Equivalent condition for a group to be cyclic [duplicate]












0












$begingroup$



This question already has an answer here:




  • Show that a group $G$ is cyclic if and only if every subgroup of $G$ has the form of $G^k$.

    2 answers




Group $G$ is cyclic $iff $



Every subgroup of $G$ is like $G^m = {g^m | gin G}$ for some $m in mathbb N$.



$G^m = {g^m | gin G}$ means $forall x in G^m, exists, g in G$ s.t. $x=g^m$.



$Rightarrow$ is trivial.$,$What about $Leftarrow$?



My idea:$,$If $G$ is infinite group, it's not very hard.



Suppose $a in G$ has infinite order, $langle arangle=G^m$ for some $m>0$, then $exists , b in G$ s.t. $a=b^m$. Then $langle brangle=G^n$ for some $n>0$, then $exists , c in G$ s.t. $b=c^n$. Since $c^m in G^m = langle arangle$, $exists ,i in mathbb N $ s.t. $c^m=a^i=b^{mi}=c^{nmi}$. $c$ clearly has infinite order, so $m=nmi$, $1=ni,n=1$.$langle brangle=G$ is cyclic.



So what if $G$ is finite?



Thanks for your time and help :)










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marked as duplicate by KReiser, Alexander Gruber abstract-algebra
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Nov 30 '18 at 3:01


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    "Any subgroup of $G$ is like $G^m = {g^m | gin G}$, $m in mathbb N$." What does that mean though? That every subgroup is cyclic?
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:02










  • $begingroup$
    @the_fox No. For example, $x in G^m$ means $exists g in G$ s.t. $g^m = x$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:09










  • $begingroup$
    Could it be that for every subgroup$H subset G$, there is $ m in mathbb{N}$ such that $H cong G^m$.
    $endgroup$
    – nafhgood
    Nov 29 '18 at 22:09










  • $begingroup$
    Then you need to specify ranges for $m$. Otherwise, for all groups $G$ and all $x in G$ it is trivially true that $x in G^1$ (just take $g=x$).
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:11












  • $begingroup$
    @the_fox For $m=1$, that's ture. But the condition is for "every" subgroup of $G$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:16


















0












$begingroup$



This question already has an answer here:




  • Show that a group $G$ is cyclic if and only if every subgroup of $G$ has the form of $G^k$.

    2 answers




Group $G$ is cyclic $iff $



Every subgroup of $G$ is like $G^m = {g^m | gin G}$ for some $m in mathbb N$.



$G^m = {g^m | gin G}$ means $forall x in G^m, exists, g in G$ s.t. $x=g^m$.



$Rightarrow$ is trivial.$,$What about $Leftarrow$?



My idea:$,$If $G$ is infinite group, it's not very hard.



Suppose $a in G$ has infinite order, $langle arangle=G^m$ for some $m>0$, then $exists , b in G$ s.t. $a=b^m$. Then $langle brangle=G^n$ for some $n>0$, then $exists , c in G$ s.t. $b=c^n$. Since $c^m in G^m = langle arangle$, $exists ,i in mathbb N $ s.t. $c^m=a^i=b^{mi}=c^{nmi}$. $c$ clearly has infinite order, so $m=nmi$, $1=ni,n=1$.$langle brangle=G$ is cyclic.



So what if $G$ is finite?



Thanks for your time and help :)










share|cite|improve this question











$endgroup$



marked as duplicate by KReiser, Alexander Gruber abstract-algebra
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Nov 30 '18 at 3:01


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 1




    $begingroup$
    "Any subgroup of $G$ is like $G^m = {g^m | gin G}$, $m in mathbb N$." What does that mean though? That every subgroup is cyclic?
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:02










  • $begingroup$
    @the_fox No. For example, $x in G^m$ means $exists g in G$ s.t. $g^m = x$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:09










  • $begingroup$
    Could it be that for every subgroup$H subset G$, there is $ m in mathbb{N}$ such that $H cong G^m$.
    $endgroup$
    – nafhgood
    Nov 29 '18 at 22:09










  • $begingroup$
    Then you need to specify ranges for $m$. Otherwise, for all groups $G$ and all $x in G$ it is trivially true that $x in G^1$ (just take $g=x$).
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:11












  • $begingroup$
    @the_fox For $m=1$, that's ture. But the condition is for "every" subgroup of $G$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:16
















0












0








0





$begingroup$



This question already has an answer here:




  • Show that a group $G$ is cyclic if and only if every subgroup of $G$ has the form of $G^k$.

    2 answers




Group $G$ is cyclic $iff $



Every subgroup of $G$ is like $G^m = {g^m | gin G}$ for some $m in mathbb N$.



$G^m = {g^m | gin G}$ means $forall x in G^m, exists, g in G$ s.t. $x=g^m$.



$Rightarrow$ is trivial.$,$What about $Leftarrow$?



My idea:$,$If $G$ is infinite group, it's not very hard.



Suppose $a in G$ has infinite order, $langle arangle=G^m$ for some $m>0$, then $exists , b in G$ s.t. $a=b^m$. Then $langle brangle=G^n$ for some $n>0$, then $exists , c in G$ s.t. $b=c^n$. Since $c^m in G^m = langle arangle$, $exists ,i in mathbb N $ s.t. $c^m=a^i=b^{mi}=c^{nmi}$. $c$ clearly has infinite order, so $m=nmi$, $1=ni,n=1$.$langle brangle=G$ is cyclic.



So what if $G$ is finite?



Thanks for your time and help :)










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Show that a group $G$ is cyclic if and only if every subgroup of $G$ has the form of $G^k$.

    2 answers




Group $G$ is cyclic $iff $



Every subgroup of $G$ is like $G^m = {g^m | gin G}$ for some $m in mathbb N$.



$G^m = {g^m | gin G}$ means $forall x in G^m, exists, g in G$ s.t. $x=g^m$.



$Rightarrow$ is trivial.$,$What about $Leftarrow$?



My idea:$,$If $G$ is infinite group, it's not very hard.



Suppose $a in G$ has infinite order, $langle arangle=G^m$ for some $m>0$, then $exists , b in G$ s.t. $a=b^m$. Then $langle brangle=G^n$ for some $n>0$, then $exists , c in G$ s.t. $b=c^n$. Since $c^m in G^m = langle arangle$, $exists ,i in mathbb N $ s.t. $c^m=a^i=b^{mi}=c^{nmi}$. $c$ clearly has infinite order, so $m=nmi$, $1=ni,n=1$.$langle brangle=G$ is cyclic.



So what if $G$ is finite?



Thanks for your time and help :)





This question already has an answer here:




  • Show that a group $G$ is cyclic if and only if every subgroup of $G$ has the form of $G^k$.

    2 answers








abstract-algebra group-theory finite-groups cyclic-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 22:25







Andrews

















asked Nov 29 '18 at 21:58









AndrewsAndrews

3691317




3691317




marked as duplicate by KReiser, Alexander Gruber abstract-algebra
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Nov 30 '18 at 3:01


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by KReiser, Alexander Gruber abstract-algebra
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Nov 30 '18 at 3:01


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1




    $begingroup$
    "Any subgroup of $G$ is like $G^m = {g^m | gin G}$, $m in mathbb N$." What does that mean though? That every subgroup is cyclic?
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:02










  • $begingroup$
    @the_fox No. For example, $x in G^m$ means $exists g in G$ s.t. $g^m = x$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:09










  • $begingroup$
    Could it be that for every subgroup$H subset G$, there is $ m in mathbb{N}$ such that $H cong G^m$.
    $endgroup$
    – nafhgood
    Nov 29 '18 at 22:09










  • $begingroup$
    Then you need to specify ranges for $m$. Otherwise, for all groups $G$ and all $x in G$ it is trivially true that $x in G^1$ (just take $g=x$).
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:11












  • $begingroup$
    @the_fox For $m=1$, that's ture. But the condition is for "every" subgroup of $G$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:16
















  • 1




    $begingroup$
    "Any subgroup of $G$ is like $G^m = {g^m | gin G}$, $m in mathbb N$." What does that mean though? That every subgroup is cyclic?
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:02










  • $begingroup$
    @the_fox No. For example, $x in G^m$ means $exists g in G$ s.t. $g^m = x$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:09










  • $begingroup$
    Could it be that for every subgroup$H subset G$, there is $ m in mathbb{N}$ such that $H cong G^m$.
    $endgroup$
    – nafhgood
    Nov 29 '18 at 22:09










  • $begingroup$
    Then you need to specify ranges for $m$. Otherwise, for all groups $G$ and all $x in G$ it is trivially true that $x in G^1$ (just take $g=x$).
    $endgroup$
    – the_fox
    Nov 29 '18 at 22:11












  • $begingroup$
    @the_fox For $m=1$, that's ture. But the condition is for "every" subgroup of $G$.
    $endgroup$
    – Andrews
    Nov 29 '18 at 22:16










1




1




$begingroup$
"Any subgroup of $G$ is like $G^m = {g^m | gin G}$, $m in mathbb N$." What does that mean though? That every subgroup is cyclic?
$endgroup$
– the_fox
Nov 29 '18 at 22:02




$begingroup$
"Any subgroup of $G$ is like $G^m = {g^m | gin G}$, $m in mathbb N$." What does that mean though? That every subgroup is cyclic?
$endgroup$
– the_fox
Nov 29 '18 at 22:02












$begingroup$
@the_fox No. For example, $x in G^m$ means $exists g in G$ s.t. $g^m = x$.
$endgroup$
– Andrews
Nov 29 '18 at 22:09




$begingroup$
@the_fox No. For example, $x in G^m$ means $exists g in G$ s.t. $g^m = x$.
$endgroup$
– Andrews
Nov 29 '18 at 22:09












$begingroup$
Could it be that for every subgroup$H subset G$, there is $ m in mathbb{N}$ such that $H cong G^m$.
$endgroup$
– nafhgood
Nov 29 '18 at 22:09




$begingroup$
Could it be that for every subgroup$H subset G$, there is $ m in mathbb{N}$ such that $H cong G^m$.
$endgroup$
– nafhgood
Nov 29 '18 at 22:09












$begingroup$
Then you need to specify ranges for $m$. Otherwise, for all groups $G$ and all $x in G$ it is trivially true that $x in G^1$ (just take $g=x$).
$endgroup$
– the_fox
Nov 29 '18 at 22:11






$begingroup$
Then you need to specify ranges for $m$. Otherwise, for all groups $G$ and all $x in G$ it is trivially true that $x in G^1$ (just take $g=x$).
$endgroup$
– the_fox
Nov 29 '18 at 22:11














$begingroup$
@the_fox For $m=1$, that's ture. But the condition is for "every" subgroup of $G$.
$endgroup$
– Andrews
Nov 29 '18 at 22:16






$begingroup$
@the_fox For $m=1$, that's ture. But the condition is for "every" subgroup of $G$.
$endgroup$
– Andrews
Nov 29 '18 at 22:16












1 Answer
1






active

oldest

votes


















0












$begingroup$

If $G^m$ forms a group, then it is easy to find that $G^m$ is a characteristic subgroup of $G$.



Suppose every subgroup of $G$ is like $G^m={g^m | gin G}$ for some $min mathbb{N}$, then every subgroup of $G$ is a characteristic subgroup. It follows that $G$ is abelian.



Suppose $G$ is finite. $Gcong mathbb{Z}_{n_1}timescdotstimesmathbb{Z}_{n_s}$ for some positive integers such that $n_1|n_2|cdots|n_s$. Pick $H = mathbb{Z}_{n_1}cong G^m$ as a subgroup of $G$ for some $minmathbb{N}$ and then the quotient group $G/Hcongmathbb{Z_2}timescdotstimesmathbb{Z}_{n_s}$. Note that the exponent of $G/H$ is $n_s$, we have $n_s|m$. However, the exponent of $G$ is also $n_s$, which implies that $g^{n_s} = 1$ and so $g^m = 1$ for all $gin G$. It follows that $G^m = {1}$. This makes $s = 1$ and thus $G$ is cyclic.






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    If $G^m$ forms a group, then it is easy to find that $G^m$ is a characteristic subgroup of $G$.



    Suppose every subgroup of $G$ is like $G^m={g^m | gin G}$ for some $min mathbb{N}$, then every subgroup of $G$ is a characteristic subgroup. It follows that $G$ is abelian.



    Suppose $G$ is finite. $Gcong mathbb{Z}_{n_1}timescdotstimesmathbb{Z}_{n_s}$ for some positive integers such that $n_1|n_2|cdots|n_s$. Pick $H = mathbb{Z}_{n_1}cong G^m$ as a subgroup of $G$ for some $minmathbb{N}$ and then the quotient group $G/Hcongmathbb{Z_2}timescdotstimesmathbb{Z}_{n_s}$. Note that the exponent of $G/H$ is $n_s$, we have $n_s|m$. However, the exponent of $G$ is also $n_s$, which implies that $g^{n_s} = 1$ and so $g^m = 1$ for all $gin G$. It follows that $G^m = {1}$. This makes $s = 1$ and thus $G$ is cyclic.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If $G^m$ forms a group, then it is easy to find that $G^m$ is a characteristic subgroup of $G$.



      Suppose every subgroup of $G$ is like $G^m={g^m | gin G}$ for some $min mathbb{N}$, then every subgroup of $G$ is a characteristic subgroup. It follows that $G$ is abelian.



      Suppose $G$ is finite. $Gcong mathbb{Z}_{n_1}timescdotstimesmathbb{Z}_{n_s}$ for some positive integers such that $n_1|n_2|cdots|n_s$. Pick $H = mathbb{Z}_{n_1}cong G^m$ as a subgroup of $G$ for some $minmathbb{N}$ and then the quotient group $G/Hcongmathbb{Z_2}timescdotstimesmathbb{Z}_{n_s}$. Note that the exponent of $G/H$ is $n_s$, we have $n_s|m$. However, the exponent of $G$ is also $n_s$, which implies that $g^{n_s} = 1$ and so $g^m = 1$ for all $gin G$. It follows that $G^m = {1}$. This makes $s = 1$ and thus $G$ is cyclic.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If $G^m$ forms a group, then it is easy to find that $G^m$ is a characteristic subgroup of $G$.



        Suppose every subgroup of $G$ is like $G^m={g^m | gin G}$ for some $min mathbb{N}$, then every subgroup of $G$ is a characteristic subgroup. It follows that $G$ is abelian.



        Suppose $G$ is finite. $Gcong mathbb{Z}_{n_1}timescdotstimesmathbb{Z}_{n_s}$ for some positive integers such that $n_1|n_2|cdots|n_s$. Pick $H = mathbb{Z}_{n_1}cong G^m$ as a subgroup of $G$ for some $minmathbb{N}$ and then the quotient group $G/Hcongmathbb{Z_2}timescdotstimesmathbb{Z}_{n_s}$. Note that the exponent of $G/H$ is $n_s$, we have $n_s|m$. However, the exponent of $G$ is also $n_s$, which implies that $g^{n_s} = 1$ and so $g^m = 1$ for all $gin G$. It follows that $G^m = {1}$. This makes $s = 1$ and thus $G$ is cyclic.






        share|cite|improve this answer









        $endgroup$



        If $G^m$ forms a group, then it is easy to find that $G^m$ is a characteristic subgroup of $G$.



        Suppose every subgroup of $G$ is like $G^m={g^m | gin G}$ for some $min mathbb{N}$, then every subgroup of $G$ is a characteristic subgroup. It follows that $G$ is abelian.



        Suppose $G$ is finite. $Gcong mathbb{Z}_{n_1}timescdotstimesmathbb{Z}_{n_s}$ for some positive integers such that $n_1|n_2|cdots|n_s$. Pick $H = mathbb{Z}_{n_1}cong G^m$ as a subgroup of $G$ for some $minmathbb{N}$ and then the quotient group $G/Hcongmathbb{Z_2}timescdotstimesmathbb{Z}_{n_s}$. Note that the exponent of $G/H$ is $n_s$, we have $n_s|m$. However, the exponent of $G$ is also $n_s$, which implies that $g^{n_s} = 1$ and so $g^m = 1$ for all $gin G$. It follows that $G^m = {1}$. This makes $s = 1$ and thus $G$ is cyclic.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 2:14









        Hongyi HuangHongyi Huang

        1




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