Does every non-elementary subgroup of the additive group of rationals contain prime multiples of elements in...
$begingroup$
Let $H$ be a non-elementary proper subgroup of $(Bbb Q,+)$, the
additive group of rational numbers. Then there exists an element $a$
in $Bbb Qsetminus H$ such that $pcdot a$ is in $H$ for some prime
number $p$.
I have not been able to prove this. I'm looking to see if someone else out there has any advice. Anything would be appreciated.
If it turns out to be false I'd be happy to see a disproof of course.
Thank you.
group-theory
$endgroup$
|
show 1 more comment
$begingroup$
Let $H$ be a non-elementary proper subgroup of $(Bbb Q,+)$, the
additive group of rational numbers. Then there exists an element $a$
in $Bbb Qsetminus H$ such that $pcdot a$ is in $H$ for some prime
number $p$.
I have not been able to prove this. I'm looking to see if someone else out there has any advice. Anything would be appreciated.
If it turns out to be false I'd be happy to see a disproof of course.
Thank you.
group-theory
$endgroup$
$begingroup$
Where did this problem arise?
$endgroup$
– Shaun
Jan 12 at 15:12
$begingroup$
@Shaun it's in a book on group theory. The author states it without proof, then uses it to prove the additive group of rationals contains only subgroups of infinite index.
$endgroup$
– Tim Eller
Jan 12 at 15:16
$begingroup$
Okay, @TimEller; which book?
$endgroup$
– Shaun
Jan 12 at 15:16
$begingroup$
@Shaun Theory of Groups by A. G. Kurosh
$endgroup$
– Tim Eller
Jan 12 at 15:17
$begingroup$
@Shaun did you edit my post? I have a question about the edit.
$endgroup$
– Tim Eller
Jan 12 at 16:28
|
show 1 more comment
$begingroup$
Let $H$ be a non-elementary proper subgroup of $(Bbb Q,+)$, the
additive group of rational numbers. Then there exists an element $a$
in $Bbb Qsetminus H$ such that $pcdot a$ is in $H$ for some prime
number $p$.
I have not been able to prove this. I'm looking to see if someone else out there has any advice. Anything would be appreciated.
If it turns out to be false I'd be happy to see a disproof of course.
Thank you.
group-theory
$endgroup$
Let $H$ be a non-elementary proper subgroup of $(Bbb Q,+)$, the
additive group of rational numbers. Then there exists an element $a$
in $Bbb Qsetminus H$ such that $pcdot a$ is in $H$ for some prime
number $p$.
I have not been able to prove this. I'm looking to see if someone else out there has any advice. Anything would be appreciated.
If it turns out to be false I'd be happy to see a disproof of course.
Thank you.
group-theory
group-theory
edited Jan 12 at 15:04
Thomas Shelby
3,0971523
3,0971523
asked Jan 12 at 14:53
Tim EllerTim Eller
112
112
$begingroup$
Where did this problem arise?
$endgroup$
– Shaun
Jan 12 at 15:12
$begingroup$
@Shaun it's in a book on group theory. The author states it without proof, then uses it to prove the additive group of rationals contains only subgroups of infinite index.
$endgroup$
– Tim Eller
Jan 12 at 15:16
$begingroup$
Okay, @TimEller; which book?
$endgroup$
– Shaun
Jan 12 at 15:16
$begingroup$
@Shaun Theory of Groups by A. G. Kurosh
$endgroup$
– Tim Eller
Jan 12 at 15:17
$begingroup$
@Shaun did you edit my post? I have a question about the edit.
$endgroup$
– Tim Eller
Jan 12 at 16:28
|
show 1 more comment
$begingroup$
Where did this problem arise?
$endgroup$
– Shaun
Jan 12 at 15:12
$begingroup$
@Shaun it's in a book on group theory. The author states it without proof, then uses it to prove the additive group of rationals contains only subgroups of infinite index.
$endgroup$
– Tim Eller
Jan 12 at 15:16
$begingroup$
Okay, @TimEller; which book?
$endgroup$
– Shaun
Jan 12 at 15:16
$begingroup$
@Shaun Theory of Groups by A. G. Kurosh
$endgroup$
– Tim Eller
Jan 12 at 15:17
$begingroup$
@Shaun did you edit my post? I have a question about the edit.
$endgroup$
– Tim Eller
Jan 12 at 16:28
$begingroup$
Where did this problem arise?
$endgroup$
– Shaun
Jan 12 at 15:12
$begingroup$
Where did this problem arise?
$endgroup$
– Shaun
Jan 12 at 15:12
$begingroup$
@Shaun it's in a book on group theory. The author states it without proof, then uses it to prove the additive group of rationals contains only subgroups of infinite index.
$endgroup$
– Tim Eller
Jan 12 at 15:16
$begingroup$
@Shaun it's in a book on group theory. The author states it without proof, then uses it to prove the additive group of rationals contains only subgroups of infinite index.
$endgroup$
– Tim Eller
Jan 12 at 15:16
$begingroup$
Okay, @TimEller; which book?
$endgroup$
– Shaun
Jan 12 at 15:16
$begingroup$
Okay, @TimEller; which book?
$endgroup$
– Shaun
Jan 12 at 15:16
$begingroup$
@Shaun Theory of Groups by A. G. Kurosh
$endgroup$
– Tim Eller
Jan 12 at 15:17
$begingroup$
@Shaun Theory of Groups by A. G. Kurosh
$endgroup$
– Tim Eller
Jan 12 at 15:17
$begingroup$
@Shaun did you edit my post? I have a question about the edit.
$endgroup$
– Tim Eller
Jan 12 at 16:28
$begingroup$
@Shaun did you edit my post? I have a question about the edit.
$endgroup$
– Tim Eller
Jan 12 at 16:28
|
show 1 more comment
1 Answer
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$begingroup$
Suppose $frac mnin mathbb Q/H$ with $gcd(m,n)=1$ and let $n=prod p_i^{a_i}$ be the prime factorization of $n$. We multiply by one of the $p_i$ after another. If we get to an element in $H$ we are done, otherwise we can assume that this process leads to an integer $r in mathbb Q/H$.
Let $sin Hcap mathbb Z$. (Note: if $frac lkin H$ then $lin Hcap mathbb Z$.) Then of course $srin H$. If $s=prod q_i^{b_i}$ is the prime factorization of $s$, then we divide $sr$ by each $q_i$ in turn until we get to an element not in $H$.
$endgroup$
$begingroup$
Thanks for answering
$endgroup$
– Tim Eller
Jan 13 at 15:47
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose $frac mnin mathbb Q/H$ with $gcd(m,n)=1$ and let $n=prod p_i^{a_i}$ be the prime factorization of $n$. We multiply by one of the $p_i$ after another. If we get to an element in $H$ we are done, otherwise we can assume that this process leads to an integer $r in mathbb Q/H$.
Let $sin Hcap mathbb Z$. (Note: if $frac lkin H$ then $lin Hcap mathbb Z$.) Then of course $srin H$. If $s=prod q_i^{b_i}$ is the prime factorization of $s$, then we divide $sr$ by each $q_i$ in turn until we get to an element not in $H$.
$endgroup$
$begingroup$
Thanks for answering
$endgroup$
– Tim Eller
Jan 13 at 15:47
add a comment |
$begingroup$
Suppose $frac mnin mathbb Q/H$ with $gcd(m,n)=1$ and let $n=prod p_i^{a_i}$ be the prime factorization of $n$. We multiply by one of the $p_i$ after another. If we get to an element in $H$ we are done, otherwise we can assume that this process leads to an integer $r in mathbb Q/H$.
Let $sin Hcap mathbb Z$. (Note: if $frac lkin H$ then $lin Hcap mathbb Z$.) Then of course $srin H$. If $s=prod q_i^{b_i}$ is the prime factorization of $s$, then we divide $sr$ by each $q_i$ in turn until we get to an element not in $H$.
$endgroup$
$begingroup$
Thanks for answering
$endgroup$
– Tim Eller
Jan 13 at 15:47
add a comment |
$begingroup$
Suppose $frac mnin mathbb Q/H$ with $gcd(m,n)=1$ and let $n=prod p_i^{a_i}$ be the prime factorization of $n$. We multiply by one of the $p_i$ after another. If we get to an element in $H$ we are done, otherwise we can assume that this process leads to an integer $r in mathbb Q/H$.
Let $sin Hcap mathbb Z$. (Note: if $frac lkin H$ then $lin Hcap mathbb Z$.) Then of course $srin H$. If $s=prod q_i^{b_i}$ is the prime factorization of $s$, then we divide $sr$ by each $q_i$ in turn until we get to an element not in $H$.
$endgroup$
Suppose $frac mnin mathbb Q/H$ with $gcd(m,n)=1$ and let $n=prod p_i^{a_i}$ be the prime factorization of $n$. We multiply by one of the $p_i$ after another. If we get to an element in $H$ we are done, otherwise we can assume that this process leads to an integer $r in mathbb Q/H$.
Let $sin Hcap mathbb Z$. (Note: if $frac lkin H$ then $lin Hcap mathbb Z$.) Then of course $srin H$. If $s=prod q_i^{b_i}$ is the prime factorization of $s$, then we divide $sr$ by each $q_i$ in turn until we get to an element not in $H$.
edited Jan 13 at 15:47


amWhy
1
1
answered Jan 12 at 15:16
lulululu
41.3k24979
41.3k24979
$begingroup$
Thanks for answering
$endgroup$
– Tim Eller
Jan 13 at 15:47
add a comment |
$begingroup$
Thanks for answering
$endgroup$
– Tim Eller
Jan 13 at 15:47
$begingroup$
Thanks for answering
$endgroup$
– Tim Eller
Jan 13 at 15:47
$begingroup$
Thanks for answering
$endgroup$
– Tim Eller
Jan 13 at 15:47
add a comment |
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$begingroup$
Where did this problem arise?
$endgroup$
– Shaun
Jan 12 at 15:12
$begingroup$
@Shaun it's in a book on group theory. The author states it without proof, then uses it to prove the additive group of rationals contains only subgroups of infinite index.
$endgroup$
– Tim Eller
Jan 12 at 15:16
$begingroup$
Okay, @TimEller; which book?
$endgroup$
– Shaun
Jan 12 at 15:16
$begingroup$
@Shaun Theory of Groups by A. G. Kurosh
$endgroup$
– Tim Eller
Jan 12 at 15:17
$begingroup$
@Shaun did you edit my post? I have a question about the edit.
$endgroup$
– Tim Eller
Jan 12 at 16:28