Mixing
Contrapositive proof: If $H$ and $K$ are nontrivial subgroups of $Bbb Q$, then $Hcap K$ is also nontrivial.
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.
I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.
$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.
The contrapositive problem can be stated as follows:
$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.
Thoughts:
To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.
My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.
A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.
Please help :)
group-theory alternative-proof rational-numbers
asked Jan 24 at 11:37
ShaunShaun
9,444113684
$endgroup$
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.
I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.
$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.
The contrapositive problem can be stated as follows:
$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.
Thoughts:
To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.
My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.
A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.
Please help :)
group-theory alternative-proof rational-numbers
asked Jan 24 at 11:37
ShaunShaun
9,444113684
$endgroup$
$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.
I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.
$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.
The contrapositive problem can be stated as follows:
$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.
Thoughts:
To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.
My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.
A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.
Please help :)
group-theory alternative-proof rational-numbers
asked Jan 24 at 11:37
ShaunShaun
9,444113684
$endgroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.
I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.
$mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $Bbb Q$ under addition. Show that $H cap K$ is a nontrivial subgroup of $Bbb Q$.
The contrapositive problem can be stated as follows:
$mathcal{C}$: Let $H$ and $K$ be subgroups of $Bbb Q$. Suppose $Hcap K={0}$ is trivial. Show that either $H$ or $K$ is trivial.
Thoughts:
To me, it seems $mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.
My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $Hcap K$ is nontrivial, contradicting the hypothesis that $Hcap K={0}$. That's just silly.
A lovely proof of the original statement $mathcal{O}$ is given here, with variations on its theme here and here.
Please help :)
group-theory alternative-proof rational-numbers
group-theory alternative-proof rational-numbers
asked Jan 24 at 11:37
ShaunShaun
9,444113684
asked Jan 24 at 11:37
ShaunShaun
9,444113684
asked Jan 24 at 11:37
ShaunShaun
9,444113684
asked Jan 24 at 11:37
ShaunShaun
9,444113684
asked Jan 24 at 11:37
ShaunShaun
9,444113684
9,444113684
$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44
add a comment |
$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44
$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44
$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.
1: If all $G_i$ have the property $P$, then so does
$$lim_{longrightarrow}G_i .$$
2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,
begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}
Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$
3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.
edited Jan 25 at 10:26
Shaun
9,444113684
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
$endgroup$
$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21
add a comment |
$begingroup$
This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$
It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55
$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09
$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17
$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27
1
$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.
1: If all $G_i$ have the property $P$, then so does
$$lim_{longrightarrow}G_i .$$
2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,
begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}
Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$
3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.
edited Jan 25 at 10:26
Shaun
9,444113684
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
$endgroup$
$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21
add a comment |
$begingroup$
Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.
1: If all $G_i$ have the property $P$, then so does
$$lim_{longrightarrow}G_i .$$
2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,
begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}
Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$
3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.
edited Jan 25 at 10:26
Shaun
9,444113684
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
$endgroup$
$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21
add a comment |
$begingroup$
Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.
1: If all $G_i$ have the property $P$, then so does
$$lim_{longrightarrow}G_i .$$
2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,
begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}
Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$
3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.
edited Jan 25 at 10:26
Shaun
9,444113684
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
$endgroup$
Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.
1: If all $G_i$ have the property $P$, then so does
$$lim_{longrightarrow}G_i .$$
2: Define $mathbb{Z}^{(n)}=mathbb{Z}$, and for $n|m$,
begin{align}
f_{n,m}:mathbb{Z}^{(n)} &longrightarrowmathbb{Z}^{(m)}, \
k &mapstofrac{m}{n}cdot k.
end{align}
Then
$$mathbb{Q}=lim_{longrightarrow}mathbb{Z}^{(n)}.$$
3: $mathbb{Z}$ has the property $P$. In fact, $Hcap K=0$ implies the injectivity of
$$mathbb{Z}longrightarrow mathbb{Z}/H times mathbb{Z}/K,$$
and thus $H=0$ or $K=0$.
edited Jan 25 at 10:26
Shaun
9,444113684
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
edited Jan 25 at 10:26
Shaun
9,444113684
edited Jan 25 at 10:26
Shaun
9,444113684
edited Jan 25 at 10:26
Shaun
9,444113684
9,444113684
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
answered Jan 25 at 10:09
Hiro WatHiro Wat
1238
1238
$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21
add a comment |
$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21
$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21
$begingroup$
This is beautiful. Thank you! :)
$endgroup$
– Shaun
Jan 25 at 10:21
add a comment |
$begingroup$
This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$
It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55
$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09
$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17
$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27
1
$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58
|
show 2 more comments
$begingroup$
This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$
It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
$endgroup$
$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55
$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09
$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17
$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27
1
$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58
|
show 2 more comments
$begingroup$
This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$
It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
$endgroup$
This follows from the fact that any two rational non-zero numbers $a/b$ and $c/d$ have a common multiple, say
$$
cb cdot frac{a}{b} = ad cdot frac{c}{d}.
$$
It will be unnatural to give a proof using abstract algebra only, as this is a simple arithmetic fact depending on specific properties of rational numbers.
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
answered Jan 24 at 12:08
Reiner MartinReiner Martin
3,509414
3,509414
$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55
$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09
$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17
$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27
1
$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58
|
show 2 more comments
$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55
$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09
$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17
$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27
1
$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58
$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55
$begingroup$
This just describes what the proofs linked to in the question for $mathcal{O}$, so it doesn't prove $mathcal{C}$ directly nor is it in the spirit of my question. (Hence the downvote.)
$endgroup$
– Shaun
Jan 24 at 15:55
$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09
$begingroup$
The property you want to prove is very specific to $mathbb{Q},$ so it is unlikely you'll find a proof using 'abstract algebra' (whatever that exactly means). The proof I give is really the core of the problem, I think.
$endgroup$
– Reiner Martin
Jan 24 at 16:09
$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17
$begingroup$
I don't think it's necessary for me, a mere PhD student, to define what exactly "abstract algebra" is. Most people with an undergraduate degree in some mathematical field has some idea what it means. Besides, you might want to revise things like a field of fractions. Also, again, please see this comment and the answer linked to in the opening post for why your answer is not helpful.
$endgroup$
– Shaun
Jan 24 at 16:17
$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27
$begingroup$
Sorry that I, a mere math PhD, could not be more helpful... Joking aside, I think the thing you want does not exist.
$endgroup$
– Reiner Martin
Jan 24 at 16:27
1
1
$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58
$begingroup$
OK, I stand corrected
$endgroup$
– Reiner Martin
Jan 25 at 10:58
|
show 2 more comments
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$begingroup$
This proves the result in the title, @DonAntonio (assuming $Bbb Q$ is indeed under addition). Here's what I want: an algebraic proof of the contrapositive $mathcal{C}$ of $mathcal{O}$.
$endgroup$
– Shaun
Jan 24 at 11:44