Why $mathbb{Z}_m times mathbb{Z}_n, , text{gcd}(m,n) >1$ isn't cyclic?
$begingroup$
I'm aware of the theorem which states that $mathbb{Z}_m times mathbb{Z}_n$ is isomorphic to $mathbb{Z}_{mn}$ (and thus cyclic) if and only if $text{gcd}(m,n)=1$.
However, in $G=mathbb{Z}_2 times mathbb{Z}_4$, where $text{gcd}(2,4)=2$, it seems to me that if we take $(0,0)$ and add to it $(1,1)$ $4$ times (least common multiple of $2,4$) then we return to $(0,0)$. That's what misled me into thinking that $G$ is cyclic.
Could you explain to me why this is a faulty syllogism and provide me with a more practical way to visualize cyclic direct products?
abstract-algebra group-theory cyclic-groups
asked Jan 3 at 10:48
LoneBoneLoneBone
968
$endgroup$
add a comment |
$begingroup$
I'm aware of the theorem which states that $mathbb{Z}_m times mathbb{Z}_n$ is isomorphic to $mathbb{Z}_{mn}$ (and thus cyclic) if and only if $text{gcd}(m,n)=1$.
However, in $G=mathbb{Z}_2 times mathbb{Z}_4$, where $text{gcd}(2,4)=2$, it seems to me that if we take $(0,0)$ and add to it $(1,1)$ $4$ times (least common multiple of $2,4$) then we return to $(0,0)$. That's what misled me into thinking that $G$ is cyclic.
Could you explain to me why this is a faulty syllogism and provide me with a more practical way to visualize cyclic direct products?
abstract-algebra group-theory cyclic-groups
asked Jan 3 at 10:48
LoneBoneLoneBone
968
$endgroup$
3
$begingroup$
Well, as a quick response, your logic is faulty because you've found an order $4$ element in an order $8$ group. If you found an order $8$ element, that'd be a different story.
$endgroup$
– Theo Bendit
Jan 3 at 10:51
1
$begingroup$
If it was cyclic you should be able to catch all elements of the group. How do you get $(1,2)$ from adding $(1,1)$ repeatedly?
$endgroup$
– John11
Jan 3 at 10:55
$begingroup$
One thing you must know to begin all reasoning: finding examples can lead and give intuition, but example of positive implication for a case has no value without any other argument related to it. It won't help you see a thing, it won't help you prove a thing. In mathematics, the examples are searched to show existence for a contradiction and/or counter-fact.
$endgroup$
– freehumorist
Jan 3 at 14:22
add a comment |
$begingroup$
I'm aware of the theorem which states that $mathbb{Z}_m times mathbb{Z}_n$ is isomorphic to $mathbb{Z}_{mn}$ (and thus cyclic) if and only if $text{gcd}(m,n)=1$.
However, in $G=mathbb{Z}_2 times mathbb{Z}_4$, where $text{gcd}(2,4)=2$, it seems to me that if we take $(0,0)$ and add to it $(1,1)$ $4$ times (least common multiple of $2,4$) then we return to $(0,0)$. That's what misled me into thinking that $G$ is cyclic.
Could you explain to me why this is a faulty syllogism and provide me with a more practical way to visualize cyclic direct products?
abstract-algebra group-theory cyclic-groups
asked Jan 3 at 10:48
LoneBoneLoneBone
968
$endgroup$
I'm aware of the theorem which states that $mathbb{Z}_m times mathbb{Z}_n$ is isomorphic to $mathbb{Z}_{mn}$ (and thus cyclic) if and only if $text{gcd}(m,n)=1$.
However, in $G=mathbb{Z}_2 times mathbb{Z}_4$, where $text{gcd}(2,4)=2$, it seems to me that if we take $(0,0)$ and add to it $(1,1)$ $4$ times (least common multiple of $2,4$) then we return to $(0,0)$. That's what misled me into thinking that $G$ is cyclic.
Could you explain to me why this is a faulty syllogism and provide me with a more practical way to visualize cyclic direct products?
abstract-algebra group-theory cyclic-groups
abstract-algebra group-theory cyclic-groups
asked Jan 3 at 10:48
LoneBoneLoneBone
968
asked Jan 3 at 10:48
LoneBoneLoneBone
968
edited Jan 3 at 13:35
LoneBone
asked Jan 3 at 10:48
LoneBoneLoneBone
968
asked Jan 3 at 10:48
LoneBoneLoneBone
968
asked Jan 3 at 10:48
LoneBoneLoneBone
968
968
3
$begingroup$
Well, as a quick response, your logic is faulty because you've found an order $4$ element in an order $8$ group. If you found an order $8$ element, that'd be a different story.
$endgroup$
– Theo Bendit
Jan 3 at 10:51
1
$begingroup$
If it was cyclic you should be able to catch all elements of the group. How do you get $(1,2)$ from adding $(1,1)$ repeatedly?
$endgroup$
– John11
Jan 3 at 10:55
$begingroup$
One thing you must know to begin all reasoning: finding examples can lead and give intuition, but example of positive implication for a case has no value without any other argument related to it. It won't help you see a thing, it won't help you prove a thing. In mathematics, the examples are searched to show existence for a contradiction and/or counter-fact.
$endgroup$
– freehumorist
Jan 3 at 14:22
add a comment |
3
$begingroup$
Well, as a quick response, your logic is faulty because you've found an order $4$ element in an order $8$ group. If you found an order $8$ element, that'd be a different story.
$endgroup$
– Theo Bendit
Jan 3 at 10:51
1
$begingroup$
If it was cyclic you should be able to catch all elements of the group. How do you get $(1,2)$ from adding $(1,1)$ repeatedly?
$endgroup$
– John11
Jan 3 at 10:55
$begingroup$
One thing you must know to begin all reasoning: finding examples can lead and give intuition, but example of positive implication for a case has no value without any other argument related to it. It won't help you see a thing, it won't help you prove a thing. In mathematics, the examples are searched to show existence for a contradiction and/or counter-fact.
$endgroup$
– freehumorist
Jan 3 at 14:22
3
3
$begingroup$
Well, as a quick response, your logic is faulty because you've found an order $4$ element in an order $8$ group. If you found an order $8$ element, that'd be a different story.
$endgroup$
– Theo Bendit
Jan 3 at 10:51
$begingroup$
Well, as a quick response, your logic is faulty because you've found an order $4$ element in an order $8$ group. If you found an order $8$ element, that'd be a different story.
$endgroup$
– Theo Bendit
Jan 3 at 10:51
1
1
$begingroup$
If it was cyclic you should be able to catch all elements of the group. How do you get $(1,2)$ from adding $(1,1)$ repeatedly?
$endgroup$
– John11
Jan 3 at 10:55
$begingroup$
If it was cyclic you should be able to catch all elements of the group. How do you get $(1,2)$ from adding $(1,1)$ repeatedly?
$endgroup$
– John11
Jan 3 at 10:55
$begingroup$
One thing you must know to begin all reasoning: finding examples can lead and give intuition, but example of positive implication for a case has no value without any other argument related to it. It won't help you see a thing, it won't help you prove a thing. In mathematics, the examples are searched to show existence for a contradiction and/or counter-fact.
$endgroup$
– freehumorist
Jan 3 at 14:22
$begingroup$
One thing you must know to begin all reasoning: finding examples can lead and give intuition, but example of positive implication for a case has no value without any other argument related to it. It won't help you see a thing, it won't help you prove a thing. In mathematics, the examples are searched to show existence for a contradiction and/or counter-fact.
$endgroup$
– freehumorist
Jan 3 at 14:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Suppose that $m$ and $n$ are not coprime. Let $d=text{gcd}(m,n)$ and
$$
a=frac{mn}{d}
$$
Then $a$ is a multiple of both $m$ and $n$. Suppose that $(r,s) in mathbb{Z}_m times mathbb{Z}_n$. We have
$$a(r,s)=(ar,as)=(0,0),
$$
since $ar$ is a multiple of $m$ and $as$ is a multiple of $n$. Thus, every element of $mathbb{Z}_m times mathbb{Z}_n$ has order at most $a$ which is less than $mn$ and so $mathbb{Z}_m times mathbb{Z}_n$ is not cyclic.
answered Jan 3 at 11:04
JevautJevaut
1,168312
$endgroup$
add a comment |
$begingroup$
In $G$, sure, adding $(1,1)$ to itself $4$ times equals $(0,0)$. But the group $G$ also contains the element $(0,3)$, and there is no amount of adding $(1,1)$ to itself that will result in $(0,3)$.
So, the subgroup $H$ of $G$, generated by $(1,1)$, is cyclic. However, $H$ only has four elements, i.e. $H={(0,0),(1,1),(0,2),(1,3)}$ while $G$ has $8$ elements.
If $G$ were cyclic, there would exist some element of $G$ which would generate all of $G$. Such an element does not exist.
answered Jan 3 at 10:57
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
By the definition of the direct product of groups $Z_m times Z_n$, the order of any element $(a,b)$ is $lcm(o(a),0(b))$ since $Z_n$ and $Z_m$ are cyclic groups so they have elements of order $n$ and $m$ respectively, so maximum order of any element of $Z_m times Z_n$ is lcm(m,n).If $m$ and $n$ are not coprime then $lcm(m,n)< mn$. So $Z_m times Z_n$ does not have any element of order $mn$. Since $Z_m times Z_n$ is a group of order $mn$ which does not have any element of order $mn$ so it cannot be cyclic. In your example also you can see $G$ does not have any element of order 8 so it cannot be cyclic.
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
$begingroup$
Nitpick: There is more than one way to define the direct product of two groups.
$endgroup$
– Shaun
Jan 3 at 11:27
$begingroup$
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
$endgroup$
– MANI SHANKAR PANDEY
Jan 3 at 15:03
add a comment |
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3 Answers
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3 Answers
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$begingroup$
Suppose that $m$ and $n$ are not coprime. Let $d=text{gcd}(m,n)$ and
$$
a=frac{mn}{d}
$$
Then $a$ is a multiple of both $m$ and $n$. Suppose that $(r,s) in mathbb{Z}_m times mathbb{Z}_n$. We have
$$a(r,s)=(ar,as)=(0,0),
$$
since $ar$ is a multiple of $m$ and $as$ is a multiple of $n$. Thus, every element of $mathbb{Z}_m times mathbb{Z}_n$ has order at most $a$ which is less than $mn$ and so $mathbb{Z}_m times mathbb{Z}_n$ is not cyclic.
answered Jan 3 at 11:04
JevautJevaut
1,168312
$endgroup$
add a comment |
$begingroup$
Suppose that $m$ and $n$ are not coprime. Let $d=text{gcd}(m,n)$ and
$$
a=frac{mn}{d}
$$
Then $a$ is a multiple of both $m$ and $n$. Suppose that $(r,s) in mathbb{Z}_m times mathbb{Z}_n$. We have
$$a(r,s)=(ar,as)=(0,0),
$$
since $ar$ is a multiple of $m$ and $as$ is a multiple of $n$. Thus, every element of $mathbb{Z}_m times mathbb{Z}_n$ has order at most $a$ which is less than $mn$ and so $mathbb{Z}_m times mathbb{Z}_n$ is not cyclic.
answered Jan 3 at 11:04
JevautJevaut
1,168312
$endgroup$
add a comment |
$begingroup$
Suppose that $m$ and $n$ are not coprime. Let $d=text{gcd}(m,n)$ and
$$
a=frac{mn}{d}
$$
Then $a$ is a multiple of both $m$ and $n$. Suppose that $(r,s) in mathbb{Z}_m times mathbb{Z}_n$. We have
$$a(r,s)=(ar,as)=(0,0),
$$
since $ar$ is a multiple of $m$ and $as$ is a multiple of $n$. Thus, every element of $mathbb{Z}_m times mathbb{Z}_n$ has order at most $a$ which is less than $mn$ and so $mathbb{Z}_m times mathbb{Z}_n$ is not cyclic.
answered Jan 3 at 11:04
JevautJevaut
1,168312
$endgroup$
Suppose that $m$ and $n$ are not coprime. Let $d=text{gcd}(m,n)$ and
$$
a=frac{mn}{d}
$$
Then $a$ is a multiple of both $m$ and $n$. Suppose that $(r,s) in mathbb{Z}_m times mathbb{Z}_n$. We have
$$a(r,s)=(ar,as)=(0,0),
$$
since $ar$ is a multiple of $m$ and $as$ is a multiple of $n$. Thus, every element of $mathbb{Z}_m times mathbb{Z}_n$ has order at most $a$ which is less than $mn$ and so $mathbb{Z}_m times mathbb{Z}_n$ is not cyclic.
answered Jan 3 at 11:04
JevautJevaut
1,168312
answered Jan 3 at 11:04
JevautJevaut
1,168312
answered Jan 3 at 11:04
JevautJevaut
1,168312
answered Jan 3 at 11:04
JevautJevaut
1,168312
1,168312
add a comment |
add a comment |
$begingroup$
In $G$, sure, adding $(1,1)$ to itself $4$ times equals $(0,0)$. But the group $G$ also contains the element $(0,3)$, and there is no amount of adding $(1,1)$ to itself that will result in $(0,3)$.
So, the subgroup $H$ of $G$, generated by $(1,1)$, is cyclic. However, $H$ only has four elements, i.e. $H={(0,0),(1,1),(0,2),(1,3)}$ while $G$ has $8$ elements.
If $G$ were cyclic, there would exist some element of $G$ which would generate all of $G$. Such an element does not exist.
answered Jan 3 at 10:57
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
In $G$, sure, adding $(1,1)$ to itself $4$ times equals $(0,0)$. But the group $G$ also contains the element $(0,3)$, and there is no amount of adding $(1,1)$ to itself that will result in $(0,3)$.
So, the subgroup $H$ of $G$, generated by $(1,1)$, is cyclic. However, $H$ only has four elements, i.e. $H={(0,0),(1,1),(0,2),(1,3)}$ while $G$ has $8$ elements.
If $G$ were cyclic, there would exist some element of $G$ which would generate all of $G$. Such an element does not exist.
answered Jan 3 at 10:57
5xum5xum
91.8k394161
$endgroup$
add a comment |
$begingroup$
In $G$, sure, adding $(1,1)$ to itself $4$ times equals $(0,0)$. But the group $G$ also contains the element $(0,3)$, and there is no amount of adding $(1,1)$ to itself that will result in $(0,3)$.
So, the subgroup $H$ of $G$, generated by $(1,1)$, is cyclic. However, $H$ only has four elements, i.e. $H={(0,0),(1,1),(0,2),(1,3)}$ while $G$ has $8$ elements.
If $G$ were cyclic, there would exist some element of $G$ which would generate all of $G$. Such an element does not exist.
answered Jan 3 at 10:57
5xum5xum
91.8k394161
$endgroup$
In $G$, sure, adding $(1,1)$ to itself $4$ times equals $(0,0)$. But the group $G$ also contains the element $(0,3)$, and there is no amount of adding $(1,1)$ to itself that will result in $(0,3)$.
So, the subgroup $H$ of $G$, generated by $(1,1)$, is cyclic. However, $H$ only has four elements, i.e. $H={(0,0),(1,1),(0,2),(1,3)}$ while $G$ has $8$ elements.
If $G$ were cyclic, there would exist some element of $G$ which would generate all of $G$. Such an element does not exist.
answered Jan 3 at 10:57
5xum5xum
91.8k394161
answered Jan 3 at 10:57
5xum5xum
91.8k394161
answered Jan 3 at 10:57
5xum5xum
91.8k394161
answered Jan 3 at 10:57
5xum5xum
91.8k394161
91.8k394161
add a comment |
add a comment |
$begingroup$
By the definition of the direct product of groups $Z_m times Z_n$, the order of any element $(a,b)$ is $lcm(o(a),0(b))$ since $Z_n$ and $Z_m$ are cyclic groups so they have elements of order $n$ and $m$ respectively, so maximum order of any element of $Z_m times Z_n$ is lcm(m,n).If $m$ and $n$ are not coprime then $lcm(m,n)< mn$. So $Z_m times Z_n$ does not have any element of order $mn$. Since $Z_m times Z_n$ is a group of order $mn$ which does not have any element of order $mn$ so it cannot be cyclic. In your example also you can see $G$ does not have any element of order 8 so it cannot be cyclic.
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
$begingroup$
Nitpick: There is more than one way to define the direct product of two groups.
$endgroup$
– Shaun
Jan 3 at 11:27
$begingroup$
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
$endgroup$
– MANI SHANKAR PANDEY
Jan 3 at 15:03
add a comment |
$begingroup$
By the definition of the direct product of groups $Z_m times Z_n$, the order of any element $(a,b)$ is $lcm(o(a),0(b))$ since $Z_n$ and $Z_m$ are cyclic groups so they have elements of order $n$ and $m$ respectively, so maximum order of any element of $Z_m times Z_n$ is lcm(m,n).If $m$ and $n$ are not coprime then $lcm(m,n)< mn$. So $Z_m times Z_n$ does not have any element of order $mn$. Since $Z_m times Z_n$ is a group of order $mn$ which does not have any element of order $mn$ so it cannot be cyclic. In your example also you can see $G$ does not have any element of order 8 so it cannot be cyclic.
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
$begingroup$
Nitpick: There is more than one way to define the direct product of two groups.
$endgroup$
– Shaun
Jan 3 at 11:27
$begingroup$
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
$endgroup$
– MANI SHANKAR PANDEY
Jan 3 at 15:03
add a comment |
$begingroup$
By the definition of the direct product of groups $Z_m times Z_n$, the order of any element $(a,b)$ is $lcm(o(a),0(b))$ since $Z_n$ and $Z_m$ are cyclic groups so they have elements of order $n$ and $m$ respectively, so maximum order of any element of $Z_m times Z_n$ is lcm(m,n).If $m$ and $n$ are not coprime then $lcm(m,n)< mn$. So $Z_m times Z_n$ does not have any element of order $mn$. Since $Z_m times Z_n$ is a group of order $mn$ which does not have any element of order $mn$ so it cannot be cyclic. In your example also you can see $G$ does not have any element of order 8 so it cannot be cyclic.
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
$endgroup$
By the definition of the direct product of groups $Z_m times Z_n$, the order of any element $(a,b)$ is $lcm(o(a),0(b))$ since $Z_n$ and $Z_m$ are cyclic groups so they have elements of order $n$ and $m$ respectively, so maximum order of any element of $Z_m times Z_n$ is lcm(m,n).If $m$ and $n$ are not coprime then $lcm(m,n)< mn$. So $Z_m times Z_n$ does not have any element of order $mn$. Since $Z_m times Z_n$ is a group of order $mn$ which does not have any element of order $mn$ so it cannot be cyclic. In your example also you can see $G$ does not have any element of order 8 so it cannot be cyclic.
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
answered Jan 3 at 11:25
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
548
548
$begingroup$
Nitpick: There is more than one way to define the direct product of two groups.
$endgroup$
– Shaun
Jan 3 at 11:27
$begingroup$
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
$endgroup$
– MANI SHANKAR PANDEY
Jan 3 at 15:03
add a comment |
$begingroup$
Nitpick: There is more than one way to define the direct product of two groups.
$endgroup$
– Shaun
Jan 3 at 11:27
$begingroup$
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
$endgroup$
– MANI SHANKAR PANDEY
Jan 3 at 15:03
$begingroup$
Nitpick: There is more than one way to define the direct product of two groups.
$endgroup$
– Shaun
Jan 3 at 11:27
$begingroup$
Nitpick: There is more than one way to define the direct product of two groups.
$endgroup$
– Shaun
Jan 3 at 11:27
$begingroup$
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
$endgroup$
– MANI SHANKAR PANDEY
Jan 3 at 15:03
$begingroup$
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
$endgroup$
– MANI SHANKAR PANDEY
Jan 3 at 15:03
add a comment |
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$begingroup$
Well, as a quick response, your logic is faulty because you've found an order $4$ element in an order $8$ group. If you found an order $8$ element, that'd be a different story.
$endgroup$
– Theo Bendit
Jan 3 at 10:51
1
$begingroup$
If it was cyclic you should be able to catch all elements of the group. How do you get $(1,2)$ from adding $(1,1)$ repeatedly?
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– John11
Jan 3 at 10:55
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One thing you must know to begin all reasoning: finding examples can lead and give intuition, but example of positive implication for a case has no value without any other argument related to it. It won't help you see a thing, it won't help you prove a thing. In mathematics, the examples are searched to show existence for a contradiction and/or counter-fact.
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– freehumorist
Jan 3 at 14:22