Why $mathbb{Z}_m times mathbb{Z}_n, , text{gcd}(m,n) >1$ isn't cyclic?
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 yesterday
LoneBone
618
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add a comment |
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 yesterday
LoneBone
618
New contributor
LoneBone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3
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.
– Theo Bendit
yesterday
1
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?
– John11
yesterday
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.
– freehumorist
20 hours ago
add a comment |
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 yesterday
LoneBone
618
New contributor
LoneBone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 yesterday
LoneBone
618
New contributor
LoneBone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
LoneBone
618
New contributor
LoneBone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 21 hours ago
asked yesterday
LoneBone
618
New contributor
LoneBone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
LoneBone
618
asked yesterday
LoneBone
618
618
New contributor
LoneBone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
LoneBone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
3
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.
– Theo Bendit
yesterday
1
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?
– John11
yesterday
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.
– freehumorist
20 hours ago
add a comment |
3
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.
– Theo Bendit
yesterday
1
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?
– John11
yesterday
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.
– freehumorist
20 hours ago
3
3
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.
– Theo Bendit
yesterday
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.
– Theo Bendit
yesterday
1
1
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?
– John11
yesterday
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?
– John11
yesterday
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.
– freehumorist
20 hours ago
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.
– freehumorist
20 hours ago
add a comment |
3 Answers
3
active
oldest
votes
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 23 hours ago
Jevaut
898111
add a comment |
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 yesterday
5xum
89.6k393161
add a comment |
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 23 hours ago
MANI SHANKAR PANDEY
387
Nitpick: There is more than one way to define the direct product of two groups.
– Shaun
23 hours ago
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
– MANI SHANKAR PANDEY
19 hours ago
add a comment |
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3 Answers
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3 Answers
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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 23 hours ago
Jevaut
898111
add a comment |
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 23 hours ago
Jevaut
898111
add a comment |
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 23 hours ago
Jevaut
898111
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 23 hours ago
Jevaut
898111
answered 23 hours ago
Jevaut
898111
answered 23 hours ago
Jevaut
898111
answered 23 hours ago
Jevaut
898111
898111
add a comment |
add a comment |
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 yesterday
5xum
89.6k393161
add a comment |
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 yesterday
5xum
89.6k393161
add a comment |
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 yesterday
5xum
89.6k393161
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 yesterday
5xum
89.6k393161
answered yesterday
5xum
89.6k393161
answered yesterday
5xum
89.6k393161
answered yesterday
5xum
89.6k393161
89.6k393161
add a comment |
add a comment |
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 23 hours ago
MANI SHANKAR PANDEY
387
Nitpick: There is more than one way to define the direct product of two groups.
– Shaun
23 hours ago
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
– MANI SHANKAR PANDEY
19 hours ago
add a comment |
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 23 hours ago
MANI SHANKAR PANDEY
387
Nitpick: There is more than one way to define the direct product of two groups.
– Shaun
23 hours ago
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
– MANI SHANKAR PANDEY
19 hours ago
add a comment |
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 23 hours ago
MANI SHANKAR PANDEY
387
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 23 hours ago
MANI SHANKAR PANDEY
387
answered 23 hours ago
MANI SHANKAR PANDEY
387
answered 23 hours ago
MANI SHANKAR PANDEY
387
answered 23 hours ago
MANI SHANKAR PANDEY
387
387
Nitpick: There is more than one way to define the direct product of two groups.
– Shaun
23 hours ago
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
– MANI SHANKAR PANDEY
19 hours ago
add a comment |
Nitpick: There is more than one way to define the direct product of two groups.
– Shaun
23 hours ago
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
– MANI SHANKAR PANDEY
19 hours ago
Nitpick: There is more than one way to define the direct product of two groups.
– Shaun
23 hours ago
Nitpick: There is more than one way to define the direct product of two groups.
– Shaun
23 hours ago
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
– MANI SHANKAR PANDEY
19 hours ago
I think that is called semidirect product. I have followed abstract algebra byDummit and Foote
– MANI SHANKAR PANDEY
19 hours ago
add a comment |
LoneBone is a new contributor. Be nice, and check out our Code of Conduct.
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3
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.
– Theo Bendit
yesterday
1
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?
– John11
yesterday
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.
– freehumorist
20 hours ago