$(mathbb{Q},+)$ isn't direct product of two non-trivial groups
$begingroup$
I know that this has been posted before. I would like to know whether my argument is correct or not. I am presenting a new argument, so this question shouldn't be marked as duplicate.
Suppose that $Q cong H times K$ such that neither H or K is trivial. This is achieved through isomorphism $psi : H times K rightarrow mathbb{Q}$. I claim that kernel $psi = Htimes {0}$. That is, if we assume that H is non-trivial.
Proof
$$smallfrac{p_{(h,k)}}{q_{(h,k)}} = psi((h,k) = psi( (h,0) + (0,k) ) = psi((h,0)) + psi((0,k)) = frac{p_{(h,0)}}{q_{(h,0)}} + frac{p_{(0,k)}}{q_{(0,k)}} = frac{p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)}}
{q_{(h,0)}q_{(0,k)}} = 0$$
$p_{(h,k)} = p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)} = 0$. This gives us
$p_{(h,0)}q_{(0,k)} = -p_{(0,k)}q_{(h,0)}$. We can choose k such that $p_{(0,k)} = 0$. Therefore $p_{(h,0)}q_{(0,k)} = 0$ implies that $p_{(h,0)} = 0$. Therefore $Htimes {0} subset operatorname{ker}(psi)$ giving us a contradiction that $psi$ is an isomorphism.
abstract-algebra proof-verification
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
$endgroup$
add a comment |
$begingroup$
I know that this has been posted before. I would like to know whether my argument is correct or not. I am presenting a new argument, so this question shouldn't be marked as duplicate.
Suppose that $Q cong H times K$ such that neither H or K is trivial. This is achieved through isomorphism $psi : H times K rightarrow mathbb{Q}$. I claim that kernel $psi = Htimes {0}$. That is, if we assume that H is non-trivial.
Proof
$$smallfrac{p_{(h,k)}}{q_{(h,k)}} = psi((h,k) = psi( (h,0) + (0,k) ) = psi((h,0)) + psi((0,k)) = frac{p_{(h,0)}}{q_{(h,0)}} + frac{p_{(0,k)}}{q_{(0,k)}} = frac{p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)}}
{q_{(h,0)}q_{(0,k)}} = 0$$
$p_{(h,k)} = p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)} = 0$. This gives us
$p_{(h,0)}q_{(0,k)} = -p_{(0,k)}q_{(h,0)}$. We can choose k such that $p_{(0,k)} = 0$. Therefore $p_{(h,0)}q_{(0,k)} = 0$ implies that $p_{(h,0)} = 0$. Therefore $Htimes {0} subset operatorname{ker}(psi)$ giving us a contradiction that $psi$ is an isomorphism.
abstract-algebra proof-verification
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
$endgroup$
$begingroup$
See Najib Idrissi's answer that the kernel is $0times H$.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:23
$begingroup$
Why is my answer marked as duplicate I am presenting new argument.
$endgroup$
– Newbie
Dec 5 '18 at 21:28
1
$begingroup$
Why is your argument new? This is exactly the answer at the duplicate that the kernel is $0times H$, so that the map cannot be an isomorphism to the direct product.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:31
$begingroup$
The argument presented in the question was based on orders of element.
$endgroup$
– Newbie
Dec 5 '18 at 21:34
add a comment |
$begingroup$
I know that this has been posted before. I would like to know whether my argument is correct or not. I am presenting a new argument, so this question shouldn't be marked as duplicate.
Suppose that $Q cong H times K$ such that neither H or K is trivial. This is achieved through isomorphism $psi : H times K rightarrow mathbb{Q}$. I claim that kernel $psi = Htimes {0}$. That is, if we assume that H is non-trivial.
Proof
$$smallfrac{p_{(h,k)}}{q_{(h,k)}} = psi((h,k) = psi( (h,0) + (0,k) ) = psi((h,0)) + psi((0,k)) = frac{p_{(h,0)}}{q_{(h,0)}} + frac{p_{(0,k)}}{q_{(0,k)}} = frac{p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)}}
{q_{(h,0)}q_{(0,k)}} = 0$$
$p_{(h,k)} = p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)} = 0$. This gives us
$p_{(h,0)}q_{(0,k)} = -p_{(0,k)}q_{(h,0)}$. We can choose k such that $p_{(0,k)} = 0$. Therefore $p_{(h,0)}q_{(0,k)} = 0$ implies that $p_{(h,0)} = 0$. Therefore $Htimes {0} subset operatorname{ker}(psi)$ giving us a contradiction that $psi$ is an isomorphism.
abstract-algebra proof-verification
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
$endgroup$
I know that this has been posted before. I would like to know whether my argument is correct or not. I am presenting a new argument, so this question shouldn't be marked as duplicate.
Suppose that $Q cong H times K$ such that neither H or K is trivial. This is achieved through isomorphism $psi : H times K rightarrow mathbb{Q}$. I claim that kernel $psi = Htimes {0}$. That is, if we assume that H is non-trivial.
Proof
$$smallfrac{p_{(h,k)}}{q_{(h,k)}} = psi((h,k) = psi( (h,0) + (0,k) ) = psi((h,0)) + psi((0,k)) = frac{p_{(h,0)}}{q_{(h,0)}} + frac{p_{(0,k)}}{q_{(0,k)}} = frac{p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)}}
{q_{(h,0)}q_{(0,k)}} = 0$$
$p_{(h,k)} = p_{(h,0)}q_{(0,k)} + p_{(0,k)}q_{(h,0)} = 0$. This gives us
$p_{(h,0)}q_{(0,k)} = -p_{(0,k)}q_{(h,0)}$. We can choose k such that $p_{(0,k)} = 0$. Therefore $p_{(h,0)}q_{(0,k)} = 0$ implies that $p_{(h,0)} = 0$. Therefore $Htimes {0} subset operatorname{ker}(psi)$ giving us a contradiction that $psi$ is an isomorphism.
abstract-algebra proof-verification
abstract-algebra proof-verification
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
edited Dec 5 '18 at 21:31
Newbie
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
asked Dec 5 '18 at 21:18
NewbieNewbie
433311
433311
$begingroup$
See Najib Idrissi's answer that the kernel is $0times H$.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:23
$begingroup$
Why is my answer marked as duplicate I am presenting new argument.
$endgroup$
– Newbie
Dec 5 '18 at 21:28
1
$begingroup$
Why is your argument new? This is exactly the answer at the duplicate that the kernel is $0times H$, so that the map cannot be an isomorphism to the direct product.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:31
$begingroup$
The argument presented in the question was based on orders of element.
$endgroup$
– Newbie
Dec 5 '18 at 21:34
add a comment |
$begingroup$
See Najib Idrissi's answer that the kernel is $0times H$.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:23
$begingroup$
Why is my answer marked as duplicate I am presenting new argument.
$endgroup$
– Newbie
Dec 5 '18 at 21:28
1
$begingroup$
Why is your argument new? This is exactly the answer at the duplicate that the kernel is $0times H$, so that the map cannot be an isomorphism to the direct product.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:31
$begingroup$
The argument presented in the question was based on orders of element.
$endgroup$
– Newbie
Dec 5 '18 at 21:34
$begingroup$
See Najib Idrissi's answer that the kernel is $0times H$.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:23
$begingroup$
See Najib Idrissi's answer that the kernel is $0times H$.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:23
$begingroup$
Why is my answer marked as duplicate I am presenting new argument.
$endgroup$
– Newbie
Dec 5 '18 at 21:28
$begingroup$
Why is my answer marked as duplicate I am presenting new argument.
$endgroup$
– Newbie
Dec 5 '18 at 21:28
1
1
$begingroup$
Why is your argument new? This is exactly the answer at the duplicate that the kernel is $0times H$, so that the map cannot be an isomorphism to the direct product.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:31
$begingroup$
Why is your argument new? This is exactly the answer at the duplicate that the kernel is $0times H$, so that the map cannot be an isomorphism to the direct product.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:31
$begingroup$
The argument presented in the question was based on orders of element.
$endgroup$
– Newbie
Dec 5 '18 at 21:34
$begingroup$
The argument presented in the question was based on orders of element.
$endgroup$
– Newbie
Dec 5 '18 at 21:34
add a comment |
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$begingroup$
See Najib Idrissi's answer that the kernel is $0times H$.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:23
$begingroup$
Why is my answer marked as duplicate I am presenting new argument.
$endgroup$
– Newbie
Dec 5 '18 at 21:28
1
$begingroup$
Why is your argument new? This is exactly the answer at the duplicate that the kernel is $0times H$, so that the map cannot be an isomorphism to the direct product.
$endgroup$
– Dietrich Burde
Dec 5 '18 at 21:31
$begingroup$
The argument presented in the question was based on orders of element.
$endgroup$
– Newbie
Dec 5 '18 at 21:34