Why $mathbb{Z}_m times mathbb{Z}_n, , text{gcd}(m,n) >1$ isn't cyclic?












2














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







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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
















2














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







share|cite|improve this question









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














2












2








2


2





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







share|cite|improve this question









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






share|cite|improve this question









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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edited 21 hours ago





















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asked yesterday









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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














  • 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










3 Answers
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3














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.






share|cite|improve this answer





























    1














    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.






    share|cite|improve this answer





























      0














      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.






      share|cite|improve this answer





















      • 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











      Your Answer





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      3 Answers
      3






      active

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      3 Answers
      3






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      3














      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.






      share|cite|improve this answer


























        3














        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.






        share|cite|improve this answer
























          3












          3








          3






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 23 hours ago









          Jevaut

          898111




          898111























              1














              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.






              share|cite|improve this answer


























                1














                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.






                share|cite|improve this answer
























                  1












                  1








                  1






                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  5xum

                  89.6k393161




                  89.6k393161























                      0














                      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.






                      share|cite|improve this answer





















                      • 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
















                      0














                      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.






                      share|cite|improve this answer





















                      • 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














                      0












                      0








                      0






                      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.






                      share|cite|improve this answer












                      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.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      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


















                      • 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










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                      Management

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                      Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
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                      Has there ever been an instance of an active nuclear power plant within or near a war zone?

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                      36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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