Mixing
Rationals as a direct sum of two proper subgroups
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Prove that the group $(mathbb{Q},+)$ is not isomorphic to the direct sum of two proper subgroups.
My approach: Suppose that $mathbb{Q}=Aoplus B$ where $A,B$ - proper subgroups of $mathbb{Q}$. By the definition it means that each $qin mathbb{Q}$ has unique representation, i.e. if $q=a_1+b_1$ and $q=a_2+b_2$ then $a_1=a_2$, $b_1=b_2$. Since $1in mathbb{Q}$ and $0in A$ then $1=0+1$ where $1in B$. Since $2=1+1in B$ hence $1=(-1)+2$ and so $-1in A$. We get two distinct representations, contradiction.
Remark: Please do not duplicate this question. I have sen some topics with this question but i did not saw the solution similar to above.
What's wrong with this solution?
group-theory
asked yesterday
ZFR
4,88831337
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up vote
0
down vote
favorite
Prove that the group $(mathbb{Q},+)$ is not isomorphic to the direct sum of two proper subgroups.
My approach: Suppose that $mathbb{Q}=Aoplus B$ where $A,B$ - proper subgroups of $mathbb{Q}$. By the definition it means that each $qin mathbb{Q}$ has unique representation, i.e. if $q=a_1+b_1$ and $q=a_2+b_2$ then $a_1=a_2$, $b_1=b_2$. Since $1in mathbb{Q}$ and $0in A$ then $1=0+1$ where $1in B$. Since $2=1+1in B$ hence $1=(-1)+2$ and so $-1in A$. We get two distinct representations, contradiction.
Remark: Please do not duplicate this question. I have sen some topics with this question but i did not saw the solution similar to above.
What's wrong with this solution?
group-theory
asked yesterday
ZFR
4,88831337
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove that the group $(mathbb{Q},+)$ is not isomorphic to the direct sum of two proper subgroups.
My approach: Suppose that $mathbb{Q}=Aoplus B$ where $A,B$ - proper subgroups of $mathbb{Q}$. By the definition it means that each $qin mathbb{Q}$ has unique representation, i.e. if $q=a_1+b_1$ and $q=a_2+b_2$ then $a_1=a_2$, $b_1=b_2$. Since $1in mathbb{Q}$ and $0in A$ then $1=0+1$ where $1in B$. Since $2=1+1in B$ hence $1=(-1)+2$ and so $-1in A$. We get two distinct representations, contradiction.
Remark: Please do not duplicate this question. I have sen some topics with this question but i did not saw the solution similar to above.
What's wrong with this solution?
group-theory
asked yesterday
ZFR
4,88831337
Prove that the group $(mathbb{Q},+)$ is not isomorphic to the direct sum of two proper subgroups.
My approach: Suppose that $mathbb{Q}=Aoplus B$ where $A,B$ - proper subgroups of $mathbb{Q}$. By the definition it means that each $qin mathbb{Q}$ has unique representation, i.e. if $q=a_1+b_1$ and $q=a_2+b_2$ then $a_1=a_2$, $b_1=b_2$. Since $1in mathbb{Q}$ and $0in A$ then $1=0+1$ where $1in B$. Since $2=1+1in B$ hence $1=(-1)+2$ and so $-1in A$. We get two distinct representations, contradiction.
Remark: Please do not duplicate this question. I have sen some topics with this question but i did not saw the solution similar to above.
What's wrong with this solution?
group-theory
group-theory
asked yesterday
ZFR
4,88831337
asked yesterday
ZFR
4,88831337
asked yesterday
ZFR
4,88831337
asked yesterday
ZFR
4,88831337
asked yesterday
ZFR
4,88831337
4,88831337
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
You assert $1 in B$, but there's no reason to think this is true.
answered yesterday
Y. Forman
11.3k423
It is by the definition of a proper subgroup, it must contain the multiplicative identity.
– Mathaddict
yesterday
Maybe because I am not understand the idea of direct sum correctly. But why it is not true? we take $1in mathbb{Q}$ and there exists unique pair of $ain A$ and $bin B$ such that $1=a+b$. Taking $a=0$ we get $b=1$. Not?
– ZFR
yesterday
@Mathaddict The question speaks of the group $mathbb Q$ under addition, so there's no multiplication to consider.
– Y. Forman
yesterday
Ahh. I guess that I got because we have no information about these $a$ and $b$, right? We have no right take value $a=0$.
– ZFR
yesterday
1
@ZFR That is exactly right.
– Y. Forman
yesterday
|
show 4 more comments
up vote
1
down vote
Following somewhat your idea, but aboiding the trap you fell into:
Assume $Bbb Q=Aoplus B$ with $A,B$ non-trivial subgroups of $Bbb Q$. Pick $0ne ain A$, $0ne bin B$. Then $a=frac nm$, $b=frac pq$ for suitable non-zero integers $n,m,p,q$. Then
$$np=underbrace{a+a+ldots+a}_{mp} =underbrace{b+b+ldots+b}_{nq}in Acap B,$$
contradiction.
answered yesterday
Hagen von Eitzen
273k21266493
That's what I want to do but as you said I fell into trap. Very good point!
– ZFR
yesterday
add a comment |
up vote
0
down vote
Here's one way to do it.
We claim $(mathbb{Q}, +)$ can not be the direct product of two non-trivial subgroups.
Suppose we two subgroups $H, K$ of $mathbb{Q}$ such that $Q cong H times K$. Then, there would be a homomorphism $h : mathbb{Q} rightarrow K$ with kernel $H$. But since any non-zero homomorphism from $(mathbb{Q}, +)$ to $(mathbb{Q}, +)$ is always bijective, we conclude $H = {0}$, which is a contradiction.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You assert $1 in B$, but there's no reason to think this is true.
answered yesterday
Y. Forman
11.3k423
It is by the definition of a proper subgroup, it must contain the multiplicative identity.
– Mathaddict
yesterday
Maybe because I am not understand the idea of direct sum correctly. But why it is not true? we take $1in mathbb{Q}$ and there exists unique pair of $ain A$ and $bin B$ such that $1=a+b$. Taking $a=0$ we get $b=1$. Not?
– ZFR
yesterday
@Mathaddict The question speaks of the group $mathbb Q$ under addition, so there's no multiplication to consider.
– Y. Forman
yesterday
Ahh. I guess that I got because we have no information about these $a$ and $b$, right? We have no right take value $a=0$.
– ZFR
yesterday
1
@ZFR That is exactly right.
– Y. Forman
yesterday
|
show 4 more comments
up vote
2
down vote
You assert $1 in B$, but there's no reason to think this is true.
answered yesterday
Y. Forman
11.3k423
It is by the definition of a proper subgroup, it must contain the multiplicative identity.
– Mathaddict
yesterday
Maybe because I am not understand the idea of direct sum correctly. But why it is not true? we take $1in mathbb{Q}$ and there exists unique pair of $ain A$ and $bin B$ such that $1=a+b$. Taking $a=0$ we get $b=1$. Not?
– ZFR
yesterday
@Mathaddict The question speaks of the group $mathbb Q$ under addition, so there's no multiplication to consider.
– Y. Forman
yesterday
Ahh. I guess that I got because we have no information about these $a$ and $b$, right? We have no right take value $a=0$.
– ZFR
yesterday
1
@ZFR That is exactly right.
– Y. Forman
yesterday
|
show 4 more comments
up vote
2
down vote
up vote
2
down vote
You assert $1 in B$, but there's no reason to think this is true.
answered yesterday
Y. Forman
11.3k423
You assert $1 in B$, but there's no reason to think this is true.
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
11.3k423
It is by the definition of a proper subgroup, it must contain the multiplicative identity.
– Mathaddict
yesterday
Maybe because I am not understand the idea of direct sum correctly. But why it is not true? we take $1in mathbb{Q}$ and there exists unique pair of $ain A$ and $bin B$ such that $1=a+b$. Taking $a=0$ we get $b=1$. Not?
– ZFR
yesterday
@Mathaddict The question speaks of the group $mathbb Q$ under addition, so there's no multiplication to consider.
– Y. Forman
yesterday
Ahh. I guess that I got because we have no information about these $a$ and $b$, right? We have no right take value $a=0$.
– ZFR
yesterday
1
@ZFR That is exactly right.
– Y. Forman
yesterday
|
show 4 more comments
It is by the definition of a proper subgroup, it must contain the multiplicative identity.
– Mathaddict
yesterday
Maybe because I am not understand the idea of direct sum correctly. But why it is not true? we take $1in mathbb{Q}$ and there exists unique pair of $ain A$ and $bin B$ such that $1=a+b$. Taking $a=0$ we get $b=1$. Not?
– ZFR
yesterday
@Mathaddict The question speaks of the group $mathbb Q$ under addition, so there's no multiplication to consider.
– Y. Forman
yesterday
Ahh. I guess that I got because we have no information about these $a$ and $b$, right? We have no right take value $a=0$.
– ZFR
yesterday
1
@ZFR That is exactly right.
– Y. Forman
yesterday
It is by the definition of a proper subgroup, it must contain the multiplicative identity.
– Mathaddict
yesterday
It is by the definition of a proper subgroup, it must contain the multiplicative identity.
– Mathaddict
yesterday
Maybe because I am not understand the idea of direct sum correctly. But why it is not true? we take $1in mathbb{Q}$ and there exists unique pair of $ain A$ and $bin B$ such that $1=a+b$. Taking $a=0$ we get $b=1$. Not?
– ZFR
yesterday
Maybe because I am not understand the idea of direct sum correctly. But why it is not true? we take $1in mathbb{Q}$ and there exists unique pair of $ain A$ and $bin B$ such that $1=a+b$. Taking $a=0$ we get $b=1$. Not?
– ZFR
yesterday
@Mathaddict The question speaks of the group $mathbb Q$ under addition, so there's no multiplication to consider.
– Y. Forman
yesterday
@Mathaddict The question speaks of the group $mathbb Q$ under addition, so there's no multiplication to consider.
– Y. Forman
yesterday
Ahh. I guess that I got because we have no information about these $a$ and $b$, right? We have no right take value $a=0$.
– ZFR
yesterday
Ahh. I guess that I got because we have no information about these $a$ and $b$, right? We have no right take value $a=0$.
– ZFR
yesterday
1
1
@ZFR That is exactly right.
– Y. Forman
yesterday
@ZFR That is exactly right.
– Y. Forman
yesterday
|
show 4 more comments
up vote
1
down vote
Following somewhat your idea, but aboiding the trap you fell into:
Assume $Bbb Q=Aoplus B$ with $A,B$ non-trivial subgroups of $Bbb Q$. Pick $0ne ain A$, $0ne bin B$. Then $a=frac nm$, $b=frac pq$ for suitable non-zero integers $n,m,p,q$. Then
$$np=underbrace{a+a+ldots+a}_{mp} =underbrace{b+b+ldots+b}_{nq}in Acap B,$$
contradiction.
answered yesterday
Hagen von Eitzen
273k21266493
That's what I want to do but as you said I fell into trap. Very good point!
– ZFR
yesterday
add a comment |
up vote
1
down vote
Following somewhat your idea, but aboiding the trap you fell into:
Assume $Bbb Q=Aoplus B$ with $A,B$ non-trivial subgroups of $Bbb Q$. Pick $0ne ain A$, $0ne bin B$. Then $a=frac nm$, $b=frac pq$ for suitable non-zero integers $n,m,p,q$. Then
$$np=underbrace{a+a+ldots+a}_{mp} =underbrace{b+b+ldots+b}_{nq}in Acap B,$$
contradiction.
answered yesterday
Hagen von Eitzen
273k21266493
That's what I want to do but as you said I fell into trap. Very good point!
– ZFR
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
Following somewhat your idea, but aboiding the trap you fell into:
Assume $Bbb Q=Aoplus B$ with $A,B$ non-trivial subgroups of $Bbb Q$. Pick $0ne ain A$, $0ne bin B$. Then $a=frac nm$, $b=frac pq$ for suitable non-zero integers $n,m,p,q$. Then
$$np=underbrace{a+a+ldots+a}_{mp} =underbrace{b+b+ldots+b}_{nq}in Acap B,$$
contradiction.
answered yesterday
Hagen von Eitzen
273k21266493
Following somewhat your idea, but aboiding the trap you fell into:
Assume $Bbb Q=Aoplus B$ with $A,B$ non-trivial subgroups of $Bbb Q$. Pick $0ne ain A$, $0ne bin B$. Then $a=frac nm$, $b=frac pq$ for suitable non-zero integers $n,m,p,q$. Then
$$np=underbrace{a+a+ldots+a}_{mp} =underbrace{b+b+ldots+b}_{nq}in Acap B,$$
contradiction.
answered yesterday
Hagen von Eitzen
273k21266493
answered yesterday
Hagen von Eitzen
273k21266493
answered yesterday
Hagen von Eitzen
273k21266493
answered yesterday
Hagen von Eitzen
273k21266493
273k21266493
That's what I want to do but as you said I fell into trap. Very good point!
– ZFR
yesterday
add a comment |
That's what I want to do but as you said I fell into trap. Very good point!
– ZFR
yesterday
That's what I want to do but as you said I fell into trap. Very good point!
– ZFR
yesterday
That's what I want to do but as you said I fell into trap. Very good point!
– ZFR
yesterday
add a comment |
up vote
0
down vote
Here's one way to do it.
We claim $(mathbb{Q}, +)$ can not be the direct product of two non-trivial subgroups.
Suppose we two subgroups $H, K$ of $mathbb{Q}$ such that $Q cong H times K$. Then, there would be a homomorphism $h : mathbb{Q} rightarrow K$ with kernel $H$. But since any non-zero homomorphism from $(mathbb{Q}, +)$ to $(mathbb{Q}, +)$ is always bijective, we conclude $H = {0}$, which is a contradiction.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
Here's one way to do it.
We claim $(mathbb{Q}, +)$ can not be the direct product of two non-trivial subgroups.
Suppose we two subgroups $H, K$ of $mathbb{Q}$ such that $Q cong H times K$. Then, there would be a homomorphism $h : mathbb{Q} rightarrow K$ with kernel $H$. But since any non-zero homomorphism from $(mathbb{Q}, +)$ to $(mathbb{Q}, +)$ is always bijective, we conclude $H = {0}$, which is a contradiction.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
Here's one way to do it.
We claim $(mathbb{Q}, +)$ can not be the direct product of two non-trivial subgroups.
Suppose we two subgroups $H, K$ of $mathbb{Q}$ such that $Q cong H times K$. Then, there would be a homomorphism $h : mathbb{Q} rightarrow K$ with kernel $H$. But since any non-zero homomorphism from $(mathbb{Q}, +)$ to $(mathbb{Q}, +)$ is always bijective, we conclude $H = {0}$, which is a contradiction.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Here's one way to do it.
We claim $(mathbb{Q}, +)$ can not be the direct product of two non-trivial subgroups.
Suppose we two subgroups $H, K$ of $mathbb{Q}$ such that $Q cong H times K$. Then, there would be a homomorphism $h : mathbb{Q} rightarrow K$ with kernel $H$. But since any non-zero homomorphism from $(mathbb{Q}, +)$ to $(mathbb{Q}, +)$ is always bijective, we conclude $H = {0}$, which is a contradiction.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Ekesh
3514
answered yesterday
Ekesh
3514
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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