Group $mathbb Q^*$ as direct product/sum The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The group $mathbbQ^times / left( mathbbQ^times right)^2$$operatornameAut(mathbb Q^*)$ =?Multiplicative group of the fraction field of a UFDShow that $mathbbQ^+/mathbbZ^+$ cannot be decomposed into the direct sum of cyclic groups.$mathbbZ^2/langle(x, y)rangle$ as a direct sum of cyclic groupDihedral group as a direct productIs any subgroup of a direct product isomorphic to a direct product of subgroups?Betti numbers and Fundamental theorem of finitely generated abelian groupsDirect sum and direct product of infinitely many abelian groups are not isomorphicFinitely generated abelian group with certain propertiesWhen is a centerless group characteristic in direct product with $mathbbZ^n$?infinite abelian group where all elements have order 1, 2, or 4$operatornameAut(mathbb Q^*)$ =?
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Group $mathbb Q^*$ as direct product/sum
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)The group $mathbbQ^times / left( mathbbQ^times right)^2$$operatornameAut(mathbb Q^*)$ =?Multiplicative group of the fraction field of a UFDShow that $mathbbQ^+/mathbbZ^+$ cannot be decomposed into the direct sum of cyclic groups.$mathbbZ^2/langle(x, y)rangle$ as a direct sum of cyclic groupDihedral group as a direct productIs any subgroup of a direct product isomorphic to a direct product of subgroups?Betti numbers and Fundamental theorem of finitely generated abelian groupsDirect sum and direct product of infinitely many abelian groups are not isomorphicFinitely generated abelian group with certain propertiesWhen is a centerless group characteristic in direct product with $mathbbZ^n$?infinite abelian group where all elements have order 1, 2, or 4$operatornameAut(mathbb Q^*)$ =?
$begingroup$
Is the group $mathbb Q^*$ (rationals without $0$ under multiplication) a direct product or a direct sum of nontrivial subgroups?
My thoughts:
Consider subgroups $langle prangle=p^kmid kin mathbb Z$ generated by a positive prime $p$ and $langle -1rangle=-1,1$.
They are normal (because $mathbb Q^*$ is abelian), intersects in $1$ and any $qin mathbb Q^*$ is uniquely written as quotient of primes' powers (finitely many).
So, I think $mathbb Q^*cong langle -1rangletimes bigoplus_plangle prangle,$ where $bigoplus$ is the direct sum.
And simply we can write $mathbb Q^*cong Bbb Z_2times bigoplus_i=1^infty Bbb Z$.
Am I right?
abstract-algebra group-theory proof-verification abelian-groups rational-numbers
edited Mar 25 at 7:51
Andrews
1,2962423
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
$endgroup$
add a comment |
$begingroup$
Is the group $mathbb Q^*$ (rationals without $0$ under multiplication) a direct product or a direct sum of nontrivial subgroups?
My thoughts:
Consider subgroups $langle prangle=p^kmid kin mathbb Z$ generated by a positive prime $p$ and $langle -1rangle=-1,1$.
They are normal (because $mathbb Q^*$ is abelian), intersects in $1$ and any $qin mathbb Q^*$ is uniquely written as quotient of primes' powers (finitely many).
So, I think $mathbb Q^*cong langle -1rangletimes bigoplus_plangle prangle,$ where $bigoplus$ is the direct sum.
And simply we can write $mathbb Q^*cong Bbb Z_2times bigoplus_i=1^infty Bbb Z$.
Am I right?
abstract-algebra group-theory proof-verification abelian-groups rational-numbers
edited Mar 25 at 7:51
Andrews
1,2962423
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
$endgroup$
4
$begingroup$
Of course, each $langle pranglecongmathbb Z$, so we might write this isomorphism using familiar groups: $$mathbb Q^*congmathbb Z_2timesbigoplus_iinmathbb Nmathbb Z$$
$endgroup$
– Karl Kronenfeld
May 10 '13 at 13:56
add a comment |
$begingroup$
Is the group $mathbb Q^*$ (rationals without $0$ under multiplication) a direct product or a direct sum of nontrivial subgroups?
My thoughts:
Consider subgroups $langle prangle=p^kmid kin mathbb Z$ generated by a positive prime $p$ and $langle -1rangle=-1,1$.
They are normal (because $mathbb Q^*$ is abelian), intersects in $1$ and any $qin mathbb Q^*$ is uniquely written as quotient of primes' powers (finitely many).
So, I think $mathbb Q^*cong langle -1rangletimes bigoplus_plangle prangle,$ where $bigoplus$ is the direct sum.
And simply we can write $mathbb Q^*cong Bbb Z_2times bigoplus_i=1^infty Bbb Z$.
Am I right?
abstract-algebra group-theory proof-verification abelian-groups rational-numbers
edited Mar 25 at 7:51
Andrews
1,2962423
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
$endgroup$
Is the group $mathbb Q^*$ (rationals without $0$ under multiplication) a direct product or a direct sum of nontrivial subgroups?
My thoughts:
Consider subgroups $langle prangle=p^kmid kin mathbb Z$ generated by a positive prime $p$ and $langle -1rangle=-1,1$.
They are normal (because $mathbb Q^*$ is abelian), intersects in $1$ and any $qin mathbb Q^*$ is uniquely written as quotient of primes' powers (finitely many).
So, I think $mathbb Q^*cong langle -1rangletimes bigoplus_plangle prangle,$ where $bigoplus$ is the direct sum.
And simply we can write $mathbb Q^*cong Bbb Z_2times bigoplus_i=1^infty Bbb Z$.
Am I right?
abstract-algebra group-theory proof-verification abelian-groups rational-numbers
abstract-algebra group-theory proof-verification abelian-groups rational-numbers
edited Mar 25 at 7:51
Andrews
1,2962423
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
edited Mar 25 at 7:51
Andrews
1,2962423
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
edited Mar 25 at 7:51
Andrews
1,2962423
edited Mar 25 at 7:51
Andrews
1,2962423
edited Mar 25 at 7:51
Andrews
1,2962423
1,2962423
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
asked May 10 '13 at 13:36
Canis LupusCanis Lupus
1,31611324
1,31611324
4
$begingroup$
Of course, each $langle pranglecongmathbb Z$, so we might write this isomorphism using familiar groups: $$mathbb Q^*congmathbb Z_2timesbigoplus_iinmathbb Nmathbb Z$$
$endgroup$
– Karl Kronenfeld
May 10 '13 at 13:56
add a comment |
4
$begingroup$
Of course, each $langle pranglecongmathbb Z$, so we might write this isomorphism using familiar groups: $$mathbb Q^*congmathbb Z_2timesbigoplus_iinmathbb Nmathbb Z$$
$endgroup$
– Karl Kronenfeld
May 10 '13 at 13:56
4
4
$begingroup$
Of course, each $langle pranglecongmathbb Z$, so we might write this isomorphism using familiar groups: $$mathbb Q^*congmathbb Z_2timesbigoplus_iinmathbb Nmathbb Z$$
$endgroup$
– Karl Kronenfeld
May 10 '13 at 13:56
$begingroup$
Of course, each $langle pranglecongmathbb Z$, so we might write this isomorphism using familiar groups: $$mathbb Q^*congmathbb Z_2timesbigoplus_iinmathbb Nmathbb Z$$
$endgroup$
– Karl Kronenfeld
May 10 '13 at 13:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, you're right.
Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?
In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form
$K^* cong newcommandZmathbbZ$
$Z/nZ oplus bigoplus_i=1^infty Z.$
Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
$endgroup$
add a comment |
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$begingroup$
Yes, you're right.
Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?
In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form
$K^* cong newcommandZmathbbZ$
$Z/nZ oplus bigoplus_i=1^infty Z.$
Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
$endgroup$
add a comment |
$begingroup$
Yes, you're right.
Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?
In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form
$K^* cong newcommandZmathbbZ$
$Z/nZ oplus bigoplus_i=1^infty Z.$
Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
$endgroup$
add a comment |
$begingroup$
Yes, you're right.
Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?
In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form
$K^* cong newcommandZmathbbZ$
$Z/nZ oplus bigoplus_i=1^infty Z.$
Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
$endgroup$
Yes, you're right.
Your statement can be generalized to the multiplicative group $K^*$ of the fraction field $K$ of a unique factorization domain $R$. Can you see how?
In fact, if I'm not mistaken it follows from this that for any number field $K$, the group $K^*$ is the product of a finite cyclic group (the group of roots of unity in $K$) with a free abelian group of countable rank, so of the form
$K^* cong newcommandZmathbbZ$
$Z/nZ oplus bigoplus_i=1^infty Z.$
Here it is not enough to take the most obvious choice of $R$, namely the full ring of integers in $K$, because this might not be a UFD. But one can always choose an $S$-integer ring (obtained from $R$ by inverting finitely many prime ideals) with this property and then apply Dirichlet's S-Unit Theorem.
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
edited May 10 '13 at 14:40
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
answered May 10 '13 at 13:52
Pete L. ClarkPete L. Clark
81.1k9162314
81.1k9162314
add a comment |
add a comment |
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$begingroup$
Of course, each $langle pranglecongmathbb Z$, so we might write this isomorphism using familiar groups: $$mathbb Q^*congmathbb Z_2timesbigoplus_iinmathbb Nmathbb Z$$
$endgroup$
– Karl Kronenfeld
May 10 '13 at 13:56