Is $(mathbb{Q},+)$ the direct product of two non-trivial subgroups?
Is this statement true or false? I am really not having any idea how to prove or a counterexample, please help.
Is $(mathbb{Q},+)$ a direct product of two non-trivial subgroups?
abstract-algebra group-theory
edited Apr 14 '13 at 13:32
Seirios
24k34599
asked Jan 14 '13 at 23:10
Markov
17.2k957178
add a comment |
Is this statement true or false? I am really not having any idea how to prove or a counterexample, please help.
Is $(mathbb{Q},+)$ a direct product of two non-trivial subgroups?
abstract-algebra group-theory
edited Apr 14 '13 at 13:32
Seirios
24k34599
asked Jan 14 '13 at 23:10
Markov
17.2k957178
add a comment |
Is this statement true or false? I am really not having any idea how to prove or a counterexample, please help.
Is $(mathbb{Q},+)$ a direct product of two non-trivial subgroups?
abstract-algebra group-theory
edited Apr 14 '13 at 13:32
Seirios
24k34599
asked Jan 14 '13 at 23:10
Markov
17.2k957178
Is this statement true or false? I am really not having any idea how to prove or a counterexample, please help.
Is $(mathbb{Q},+)$ a direct product of two non-trivial subgroups?
abstract-algebra group-theory
abstract-algebra group-theory
edited Apr 14 '13 at 13:32
Seirios
24k34599
asked Jan 14 '13 at 23:10
Markov
17.2k957178
edited Apr 14 '13 at 13:32
Seirios
24k34599
asked Jan 14 '13 at 23:10
Markov
17.2k957178
edited Apr 14 '13 at 13:32
Seirios
24k34599
edited Apr 14 '13 at 13:32
Seirios
24k34599
edited Apr 14 '13 at 13:32
Seirios
24k34599
24k34599
asked Jan 14 '13 at 23:10
Markov
17.2k957178
asked Jan 14 '13 at 23:10
Markov
17.2k957178
asked Jan 14 '13 at 23:10
Markov
17.2k957178
17.2k957178
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
HINT: If $Bbb Q=Htimes G$, then there is an obvious homomorphism $h:Bbb Qto G$ with kernel $H$. Now consider this earlier question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
1
$H={0}$ only as any group homomorphism is bijection.hence the statement is false. am I right?
– Markov
Jan 14 '13 at 23:40
@Panu: That’s right.
– Brian M. Scott
Jan 15 '13 at 0:52
add a comment |
If $G$ is an abelian group and $Gsimeq G_1times G_2$, then there exist $H_1,H_2le G$ two subgroups such that $G=H_1+H_2$ and $H_1cap H_2={0}$ (and viceversa!). Furthermore, $H_i$ can be chosen isomorphic to $G_i$, $i=1,2$.
If $H_ilemathbb Q$, $H_ineq {0}$, then $H_icapmathbb Zneq {0}$ (why?), and therefore $(H_1capmathbb Z)cap(H_2capmathbb Z)neq {0}$. In particular, $H_1cap H_2neq{0}$.
add a comment |
Let $H,K leq mathbb{Q}$ be two subgroups such that $mathbb{Q} simeq H times K$.
Notice that $H$ is divisible: If $h in H$ and $n in mathbb{N}^*$, there exist $(h_0,k_0) in H times K$ such that $displaystyle frac{h}{n}=h_0+k_0$; so $nk_0 = h-nh_0 in H cap K={0}$, hence $k_0=0$ and $h/n in H$.
If there exists $displaystyle h:=frac{p}{q} in H backslash {0}$, then $displaystyle 1=q cdot frac{h}{p} in H$ ($H$ is divisible). So $mathbb{Z} subset H$, and because $H$ is divisible, $mathbb{Q} subset H$.
Therefore, $H= mathbb{Q}$ and $K= {0}$.
answered Apr 14 '13 at 13:31
Seirios
24k34599
add a comment |
Direct product decompositions $A cong B times C$ of an abelian group are in natural bijection with idempotent endomorphisms $e : A to A$ of $A$, where the idempotent corresponding to $A cong B times C$ is the projection onto $B$, so that $B = text{im}(e)$ and $C = text{ker}(e)$. So if you understand the ring of endomorphisms of $A$ you can compute all idempotents in it, and if there aren't any nontrivial ones then $A$ doesn't admit any nontrivial decompositions.
The endomorphism ring of $mathbb{Q}$ is $mathbb{Q}$ (exercise), which has no nontrivial idempotents.
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT: If $Bbb Q=Htimes G$, then there is an obvious homomorphism $h:Bbb Qto G$ with kernel $H$. Now consider this earlier question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
1
$H={0}$ only as any group homomorphism is bijection.hence the statement is false. am I right?
– Markov
Jan 14 '13 at 23:40
@Panu: That’s right.
– Brian M. Scott
Jan 15 '13 at 0:52
add a comment |
HINT: If $Bbb Q=Htimes G$, then there is an obvious homomorphism $h:Bbb Qto G$ with kernel $H$. Now consider this earlier question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
1
$H={0}$ only as any group homomorphism is bijection.hence the statement is false. am I right?
– Markov
Jan 14 '13 at 23:40
@Panu: That’s right.
– Brian M. Scott
Jan 15 '13 at 0:52
add a comment |
HINT: If $Bbb Q=Htimes G$, then there is an obvious homomorphism $h:Bbb Qto G$ with kernel $H$. Now consider this earlier question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
HINT: If $Bbb Q=Htimes G$, then there is an obvious homomorphism $h:Bbb Qto G$ with kernel $H$. Now consider this earlier question.
edited Apr 13 '17 at 12:20
Community♦
1
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
answered Jan 14 '13 at 23:17
Brian M. Scott
455k38505907
455k38505907
1
$H={0}$ only as any group homomorphism is bijection.hence the statement is false. am I right?
– Markov
Jan 14 '13 at 23:40
@Panu: That’s right.
– Brian M. Scott
Jan 15 '13 at 0:52
add a comment |
1
$H={0}$ only as any group homomorphism is bijection.hence the statement is false. am I right?
– Markov
Jan 14 '13 at 23:40
@Panu: That’s right.
– Brian M. Scott
Jan 15 '13 at 0:52
1
1
$H={0}$ only as any group homomorphism is bijection.hence the statement is false. am I right?
– Markov
Jan 14 '13 at 23:40
$H={0}$ only as any group homomorphism is bijection.hence the statement is false. am I right?
– Markov
Jan 14 '13 at 23:40
@Panu: That’s right.
– Brian M. Scott
Jan 15 '13 at 0:52
@Panu: That’s right.
– Brian M. Scott
Jan 15 '13 at 0:52
add a comment |
If $G$ is an abelian group and $Gsimeq G_1times G_2$, then there exist $H_1,H_2le G$ two subgroups such that $G=H_1+H_2$ and $H_1cap H_2={0}$ (and viceversa!). Furthermore, $H_i$ can be chosen isomorphic to $G_i$, $i=1,2$.
If $H_ilemathbb Q$, $H_ineq {0}$, then $H_icapmathbb Zneq {0}$ (why?), and therefore $(H_1capmathbb Z)cap(H_2capmathbb Z)neq {0}$. In particular, $H_1cap H_2neq{0}$.
add a comment |
If $G$ is an abelian group and $Gsimeq G_1times G_2$, then there exist $H_1,H_2le G$ two subgroups such that $G=H_1+H_2$ and $H_1cap H_2={0}$ (and viceversa!). Furthermore, $H_i$ can be chosen isomorphic to $G_i$, $i=1,2$.
If $H_ilemathbb Q$, $H_ineq {0}$, then $H_icapmathbb Zneq {0}$ (why?), and therefore $(H_1capmathbb Z)cap(H_2capmathbb Z)neq {0}$. In particular, $H_1cap H_2neq{0}$.
add a comment |
If $G$ is an abelian group and $Gsimeq G_1times G_2$, then there exist $H_1,H_2le G$ two subgroups such that $G=H_1+H_2$ and $H_1cap H_2={0}$ (and viceversa!). Furthermore, $H_i$ can be chosen isomorphic to $G_i$, $i=1,2$.
If $H_ilemathbb Q$, $H_ineq {0}$, then $H_icapmathbb Zneq {0}$ (why?), and therefore $(H_1capmathbb Z)cap(H_2capmathbb Z)neq {0}$. In particular, $H_1cap H_2neq{0}$.
If $G$ is an abelian group and $Gsimeq G_1times G_2$, then there exist $H_1,H_2le G$ two subgroups such that $G=H_1+H_2$ and $H_1cap H_2={0}$ (and viceversa!). Furthermore, $H_i$ can be chosen isomorphic to $G_i$, $i=1,2$.
If $H_ilemathbb Q$, $H_ineq {0}$, then $H_icapmathbb Zneq {0}$ (why?), and therefore $(H_1capmathbb Z)cap(H_2capmathbb Z)neq {0}$. In particular, $H_1cap H_2neq{0}$.
answered Jan 18 '13 at 9:42
user26857
add a comment |
add a comment |
Let $H,K leq mathbb{Q}$ be two subgroups such that $mathbb{Q} simeq H times K$.
Notice that $H$ is divisible: If $h in H$ and $n in mathbb{N}^*$, there exist $(h_0,k_0) in H times K$ such that $displaystyle frac{h}{n}=h_0+k_0$; so $nk_0 = h-nh_0 in H cap K={0}$, hence $k_0=0$ and $h/n in H$.
If there exists $displaystyle h:=frac{p}{q} in H backslash {0}$, then $displaystyle 1=q cdot frac{h}{p} in H$ ($H$ is divisible). So $mathbb{Z} subset H$, and because $H$ is divisible, $mathbb{Q} subset H$.
Therefore, $H= mathbb{Q}$ and $K= {0}$.
answered Apr 14 '13 at 13:31
Seirios
24k34599
add a comment |
Let $H,K leq mathbb{Q}$ be two subgroups such that $mathbb{Q} simeq H times K$.
Notice that $H$ is divisible: If $h in H$ and $n in mathbb{N}^*$, there exist $(h_0,k_0) in H times K$ such that $displaystyle frac{h}{n}=h_0+k_0$; so $nk_0 = h-nh_0 in H cap K={0}$, hence $k_0=0$ and $h/n in H$.
If there exists $displaystyle h:=frac{p}{q} in H backslash {0}$, then $displaystyle 1=q cdot frac{h}{p} in H$ ($H$ is divisible). So $mathbb{Z} subset H$, and because $H$ is divisible, $mathbb{Q} subset H$.
Therefore, $H= mathbb{Q}$ and $K= {0}$.
answered Apr 14 '13 at 13:31
Seirios
24k34599
add a comment |
Let $H,K leq mathbb{Q}$ be two subgroups such that $mathbb{Q} simeq H times K$.
Notice that $H$ is divisible: If $h in H$ and $n in mathbb{N}^*$, there exist $(h_0,k_0) in H times K$ such that $displaystyle frac{h}{n}=h_0+k_0$; so $nk_0 = h-nh_0 in H cap K={0}$, hence $k_0=0$ and $h/n in H$.
If there exists $displaystyle h:=frac{p}{q} in H backslash {0}$, then $displaystyle 1=q cdot frac{h}{p} in H$ ($H$ is divisible). So $mathbb{Z} subset H$, and because $H$ is divisible, $mathbb{Q} subset H$.
Therefore, $H= mathbb{Q}$ and $K= {0}$.
answered Apr 14 '13 at 13:31
Seirios
24k34599
Let $H,K leq mathbb{Q}$ be two subgroups such that $mathbb{Q} simeq H times K$.
Notice that $H$ is divisible: If $h in H$ and $n in mathbb{N}^*$, there exist $(h_0,k_0) in H times K$ such that $displaystyle frac{h}{n}=h_0+k_0$; so $nk_0 = h-nh_0 in H cap K={0}$, hence $k_0=0$ and $h/n in H$.
If there exists $displaystyle h:=frac{p}{q} in H backslash {0}$, then $displaystyle 1=q cdot frac{h}{p} in H$ ($H$ is divisible). So $mathbb{Z} subset H$, and because $H$ is divisible, $mathbb{Q} subset H$.
Therefore, $H= mathbb{Q}$ and $K= {0}$.
answered Apr 14 '13 at 13:31
Seirios
24k34599
answered Apr 14 '13 at 13:31
Seirios
24k34599
answered Apr 14 '13 at 13:31
Seirios
24k34599
answered Apr 14 '13 at 13:31
Seirios
24k34599
24k34599
add a comment |
add a comment |
Direct product decompositions $A cong B times C$ of an abelian group are in natural bijection with idempotent endomorphisms $e : A to A$ of $A$, where the idempotent corresponding to $A cong B times C$ is the projection onto $B$, so that $B = text{im}(e)$ and $C = text{ker}(e)$. So if you understand the ring of endomorphisms of $A$ you can compute all idempotents in it, and if there aren't any nontrivial ones then $A$ doesn't admit any nontrivial decompositions.
The endomorphism ring of $mathbb{Q}$ is $mathbb{Q}$ (exercise), which has no nontrivial idempotents.
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
add a comment |
Direct product decompositions $A cong B times C$ of an abelian group are in natural bijection with idempotent endomorphisms $e : A to A$ of $A$, where the idempotent corresponding to $A cong B times C$ is the projection onto $B$, so that $B = text{im}(e)$ and $C = text{ker}(e)$. So if you understand the ring of endomorphisms of $A$ you can compute all idempotents in it, and if there aren't any nontrivial ones then $A$ doesn't admit any nontrivial decompositions.
The endomorphism ring of $mathbb{Q}$ is $mathbb{Q}$ (exercise), which has no nontrivial idempotents.
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
add a comment |
Direct product decompositions $A cong B times C$ of an abelian group are in natural bijection with idempotent endomorphisms $e : A to A$ of $A$, where the idempotent corresponding to $A cong B times C$ is the projection onto $B$, so that $B = text{im}(e)$ and $C = text{ker}(e)$. So if you understand the ring of endomorphisms of $A$ you can compute all idempotents in it, and if there aren't any nontrivial ones then $A$ doesn't admit any nontrivial decompositions.
The endomorphism ring of $mathbb{Q}$ is $mathbb{Q}$ (exercise), which has no nontrivial idempotents.
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
Direct product decompositions $A cong B times C$ of an abelian group are in natural bijection with idempotent endomorphisms $e : A to A$ of $A$, where the idempotent corresponding to $A cong B times C$ is the projection onto $B$, so that $B = text{im}(e)$ and $C = text{ker}(e)$. So if you understand the ring of endomorphisms of $A$ you can compute all idempotents in it, and if there aren't any nontrivial ones then $A$ doesn't admit any nontrivial decompositions.
The endomorphism ring of $mathbb{Q}$ is $mathbb{Q}$ (exercise), which has no nontrivial idempotents.
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
answered Dec 11 '18 at 9:11
Qiaochu Yuan
277k32581919
277k32581919
add a comment |
add a comment |
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