this book:
On p.80
Problem 8:
Show that a simple abelian group $G neq {e}$ is a cyclic group whose order is prime.
Did the author intend the following problem?
If this problem were the following, I could solve it:
Problem 8':
Show that a simple finite abelian group $G neq {e}$ is a cyclic group whose order is prime.
Proof:
Because $G neq {e}$, there is an element $g in G$ which is not equal to $e$.
$H := {g^i | i in mathbb{Z}}$ is a subgroup of $G$ and $G$ is abelian.
So $H$ is a normal subgroup of $G$.
And $G$ is simple.
And $H ni g ne e$.
So $H = G$.
Let $n := #H = #G$.
Then, $n$ is prime.
If $n$ is not prime, then we can write $n = d d'$, $1 < d < n$, $1 < d' < n$.
Then $H' := {(g^d)^i | i in mathbb{Z}}$ is a subgroup of $G$ whose order is $d'$.
So, $H'$ is neither ${e}$ nor $G$.
Becasue $G$ is abelian, $H'$ is a normal subgroup of $G$.
And $G$ is simple.
This is a contradiction.
abstract-algebra group-theory simple-groups
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
$endgroup$
I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.
There is the following problem in this book:
On p.80
Problem 8:
Show that a simple abelian group $G neq {e}$ is a cyclic group whose order is prime.
Did the author intend the following problem?
If this problem were the following, I could solve it:
Problem 8':
Show that a simple finite abelian group $G neq {e}$ is a cyclic group whose order is prime.
Proof:
Because $G neq {e}$, there is an element $g in G$ which is not equal to $e$.
$H := {g^i | i in mathbb{Z}}$ is a subgroup of $G$ and $G$ is abelian.
So $H$ is a normal subgroup of $G$.
And $G$ is simple.
And $H ni g ne e$.
So $H = G$.
Let $n := #H = #G$.
Then, $n$ is prime.
If $n$ is not prime, then we can write $n = d d'$, $1 < d < n$, $1 < d' < n$.
Then $H' := {(g^d)^i | i in mathbb{Z}}$ is a subgroup of $G$ whose order is $d'$.
So, $H'$ is neither ${e}$ nor $G$.
Becasue $G$ is abelian, $H'$ is a normal subgroup of $G$.
And $G$ is simple.
This is a contradiction.
abstract-algebra group-theory simple-groups
abstract-algebra group-theory simple-groups
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
edited Feb 1 at 9:22
tchappy ha
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
asked Feb 1 at 9:18
tchappy hatchappy ha
783412
783412
4
$begingroup$
No, the problem as stated in the book, is correct. You are supposed to prove that any simple abelian group is finite. It is not difficult, so you should try and do it yourself.
$endgroup$
– Derek Holt
Feb 1 at 9:20
$begingroup$
Thank you very very much, for your answer. Derek Holt. I will try.
$endgroup$
– tchappy ha
Feb 1 at 9:21
add a comment |
4
$begingroup$
No, the problem as stated in the book, is correct. You are supposed to prove that any simple abelian group is finite. It is not difficult, so you should try and do it yourself.
$endgroup$
– Derek Holt
Feb 1 at 9:20
$begingroup$
Thank you very very much, for your answer. Derek Holt. I will try.
$endgroup$
– tchappy ha
Feb 1 at 9:21
4
4
$begingroup$
No, the problem as stated in the book, is correct. You are supposed to prove that any simple abelian group is finite. It is not difficult, so you should try and do it yourself.
$endgroup$
– Derek Holt
Feb 1 at 9:20
$begingroup$
No, the problem as stated in the book, is correct. You are supposed to prove that any simple abelian group is finite. It is not difficult, so you should try and do it yourself.
$endgroup$
– Derek Holt
Feb 1 at 9:20
$begingroup$
Thank you very very much, for your answer. Derek Holt. I will try.
$endgroup$
– tchappy ha
Feb 1 at 9:21
$begingroup$
Thank you very very much, for your answer. Derek Holt. I will try.
$endgroup$
– tchappy ha
Feb 1 at 9:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Suppose that $G$ is simple Abelian, if $x$ in $G$ has an infinite order the group generate by $2x$ is a proper normal subgroup since it does not contain $x$.
If $x$ has a finite order, there exists $y$ not in $H$ the group generated by $x$, and $H$ is normal and proper.
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
$begingroup$
Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x in G-{0}$. Then, $G = {i x | i in mathbb{Z}}$ because $G$ is simple. So $x$ has an infinite order. ${i 2 x | i in mathbb{Z}}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction.
$endgroup$
– tchappy ha
Feb 1 at 10:26
$begingroup$
Sorry, I understand your answer now.
$endgroup$
– tchappy ha
Feb 1 at 10:32
2
$begingroup$
Suppose that $G$ is simple infinite Abelian. If $x in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide.
$endgroup$
– tchappy ha
Feb 1 at 10:37
add a comment |
$begingroup$
Since $G$ is abelian, all the subgroups are normal. So $G$ simple abelian means there are no proper subgroups.
So take $xneq e in G$.
Claim: $langle xrangle =G$.
If not, $langle xrangle $ is a proper normal subgroup. Contradiction.
So $G$ is cyclic, and contains no proper subgroups.
Claim: $mid Gmid= p$, where $p$ is prime.
Suppose not. Then either $mid Gmid=n$, where $n=rs$ for some $r,sneq1$, or $G$ is infinite. Take $xneq e$. In the first case, $langle x^rrangle le G$ is proper. Contradiction. In the second case, $langle x^2ranglele G $ is proper (this requires a little argument which I will leave to you). Also a contradiction.
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
Thank you very much, for your answer, Chris Custer.
$endgroup$
– tchappy ha
Feb 1 at 11:28
add a comment |
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2 Answers
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$begingroup$
Suppose that $G$ is simple Abelian, if $x$ in $G$ has an infinite order the group generate by $2x$ is a proper normal subgroup since it does not contain $x$.
If $x$ has a finite order, there exists $y$ not in $H$ the group generated by $x$, and $H$ is normal and proper.
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
$begingroup$
Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x in G-{0}$. Then, $G = {i x | i in mathbb{Z}}$ because $G$ is simple. So $x$ has an infinite order. ${i 2 x | i in mathbb{Z}}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction.
$endgroup$
– tchappy ha
Feb 1 at 10:26
$begingroup$
Sorry, I understand your answer now.
$endgroup$
– tchappy ha
Feb 1 at 10:32
2
$begingroup$
Suppose that $G$ is simple infinite Abelian. If $x in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide.
$endgroup$
– tchappy ha
Feb 1 at 10:37
add a comment |
$begingroup$
Suppose that $G$ is simple Abelian, if $x$ in $G$ has an infinite order the group generate by $2x$ is a proper normal subgroup since it does not contain $x$.
If $x$ has a finite order, there exists $y$ not in $H$ the group generated by $x$, and $H$ is normal and proper.
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
$begingroup$
Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x in G-{0}$. Then, $G = {i x | i in mathbb{Z}}$ because $G$ is simple. So $x$ has an infinite order. ${i 2 x | i in mathbb{Z}}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction.
$endgroup$
– tchappy ha
Feb 1 at 10:26
$begingroup$
Sorry, I understand your answer now.
$endgroup$
– tchappy ha
Feb 1 at 10:32
2
$begingroup$
Suppose that $G$ is simple infinite Abelian. If $x in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide.
$endgroup$
– tchappy ha
Feb 1 at 10:37
add a comment |
$begingroup$
Suppose that $G$ is simple Abelian, if $x$ in $G$ has an infinite order the group generate by $2x$ is a proper normal subgroup since it does not contain $x$.
If $x$ has a finite order, there exists $y$ not in $H$ the group generated by $x$, and $H$ is normal and proper.
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
Suppose that $G$ is simple Abelian, if $x$ in $G$ has an infinite order the group generate by $2x$ is a proper normal subgroup since it does not contain $x$.
If $x$ has a finite order, there exists $y$ not in $H$ the group generated by $x$, and $H$ is normal and proper.
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
answered Feb 1 at 9:26
Tsemo AristideTsemo Aristide
60.4k11446
60.4k11446
$begingroup$
Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x in G-{0}$. Then, $G = {i x | i in mathbb{Z}}$ because $G$ is simple. So $x$ has an infinite order. ${i 2 x | i in mathbb{Z}}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction.
$endgroup$
– tchappy ha
Feb 1 at 10:26
$begingroup$
Sorry, I understand your answer now.
$endgroup$
– tchappy ha
Feb 1 at 10:32
2
$begingroup$
Suppose that $G$ is simple infinite Abelian. If $x in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide.
$endgroup$
– tchappy ha
Feb 1 at 10:37
add a comment |
$begingroup$
Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x in G-{0}$. Then, $G = {i x | i in mathbb{Z}}$ because $G$ is simple. So $x$ has an infinite order. ${i 2 x | i in mathbb{Z}}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction.
$endgroup$
– tchappy ha
Feb 1 at 10:26
$begingroup$
Sorry, I understand your answer now.
$endgroup$
– tchappy ha
Feb 1 at 10:32
2
$begingroup$
Suppose that $G$ is simple infinite Abelian. If $x in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide.
$endgroup$
– tchappy ha
Feb 1 at 10:37
$begingroup$
Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x in G-{0}$. Then, $G = {i x | i in mathbb{Z}}$ because $G$ is simple. So $x$ has an infinite order. ${i 2 x | i in mathbb{Z}}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction.
$endgroup$
– tchappy ha
Feb 1 at 10:26
$begingroup$
Thank you very much, Tsemo Aristide. But I cannot understand your answer. Suppose $G$ is simple infinite Abelian and $x in G-{0}$. Then, $G = {i x | i in mathbb{Z}}$ because $G$ is simple. So $x$ has an infinite order. ${i 2 x | i in mathbb{Z}}$ is a proper normal subgroup of $G$ since it does not contain $x$. But $G$ is simple. This is a contradiction.
$endgroup$
– tchappy ha
Feb 1 at 10:26
$begingroup$
Sorry, I understand your answer now.
$endgroup$
– tchappy ha
Feb 1 at 10:32
$begingroup$
Sorry, I understand your answer now.
$endgroup$
– tchappy ha
Feb 1 at 10:32
2
2
$begingroup$
Suppose that $G$ is simple infinite Abelian. If $x in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide.
$endgroup$
– tchappy ha
Feb 1 at 10:37
$begingroup$
Suppose that $G$ is simple infinite Abelian. If $x in G$, $x$ has an infinite order or a finite order. If $x$ has an infinite order, then the group generated by $2x$ is a proper normal subgroup since it does not contain $x$. This is a contradiction. If $x$ has a finite order, then there exists $y$ not in the group generated by $x$, because $G$ is infinite. So $H$ is normal and proper. This is a contradiction. Thank you very very much, Tsemo Aristide.
$endgroup$
– tchappy ha
Feb 1 at 10:37
add a comment |
$begingroup$
Since $G$ is abelian, all the subgroups are normal. So $G$ simple abelian means there are no proper subgroups.
So take $xneq e in G$.
Claim: $langle xrangle =G$.
If not, $langle xrangle $ is a proper normal subgroup. Contradiction.
So $G$ is cyclic, and contains no proper subgroups.
Claim: $mid Gmid= p$, where $p$ is prime.
Suppose not. Then either $mid Gmid=n$, where $n=rs$ for some $r,sneq1$, or $G$ is infinite. Take $xneq e$. In the first case, $langle x^rrangle le G$ is proper. Contradiction. In the second case, $langle x^2ranglele G $ is proper (this requires a little argument which I will leave to you). Also a contradiction.
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
Thank you very much, for your answer, Chris Custer.
$endgroup$
– tchappy ha
Feb 1 at 11:28
add a comment |
$begingroup$
Since $G$ is abelian, all the subgroups are normal. So $G$ simple abelian means there are no proper subgroups.
So take $xneq e in G$.
Claim: $langle xrangle =G$.
If not, $langle xrangle $ is a proper normal subgroup. Contradiction.
So $G$ is cyclic, and contains no proper subgroups.
Claim: $mid Gmid= p$, where $p$ is prime.
Suppose not. Then either $mid Gmid=n$, where $n=rs$ for some $r,sneq1$, or $G$ is infinite. Take $xneq e$. In the first case, $langle x^rrangle le G$ is proper. Contradiction. In the second case, $langle x^2ranglele G $ is proper (this requires a little argument which I will leave to you). Also a contradiction.
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
$endgroup$
$begingroup$
Thank you very much, for your answer, Chris Custer.
$endgroup$
– tchappy ha
Feb 1 at 11:28
add a comment |
$begingroup$
Since $G$ is abelian, all the subgroups are normal. So $G$ simple abelian means there are no proper subgroups.
So take $xneq e in G$.
Claim: $langle xrangle =G$.
If not, $langle xrangle $ is a proper normal subgroup. Contradiction.
So $G$ is cyclic, and contains no proper subgroups.
Claim: $mid Gmid= p$, where $p$ is prime.
Suppose not. Then either $mid Gmid=n$, where $n=rs$ for some $r,sneq1$, or $G$ is infinite. Take $xneq e$. In the first case, $langle x^rrangle le G$ is proper. Contradiction. In the second case, $langle x^2ranglele G $ is proper (this requires a little argument which I will leave to you). Also a contradiction.
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
$endgroup$
Since $G$ is abelian, all the subgroups are normal. So $G$ simple abelian means there are no proper subgroups.
So take $xneq e in G$.
Claim: $langle xrangle =G$.
If not, $langle xrangle $ is a proper normal subgroup. Contradiction.
So $G$ is cyclic, and contains no proper subgroups.
Claim: $mid Gmid= p$, where $p$ is prime.
Suppose not. Then either $mid Gmid=n$, where $n=rs$ for some $r,sneq1$, or $G$ is infinite. Take $xneq e$. In the first case, $langle x^rrangle le G$ is proper. Contradiction. In the second case, $langle x^2ranglele G $ is proper (this requires a little argument which I will leave to you). Also a contradiction.
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
edited Feb 1 at 11:29
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
answered Feb 1 at 10:55
Chris CusterChris Custer
14.3k3827
14.3k3827
$begingroup$
Thank you very much, for your answer, Chris Custer.
$endgroup$
– tchappy ha
Feb 1 at 11:28
add a comment |
$begingroup$
Thank you very much, for your answer, Chris Custer.
$endgroup$
– tchappy ha
Feb 1 at 11:28
$begingroup$
Thank you very much, for your answer, Chris Custer.
$endgroup$
– tchappy ha
Feb 1 at 11:28
$begingroup$
Thank you very much, for your answer, Chris Custer.
$endgroup$
– tchappy ha
Feb 1 at 11:28
add a comment |
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4
$begingroup$
No, the problem as stated in the book, is correct. You are supposed to prove that any simple abelian group is finite. It is not difficult, so you should try and do it yourself.
$endgroup$
– Derek Holt
Feb 1 at 9:20
$begingroup$
Thank you very very much, for your answer. Derek Holt. I will try.
$endgroup$
– tchappy ha
Feb 1 at 9:21