Mixing
A group $G$ with the following property …
$begingroup$
Let $G$ be a group with the following property : Given any positive integers $m,n$ and $r$, there exists elements $g$ and $h$ in $G$ such that $o(g) = m$, $o(h) = n$ and $o(gcdot h) = r$ ; where $o(cdot)$ denotes an order of an element in $G$. Then which of the following are necessarily $textbf{true}$ ?
(1.) $G$ has to be an infinite group.
(2.) $G$ cannot be a cyclic group.
(3.) $G$ has infinitely many cyclic subgroups.
(4.) $G$ has to be a non-abelian group.
I even don't know how to start. Please can anyone give me an idea or a hint ?
Thanks in advance.
group-theory
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group with the following property : Given any positive integers $m,n$ and $r$, there exists elements $g$ and $h$ in $G$ such that $o(g) = m$, $o(h) = n$ and $o(gcdot h) = r$ ; where $o(cdot)$ denotes an order of an element in $G$. Then which of the following are necessarily $textbf{true}$ ?
(1.) $G$ has to be an infinite group.
(2.) $G$ cannot be a cyclic group.
(3.) $G$ has infinitely many cyclic subgroups.
(4.) $G$ has to be a non-abelian group.
I even don't know how to start. Please can anyone give me an idea or a hint ?
Thanks in advance.
group-theory
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
$endgroup$
$begingroup$
Can $G$ be finite? What happens if m exceeds the order of the group? If the group is commutative, then think about the relation between the order of two elements and that of their product.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 8:26
$begingroup$
If you determine the answer to the problem (1.), which thanks to the comment above should not pose a huge problem, then think about "how does the infinite cyclic group look like" and whether is it possible to have an element of finite order in such group.
$endgroup$
– I_Really_Want_To_Heal_Myself
Jan 17 at 8:39
$begingroup$
@UDAY PATEL Does the answer help you? If so, please accept it. If not, tell me how to improve it.
$endgroup$
– James
Jan 18 at 6:31
add a comment |
$begingroup$
Let $G$ be a group with the following property : Given any positive integers $m,n$ and $r$, there exists elements $g$ and $h$ in $G$ such that $o(g) = m$, $o(h) = n$ and $o(gcdot h) = r$ ; where $o(cdot)$ denotes an order of an element in $G$. Then which of the following are necessarily $textbf{true}$ ?
(1.) $G$ has to be an infinite group.
(2.) $G$ cannot be a cyclic group.
(3.) $G$ has infinitely many cyclic subgroups.
(4.) $G$ has to be a non-abelian group.
I even don't know how to start. Please can anyone give me an idea or a hint ?
Thanks in advance.
group-theory
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
$endgroup$
Let $G$ be a group with the following property : Given any positive integers $m,n$ and $r$, there exists elements $g$ and $h$ in $G$ such that $o(g) = m$, $o(h) = n$ and $o(gcdot h) = r$ ; where $o(cdot)$ denotes an order of an element in $G$. Then which of the following are necessarily $textbf{true}$ ?
(1.) $G$ has to be an infinite group.
(2.) $G$ cannot be a cyclic group.
(3.) $G$ has infinitely many cyclic subgroups.
(4.) $G$ has to be a non-abelian group.
I even don't know how to start. Please can anyone give me an idea or a hint ?
Thanks in advance.
group-theory
group-theory
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
asked Jan 17 at 8:23
UDAY PATELUDAY PATEL
236213
236213
$begingroup$
Can $G$ be finite? What happens if m exceeds the order of the group? If the group is commutative, then think about the relation between the order of two elements and that of their product.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 8:26
$begingroup$
If you determine the answer to the problem (1.), which thanks to the comment above should not pose a huge problem, then think about "how does the infinite cyclic group look like" and whether is it possible to have an element of finite order in such group.
$endgroup$
– I_Really_Want_To_Heal_Myself
Jan 17 at 8:39
$begingroup$
@UDAY PATEL Does the answer help you? If so, please accept it. If not, tell me how to improve it.
$endgroup$
– James
Jan 18 at 6:31
add a comment |
$begingroup$
Can $G$ be finite? What happens if m exceeds the order of the group? If the group is commutative, then think about the relation between the order of two elements and that of their product.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 8:26
$begingroup$
If you determine the answer to the problem (1.), which thanks to the comment above should not pose a huge problem, then think about "how does the infinite cyclic group look like" and whether is it possible to have an element of finite order in such group.
$endgroup$
– I_Really_Want_To_Heal_Myself
Jan 17 at 8:39
$begingroup$
@UDAY PATEL Does the answer help you? If so, please accept it. If not, tell me how to improve it.
$endgroup$
– James
Jan 18 at 6:31
$begingroup$
Can $G$ be finite? What happens if m exceeds the order of the group? If the group is commutative, then think about the relation between the order of two elements and that of their product.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 8:26
$begingroup$
Can $G$ be finite? What happens if m exceeds the order of the group? If the group is commutative, then think about the relation between the order of two elements and that of their product.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 8:26
$begingroup$
If you determine the answer to the problem (1.), which thanks to the comment above should not pose a huge problem, then think about "how does the infinite cyclic group look like" and whether is it possible to have an element of finite order in such group.
$endgroup$
– I_Really_Want_To_Heal_Myself
Jan 17 at 8:39
$begingroup$
If you determine the answer to the problem (1.), which thanks to the comment above should not pose a huge problem, then think about "how does the infinite cyclic group look like" and whether is it possible to have an element of finite order in such group.
$endgroup$
– I_Really_Want_To_Heal_Myself
Jan 17 at 8:39
$begingroup$
@UDAY PATEL Does the answer help you? If so, please accept it. If not, tell me how to improve it.
$endgroup$
– James
Jan 18 at 6:31
$begingroup$
@UDAY PATEL Does the answer help you? If so, please accept it. If not, tell me how to improve it.
$endgroup$
– James
Jan 18 at 6:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The order of a group element $g$ is defined to be minimum $k$ such that $g^k=e$, where $e$ denote the neutral element. The condition that you have elements of arbitrary orders gives you that the group can't be cyclic (because then it would necessarily have finite order). Thus, 2.) is always true.
Also since the order is unique and you need to have at least one element for every integer that hast his integer as order, the group cannot be finite and 1.) is true.
Finally, you may look at the orbit $Gg$ of any element $g$ of order $n$. This is a subgroup of $G$ and cyclic of order $n$. Thus, 3.) is also true.
Concerning 4.): Using Order of product of two elements in a group we know that in an Abelian group the order of the product of two elements has to be divisible by the order of the product. Thus assume such a group exists and that it is Abelian. Then let $g,h$ be group elements of orders $n,m$ respectively, and take $r$ such that $rnmid mn$. Then, $r=o(gh)mid o(g)o(h)=mn$ which contradicts $rnmid mn$. Thus the group cannot be Abelian and 4.) is true.
EDIT: As is discussed in the comments under this answer, such a group actually doesn't exist.
answered Jan 17 at 8:38
JamesJames
1,594217
$endgroup$
$begingroup$
Probably 4) is also true, since for commuting $a,b$ the order of $ab$ has to be a divisor of the product of the orders of $a$ and of $b$.
$endgroup$
– Jens Schwaiger
Jan 17 at 8:43
$begingroup$
@JensSchwaiger But then take $m,n,r$ such thet $r$ does not divide $mn$ and you obtain a counterexample. Because $r=o(mn)|o(a)o(b)=mn$ (as you just wrote) but we took $r$ with $rnmid mn$. Note that the assertion from the op was that we can choose $m,n,r$ arbitrary.
$endgroup$
– James
Jan 17 at 8:57
1
$begingroup$
An additional task would be to find a groups with the property asked by the OP.
$endgroup$
– Jens Schwaiger
Jan 17 at 9:12
1
$begingroup$
In the link I mentioned above a finite example $G_{m,n,r}$ is constructed for each choice of $m, n r$, that is, in $G_{m,n,r}$ there are elements $g, h$ of order $m, n$ such that $g h$ has order $r$. Now take the direct product of all the $G_{m,n,r}$. This will have the required property.
$endgroup$
– Andreas Caranti
Jan 17 at 13:15
1
$begingroup$
If seems that $r=1$ is only possible when $n=m$ since $o(gh)=1$ implies $gh=e$ or $h=g^{-1}$. And this implies $o(h)=o(g^{-1})=o(g)$. Therefore there is no groups satisfying the required property.
$endgroup$
– Jens Schwaiger
Jan 18 at 7:54
|
show 8 more comments
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$begingroup$
The order of a group element $g$ is defined to be minimum $k$ such that $g^k=e$, where $e$ denote the neutral element. The condition that you have elements of arbitrary orders gives you that the group can't be cyclic (because then it would necessarily have finite order). Thus, 2.) is always true.
Also since the order is unique and you need to have at least one element for every integer that hast his integer as order, the group cannot be finite and 1.) is true.
Finally, you may look at the orbit $Gg$ of any element $g$ of order $n$. This is a subgroup of $G$ and cyclic of order $n$. Thus, 3.) is also true.
Concerning 4.): Using Order of product of two elements in a group we know that in an Abelian group the order of the product of two elements has to be divisible by the order of the product. Thus assume such a group exists and that it is Abelian. Then let $g,h$ be group elements of orders $n,m$ respectively, and take $r$ such that $rnmid mn$. Then, $r=o(gh)mid o(g)o(h)=mn$ which contradicts $rnmid mn$. Thus the group cannot be Abelian and 4.) is true.
EDIT: As is discussed in the comments under this answer, such a group actually doesn't exist.
answered Jan 17 at 8:38
JamesJames
1,594217
$endgroup$
$begingroup$
Probably 4) is also true, since for commuting $a,b$ the order of $ab$ has to be a divisor of the product of the orders of $a$ and of $b$.
$endgroup$
– Jens Schwaiger
Jan 17 at 8:43
$begingroup$
@JensSchwaiger But then take $m,n,r$ such thet $r$ does not divide $mn$ and you obtain a counterexample. Because $r=o(mn)|o(a)o(b)=mn$ (as you just wrote) but we took $r$ with $rnmid mn$. Note that the assertion from the op was that we can choose $m,n,r$ arbitrary.
$endgroup$
– James
Jan 17 at 8:57
1
$begingroup$
An additional task would be to find a groups with the property asked by the OP.
$endgroup$
– Jens Schwaiger
Jan 17 at 9:12
1
$begingroup$
In the link I mentioned above a finite example $G_{m,n,r}$ is constructed for each choice of $m, n r$, that is, in $G_{m,n,r}$ there are elements $g, h$ of order $m, n$ such that $g h$ has order $r$. Now take the direct product of all the $G_{m,n,r}$. This will have the required property.
$endgroup$
– Andreas Caranti
Jan 17 at 13:15
1
$begingroup$
If seems that $r=1$ is only possible when $n=m$ since $o(gh)=1$ implies $gh=e$ or $h=g^{-1}$. And this implies $o(h)=o(g^{-1})=o(g)$. Therefore there is no groups satisfying the required property.
$endgroup$
– Jens Schwaiger
Jan 18 at 7:54
|
show 8 more comments
$begingroup$
The order of a group element $g$ is defined to be minimum $k$ such that $g^k=e$, where $e$ denote the neutral element. The condition that you have elements of arbitrary orders gives you that the group can't be cyclic (because then it would necessarily have finite order). Thus, 2.) is always true.
Also since the order is unique and you need to have at least one element for every integer that hast his integer as order, the group cannot be finite and 1.) is true.
Finally, you may look at the orbit $Gg$ of any element $g$ of order $n$. This is a subgroup of $G$ and cyclic of order $n$. Thus, 3.) is also true.
Concerning 4.): Using Order of product of two elements in a group we know that in an Abelian group the order of the product of two elements has to be divisible by the order of the product. Thus assume such a group exists and that it is Abelian. Then let $g,h$ be group elements of orders $n,m$ respectively, and take $r$ such that $rnmid mn$. Then, $r=o(gh)mid o(g)o(h)=mn$ which contradicts $rnmid mn$. Thus the group cannot be Abelian and 4.) is true.
EDIT: As is discussed in the comments under this answer, such a group actually doesn't exist.
answered Jan 17 at 8:38
JamesJames
1,594217
$endgroup$
$begingroup$
Probably 4) is also true, since for commuting $a,b$ the order of $ab$ has to be a divisor of the product of the orders of $a$ and of $b$.
$endgroup$
– Jens Schwaiger
Jan 17 at 8:43
$begingroup$
@JensSchwaiger But then take $m,n,r$ such thet $r$ does not divide $mn$ and you obtain a counterexample. Because $r=o(mn)|o(a)o(b)=mn$ (as you just wrote) but we took $r$ with $rnmid mn$. Note that the assertion from the op was that we can choose $m,n,r$ arbitrary.
$endgroup$
– James
Jan 17 at 8:57
1
$begingroup$
An additional task would be to find a groups with the property asked by the OP.
$endgroup$
– Jens Schwaiger
Jan 17 at 9:12
1
$begingroup$
In the link I mentioned above a finite example $G_{m,n,r}$ is constructed for each choice of $m, n r$, that is, in $G_{m,n,r}$ there are elements $g, h$ of order $m, n$ such that $g h$ has order $r$. Now take the direct product of all the $G_{m,n,r}$. This will have the required property.
$endgroup$
– Andreas Caranti
Jan 17 at 13:15
1
$begingroup$
If seems that $r=1$ is only possible when $n=m$ since $o(gh)=1$ implies $gh=e$ or $h=g^{-1}$. And this implies $o(h)=o(g^{-1})=o(g)$. Therefore there is no groups satisfying the required property.
$endgroup$
– Jens Schwaiger
Jan 18 at 7:54
|
show 8 more comments
$begingroup$
The order of a group element $g$ is defined to be minimum $k$ such that $g^k=e$, where $e$ denote the neutral element. The condition that you have elements of arbitrary orders gives you that the group can't be cyclic (because then it would necessarily have finite order). Thus, 2.) is always true.
Also since the order is unique and you need to have at least one element for every integer that hast his integer as order, the group cannot be finite and 1.) is true.
Finally, you may look at the orbit $Gg$ of any element $g$ of order $n$. This is a subgroup of $G$ and cyclic of order $n$. Thus, 3.) is also true.
Concerning 4.): Using Order of product of two elements in a group we know that in an Abelian group the order of the product of two elements has to be divisible by the order of the product. Thus assume such a group exists and that it is Abelian. Then let $g,h$ be group elements of orders $n,m$ respectively, and take $r$ such that $rnmid mn$. Then, $r=o(gh)mid o(g)o(h)=mn$ which contradicts $rnmid mn$. Thus the group cannot be Abelian and 4.) is true.
EDIT: As is discussed in the comments under this answer, such a group actually doesn't exist.
answered Jan 17 at 8:38
JamesJames
1,594217
$endgroup$
The order of a group element $g$ is defined to be minimum $k$ such that $g^k=e$, where $e$ denote the neutral element. The condition that you have elements of arbitrary orders gives you that the group can't be cyclic (because then it would necessarily have finite order). Thus, 2.) is always true.
Also since the order is unique and you need to have at least one element for every integer that hast his integer as order, the group cannot be finite and 1.) is true.
Finally, you may look at the orbit $Gg$ of any element $g$ of order $n$. This is a subgroup of $G$ and cyclic of order $n$. Thus, 3.) is also true.
Concerning 4.): Using Order of product of two elements in a group we know that in an Abelian group the order of the product of two elements has to be divisible by the order of the product. Thus assume such a group exists and that it is Abelian. Then let $g,h$ be group elements of orders $n,m$ respectively, and take $r$ such that $rnmid mn$. Then, $r=o(gh)mid o(g)o(h)=mn$ which contradicts $rnmid mn$. Thus the group cannot be Abelian and 4.) is true.
EDIT: As is discussed in the comments under this answer, such a group actually doesn't exist.
answered Jan 17 at 8:38
JamesJames
1,594217
edited Jan 23 at 10:50
answered Jan 17 at 8:38
JamesJames
1,594217
answered Jan 17 at 8:38
JamesJames
1,594217
answered Jan 17 at 8:38
JamesJames
1,594217
1,594217
$begingroup$
Probably 4) is also true, since for commuting $a,b$ the order of $ab$ has to be a divisor of the product of the orders of $a$ and of $b$.
$endgroup$
– Jens Schwaiger
Jan 17 at 8:43
$begingroup$
@JensSchwaiger But then take $m,n,r$ such thet $r$ does not divide $mn$ and you obtain a counterexample. Because $r=o(mn)|o(a)o(b)=mn$ (as you just wrote) but we took $r$ with $rnmid mn$. Note that the assertion from the op was that we can choose $m,n,r$ arbitrary.
$endgroup$
– James
Jan 17 at 8:57
1
$begingroup$
An additional task would be to find a groups with the property asked by the OP.
$endgroup$
– Jens Schwaiger
Jan 17 at 9:12
1
$begingroup$
In the link I mentioned above a finite example $G_{m,n,r}$ is constructed for each choice of $m, n r$, that is, in $G_{m,n,r}$ there are elements $g, h$ of order $m, n$ such that $g h$ has order $r$. Now take the direct product of all the $G_{m,n,r}$. This will have the required property.
$endgroup$
– Andreas Caranti
Jan 17 at 13:15
1
$begingroup$
If seems that $r=1$ is only possible when $n=m$ since $o(gh)=1$ implies $gh=e$ or $h=g^{-1}$. And this implies $o(h)=o(g^{-1})=o(g)$. Therefore there is no groups satisfying the required property.
$endgroup$
– Jens Schwaiger
Jan 18 at 7:54
|
show 8 more comments
$begingroup$
Probably 4) is also true, since for commuting $a,b$ the order of $ab$ has to be a divisor of the product of the orders of $a$ and of $b$.
$endgroup$
– Jens Schwaiger
Jan 17 at 8:43
$begingroup$
@JensSchwaiger But then take $m,n,r$ such thet $r$ does not divide $mn$ and you obtain a counterexample. Because $r=o(mn)|o(a)o(b)=mn$ (as you just wrote) but we took $r$ with $rnmid mn$. Note that the assertion from the op was that we can choose $m,n,r$ arbitrary.
$endgroup$
– James
Jan 17 at 8:57
1
$begingroup$
An additional task would be to find a groups with the property asked by the OP.
$endgroup$
– Jens Schwaiger
Jan 17 at 9:12
1
$begingroup$
In the link I mentioned above a finite example $G_{m,n,r}$ is constructed for each choice of $m, n r$, that is, in $G_{m,n,r}$ there are elements $g, h$ of order $m, n$ such that $g h$ has order $r$. Now take the direct product of all the $G_{m,n,r}$. This will have the required property.
$endgroup$
– Andreas Caranti
Jan 17 at 13:15
1
$begingroup$
If seems that $r=1$ is only possible when $n=m$ since $o(gh)=1$ implies $gh=e$ or $h=g^{-1}$. And this implies $o(h)=o(g^{-1})=o(g)$. Therefore there is no groups satisfying the required property.
$endgroup$
– Jens Schwaiger
Jan 18 at 7:54
$begingroup$
Probably 4) is also true, since for commuting $a,b$ the order of $ab$ has to be a divisor of the product of the orders of $a$ and of $b$.
$endgroup$
– Jens Schwaiger
Jan 17 at 8:43
$begingroup$
Probably 4) is also true, since for commuting $a,b$ the order of $ab$ has to be a divisor of the product of the orders of $a$ and of $b$.
$endgroup$
– Jens Schwaiger
Jan 17 at 8:43
$begingroup$
@JensSchwaiger But then take $m,n,r$ such thet $r$ does not divide $mn$ and you obtain a counterexample. Because $r=o(mn)|o(a)o(b)=mn$ (as you just wrote) but we took $r$ with $rnmid mn$. Note that the assertion from the op was that we can choose $m,n,r$ arbitrary.
$endgroup$
– James
Jan 17 at 8:57
$begingroup$
@JensSchwaiger But then take $m,n,r$ such thet $r$ does not divide $mn$ and you obtain a counterexample. Because $r=o(mn)|o(a)o(b)=mn$ (as you just wrote) but we took $r$ with $rnmid mn$. Note that the assertion from the op was that we can choose $m,n,r$ arbitrary.
$endgroup$
– James
Jan 17 at 8:57
1
1
$begingroup$
An additional task would be to find a groups with the property asked by the OP.
$endgroup$
– Jens Schwaiger
Jan 17 at 9:12
$begingroup$
An additional task would be to find a groups with the property asked by the OP.
$endgroup$
– Jens Schwaiger
Jan 17 at 9:12
1
1
$begingroup$
In the link I mentioned above a finite example $G_{m,n,r}$ is constructed for each choice of $m, n r$, that is, in $G_{m,n,r}$ there are elements $g, h$ of order $m, n$ such that $g h$ has order $r$. Now take the direct product of all the $G_{m,n,r}$. This will have the required property.
$endgroup$
– Andreas Caranti
Jan 17 at 13:15
$begingroup$
In the link I mentioned above a finite example $G_{m,n,r}$ is constructed for each choice of $m, n r$, that is, in $G_{m,n,r}$ there are elements $g, h$ of order $m, n$ such that $g h$ has order $r$. Now take the direct product of all the $G_{m,n,r}$. This will have the required property.
$endgroup$
– Andreas Caranti
Jan 17 at 13:15
1
1
$begingroup$
If seems that $r=1$ is only possible when $n=m$ since $o(gh)=1$ implies $gh=e$ or $h=g^{-1}$. And this implies $o(h)=o(g^{-1})=o(g)$. Therefore there is no groups satisfying the required property.
$endgroup$
– Jens Schwaiger
Jan 18 at 7:54
$begingroup$
If seems that $r=1$ is only possible when $n=m$ since $o(gh)=1$ implies $gh=e$ or $h=g^{-1}$. And this implies $o(h)=o(g^{-1})=o(g)$. Therefore there is no groups satisfying the required property.
$endgroup$
– Jens Schwaiger
Jan 18 at 7:54
|
show 8 more comments
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$begingroup$
Can $G$ be finite? What happens if m exceeds the order of the group? If the group is commutative, then think about the relation between the order of two elements and that of their product.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 17 at 8:26
$begingroup$
If you determine the answer to the problem (1.), which thanks to the comment above should not pose a huge problem, then think about "how does the infinite cyclic group look like" and whether is it possible to have an element of finite order in such group.
$endgroup$
– I_Really_Want_To_Heal_Myself
Jan 17 at 8:39
$begingroup$
@UDAY PATEL Does the answer help you? If so, please accept it. If not, tell me how to improve it.
$endgroup$
– James
Jan 18 at 6:31