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Is every pogroup generated by its own prime elements isomorphic (as pogroup) to some $mathbbZ^(I)$?
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A pogroup (partially ordered group) is a group with a partial order $leq$ such that if $xleq y$ then $zxleq zy$ and $xzleq yz$.
A prime element of a pogroup is an element $x$ such that $x>e$ ($e$ is the neutral element), and for any elements $ygeq e$ and $zgeq e$, if $xleq yz$, then $xleq y$ or $xleq z$.
If the pogroup $X$ is generated by its prime elements and is commutative, it is isomorphic to some $mathbbZ^(I)$. In fact, denoting the group additively, let $I$ be the set of all prime elements, and let $F:mathbbZ^(I)rightarrow X$ given by $F(n)=sum_iin In_ii$, then for $ninmathbbZ^(I)$, if $F(n)geq 0$, then $sum_iin In_iigeq 0$, then let $I^+=iin I:n_i>0$ and $I^-=iin I:n_i<0$, then $sum_iin I^+n_iigeqsum_iin I^-(-n_i)i$, so that for any $ain I^-$, then $aleqsum_iin I^+(-n_i)i$, then there is a $bin I^+$ such that $aleq b$, so $b-ageq0$, but $bleq(b-a)+a$, and $b-a<b$, so $bleq a$, so $a=b$, contradiction; therefore $I^-=emptyset$, so $forall iin I:n_igeq 0$, so $ngeq 0$. Therefore, it is easy to see that $F$ is an isomorphism of pogroups from $mathbbZ^(I)$ to $X$ (because $X$ is generated by its prime elements).
If we drop the commutativity condition, I conjecture that this is no longer true. I tried to consider a free group $X$ over a set $E$ of more than one generators. I considered the set $P$ of products (including the empty one) of conjugates of generators. It is easy to see that $ein P$, that all generators are in $P$, that $PPsubseteq P$ and tha $x^-1Pxsubseteq P$ for all $xin X$. Therefore the relation "$xleq y$" given by $x^-1yin P$ is a preorder in $X$ such that $xleq y$ implies $zxleq zy$ and $xzleq yz$. For $xin X$, if $xin P$ and $x^-1in P$, then $x=prod_i<ma_i$ and $x^-1=prod_j<nb_j$, where $a_i$ and $b_j$ are conjugates of generators, then $e=(prod_i<ma_i)(prod_j<nb_j)$, but there is a homomorphism $f:XrightarrowmathbbZ$ such that $forall gin E:f(g)=1$, then $f(e)=(sum_i<mf(a_i))+(sum_j<nf(b_j))$, then $0=m+n$, so $m=0$, so $x=e$. Therefore $leq$ is an order compatible with the group structure over $X$. Clearly $X$ is not commutative, so that it cannot be isomorphic to any $mathbbZ^(I)$. I conjecture the generators are prime, but I still do not know how to decide this.
group-theory order-theory ordered-groups
asked Jul 28 at 4:57
Daniel Kawai
689
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A pogroup (partially ordered group) is a group with a partial order $leq$ such that if $xleq y$ then $zxleq zy$ and $xzleq yz$.
A prime element of a pogroup is an element $x$ such that $x>e$ ($e$ is the neutral element), and for any elements $ygeq e$ and $zgeq e$, if $xleq yz$, then $xleq y$ or $xleq z$.
If the pogroup $X$ is generated by its prime elements and is commutative, it is isomorphic to some $mathbbZ^(I)$. In fact, denoting the group additively, let $I$ be the set of all prime elements, and let $F:mathbbZ^(I)rightarrow X$ given by $F(n)=sum_iin In_ii$, then for $ninmathbbZ^(I)$, if $F(n)geq 0$, then $sum_iin In_iigeq 0$, then let $I^+=iin I:n_i>0$ and $I^-=iin I:n_i<0$, then $sum_iin I^+n_iigeqsum_iin I^-(-n_i)i$, so that for any $ain I^-$, then $aleqsum_iin I^+(-n_i)i$, then there is a $bin I^+$ such that $aleq b$, so $b-ageq0$, but $bleq(b-a)+a$, and $b-a<b$, so $bleq a$, so $a=b$, contradiction; therefore $I^-=emptyset$, so $forall iin I:n_igeq 0$, so $ngeq 0$. Therefore, it is easy to see that $F$ is an isomorphism of pogroups from $mathbbZ^(I)$ to $X$ (because $X$ is generated by its prime elements).
If we drop the commutativity condition, I conjecture that this is no longer true. I tried to consider a free group $X$ over a set $E$ of more than one generators. I considered the set $P$ of products (including the empty one) of conjugates of generators. It is easy to see that $ein P$, that all generators are in $P$, that $PPsubseteq P$ and tha $x^-1Pxsubseteq P$ for all $xin X$. Therefore the relation "$xleq y$" given by $x^-1yin P$ is a preorder in $X$ such that $xleq y$ implies $zxleq zy$ and $xzleq yz$. For $xin X$, if $xin P$ and $x^-1in P$, then $x=prod_i<ma_i$ and $x^-1=prod_j<nb_j$, where $a_i$ and $b_j$ are conjugates of generators, then $e=(prod_i<ma_i)(prod_j<nb_j)$, but there is a homomorphism $f:XrightarrowmathbbZ$ such that $forall gin E:f(g)=1$, then $f(e)=(sum_i<mf(a_i))+(sum_j<nf(b_j))$, then $0=m+n$, so $m=0$, so $x=e$. Therefore $leq$ is an order compatible with the group structure over $X$. Clearly $X$ is not commutative, so that it cannot be isomorphic to any $mathbbZ^(I)$. I conjecture the generators are prime, but I still do not know how to decide this.
group-theory order-theory ordered-groups
asked Jul 28 at 4:57
Daniel Kawai
689
add a comment |Â
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favorite
up vote
0
down vote
favorite
A pogroup (partially ordered group) is a group with a partial order $leq$ such that if $xleq y$ then $zxleq zy$ and $xzleq yz$.
A prime element of a pogroup is an element $x$ such that $x>e$ ($e$ is the neutral element), and for any elements $ygeq e$ and $zgeq e$, if $xleq yz$, then $xleq y$ or $xleq z$.
If the pogroup $X$ is generated by its prime elements and is commutative, it is isomorphic to some $mathbbZ^(I)$. In fact, denoting the group additively, let $I$ be the set of all prime elements, and let $F:mathbbZ^(I)rightarrow X$ given by $F(n)=sum_iin In_ii$, then for $ninmathbbZ^(I)$, if $F(n)geq 0$, then $sum_iin In_iigeq 0$, then let $I^+=iin I:n_i>0$ and $I^-=iin I:n_i<0$, then $sum_iin I^+n_iigeqsum_iin I^-(-n_i)i$, so that for any $ain I^-$, then $aleqsum_iin I^+(-n_i)i$, then there is a $bin I^+$ such that $aleq b$, so $b-ageq0$, but $bleq(b-a)+a$, and $b-a<b$, so $bleq a$, so $a=b$, contradiction; therefore $I^-=emptyset$, so $forall iin I:n_igeq 0$, so $ngeq 0$. Therefore, it is easy to see that $F$ is an isomorphism of pogroups from $mathbbZ^(I)$ to $X$ (because $X$ is generated by its prime elements).
If we drop the commutativity condition, I conjecture that this is no longer true. I tried to consider a free group $X$ over a set $E$ of more than one generators. I considered the set $P$ of products (including the empty one) of conjugates of generators. It is easy to see that $ein P$, that all generators are in $P$, that $PPsubseteq P$ and tha $x^-1Pxsubseteq P$ for all $xin X$. Therefore the relation "$xleq y$" given by $x^-1yin P$ is a preorder in $X$ such that $xleq y$ implies $zxleq zy$ and $xzleq yz$. For $xin X$, if $xin P$ and $x^-1in P$, then $x=prod_i<ma_i$ and $x^-1=prod_j<nb_j$, where $a_i$ and $b_j$ are conjugates of generators, then $e=(prod_i<ma_i)(prod_j<nb_j)$, but there is a homomorphism $f:XrightarrowmathbbZ$ such that $forall gin E:f(g)=1$, then $f(e)=(sum_i<mf(a_i))+(sum_j<nf(b_j))$, then $0=m+n$, so $m=0$, so $x=e$. Therefore $leq$ is an order compatible with the group structure over $X$. Clearly $X$ is not commutative, so that it cannot be isomorphic to any $mathbbZ^(I)$. I conjecture the generators are prime, but I still do not know how to decide this.
group-theory order-theory ordered-groups
asked Jul 28 at 4:57
Daniel Kawai
689
A pogroup (partially ordered group) is a group with a partial order $leq$ such that if $xleq y$ then $zxleq zy$ and $xzleq yz$.
A prime element of a pogroup is an element $x$ such that $x>e$ ($e$ is the neutral element), and for any elements $ygeq e$ and $zgeq e$, if $xleq yz$, then $xleq y$ or $xleq z$.
If the pogroup $X$ is generated by its prime elements and is commutative, it is isomorphic to some $mathbbZ^(I)$. In fact, denoting the group additively, let $I$ be the set of all prime elements, and let $F:mathbbZ^(I)rightarrow X$ given by $F(n)=sum_iin In_ii$, then for $ninmathbbZ^(I)$, if $F(n)geq 0$, then $sum_iin In_iigeq 0$, then let $I^+=iin I:n_i>0$ and $I^-=iin I:n_i<0$, then $sum_iin I^+n_iigeqsum_iin I^-(-n_i)i$, so that for any $ain I^-$, then $aleqsum_iin I^+(-n_i)i$, then there is a $bin I^+$ such that $aleq b$, so $b-ageq0$, but $bleq(b-a)+a$, and $b-a<b$, so $bleq a$, so $a=b$, contradiction; therefore $I^-=emptyset$, so $forall iin I:n_igeq 0$, so $ngeq 0$. Therefore, it is easy to see that $F$ is an isomorphism of pogroups from $mathbbZ^(I)$ to $X$ (because $X$ is generated by its prime elements).
If we drop the commutativity condition, I conjecture that this is no longer true. I tried to consider a free group $X$ over a set $E$ of more than one generators. I considered the set $P$ of products (including the empty one) of conjugates of generators. It is easy to see that $ein P$, that all generators are in $P$, that $PPsubseteq P$ and tha $x^-1Pxsubseteq P$ for all $xin X$. Therefore the relation "$xleq y$" given by $x^-1yin P$ is a preorder in $X$ such that $xleq y$ implies $zxleq zy$ and $xzleq yz$. For $xin X$, if $xin P$ and $x^-1in P$, then $x=prod_i<ma_i$ and $x^-1=prod_j<nb_j$, where $a_i$ and $b_j$ are conjugates of generators, then $e=(prod_i<ma_i)(prod_j<nb_j)$, but there is a homomorphism $f:XrightarrowmathbbZ$ such that $forall gin E:f(g)=1$, then $f(e)=(sum_i<mf(a_i))+(sum_j<nf(b_j))$, then $0=m+n$, so $m=0$, so $x=e$. Therefore $leq$ is an order compatible with the group structure over $X$. Clearly $X$ is not commutative, so that it cannot be isomorphic to any $mathbbZ^(I)$. I conjecture the generators are prime, but I still do not know how to decide this.
group-theory order-theory ordered-groups
asked Jul 28 at 4:57
Daniel Kawai
689
asked Jul 28 at 4:57
Daniel Kawai
689
asked Jul 28 at 4:57
Daniel Kawai
689
asked Jul 28 at 4:57
Daniel Kawai
689
689
add a comment |Â
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