Let $G$ be a finite abelian group with identity $e$. Prove that $exists$ an element $x in G$ such that ord y...












0












$begingroup$


My attempt:



I think somehow we have to show that if $|x|=operatorname{lcm}{|x_i|: x_i in G}$ (as the group is finite abelian) then such an element actually lies in $G-{e}$. But I don't know how to show that such an element indeed lies in $G$.



Thanks for any help!






abstract-algebra group-theory finite-groups







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$endgroup$












  • $begingroup$
    Do you know the structure theorem for finite Abelian groups?
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 4:42










  • $begingroup$
    yeah i do know!
    $endgroup$
    – Infinity
    Jan 7 at 4:43






  • 1




    $begingroup$
    This is going to be false in the case that $G={e}$, for the trivial fact that any statement of the form "$exists x in varnothing ....$" is always false.
    $endgroup$
    – MathematicsStudent1122
    Jan 7 at 4:50












  • $begingroup$
    Pick an element of maxmial order ...
    $endgroup$
    – Hagen von Eitzen
    Jan 7 at 5:25
















0












$begingroup$


My attempt:



I think somehow we have to show that if $|x|=operatorname{lcm}{|x_i|: x_i in G}$ (as the group is finite abelian) then such an element actually lies in $G-{e}$. But I don't know how to show that such an element indeed lies in $G$.



Thanks for any help!






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know the structure theorem for finite Abelian groups?
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 4:42










  • $begingroup$
    yeah i do know!
    $endgroup$
    – Infinity
    Jan 7 at 4:43






  • 1




    $begingroup$
    This is going to be false in the case that $G={e}$, for the trivial fact that any statement of the form "$exists x in varnothing ....$" is always false.
    $endgroup$
    – MathematicsStudent1122
    Jan 7 at 4:50












  • $begingroup$
    Pick an element of maxmial order ...
    $endgroup$
    – Hagen von Eitzen
    Jan 7 at 5:25














0












0








0





$begingroup$


My attempt:



I think somehow we have to show that if $|x|=operatorname{lcm}{|x_i|: x_i in G}$ (as the group is finite abelian) then such an element actually lies in $G-{e}$. But I don't know how to show that such an element indeed lies in $G$.



Thanks for any help!






abstract-algebra group-theory finite-groups







share|cite|improve this question











$endgroup$




My attempt:



I think somehow we have to show that if $|x|=operatorname{lcm}{|x_i|: x_i in G}$ (as the group is finite abelian) then such an element actually lies in $G-{e}$. But I don't know how to show that such an element indeed lies in $G$.



Thanks for any help!






abstract-algebra group-theory finite-groups




abstract-algebra group-theory finite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 7:12









CHOUDHARY bhim sen

1239




1239










asked Jan 7 at 4:34









InfinityInfinity

321112




321112












  • $begingroup$
    Do you know the structure theorem for finite Abelian groups?
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 4:42










  • $begingroup$
    yeah i do know!
    $endgroup$
    – Infinity
    Jan 7 at 4:43






  • 1




    $begingroup$
    This is going to be false in the case that $G={e}$, for the trivial fact that any statement of the form "$exists x in varnothing ....$" is always false.
    $endgroup$
    – MathematicsStudent1122
    Jan 7 at 4:50












  • $begingroup$
    Pick an element of maxmial order ...
    $endgroup$
    – Hagen von Eitzen
    Jan 7 at 5:25


















  • $begingroup$
    Do you know the structure theorem for finite Abelian groups?
    $endgroup$
    – Lord Shark the Unknown
    Jan 7 at 4:42










  • $begingroup$
    yeah i do know!
    $endgroup$
    – Infinity
    Jan 7 at 4:43






  • 1




    $begingroup$
    This is going to be false in the case that $G={e}$, for the trivial fact that any statement of the form "$exists x in varnothing ....$" is always false.
    $endgroup$
    – MathematicsStudent1122
    Jan 7 at 4:50












  • $begingroup$
    Pick an element of maxmial order ...
    $endgroup$
    – Hagen von Eitzen
    Jan 7 at 5:25
















$begingroup$
Do you know the structure theorem for finite Abelian groups?
$endgroup$
– Lord Shark the Unknown
Jan 7 at 4:42




$begingroup$
Do you know the structure theorem for finite Abelian groups?
$endgroup$
– Lord Shark the Unknown
Jan 7 at 4:42












$begingroup$
yeah i do know!
$endgroup$
– Infinity
Jan 7 at 4:43




$begingroup$
yeah i do know!
$endgroup$
– Infinity
Jan 7 at 4:43




1




1




$begingroup$
This is going to be false in the case that $G={e}$, for the trivial fact that any statement of the form "$exists x in varnothing ....$" is always false.
$endgroup$
– MathematicsStudent1122
Jan 7 at 4:50






$begingroup$
This is going to be false in the case that $G={e}$, for the trivial fact that any statement of the form "$exists x in varnothing ....$" is always false.
$endgroup$
– MathematicsStudent1122
Jan 7 at 4:50














$begingroup$
Pick an element of maxmial order ...
$endgroup$
– Hagen von Eitzen
Jan 7 at 5:25




$begingroup$
Pick an element of maxmial order ...
$endgroup$
– Hagen von Eitzen
Jan 7 at 5:25










1 Answer
1






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1












$begingroup$

If $G = bigoplus_{1 leq i leq k} mathbb{Z}_{n_i}$ with $n_i | n_{i+1}$, consider the element $(0,0, cdots ,1)$ with order $n_k$. It is clear that $exp G = n_k$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah okay..Thanks!
    $endgroup$
    – Infinity
    Jan 7 at 4:52











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

If $G = bigoplus_{1 leq i leq k} mathbb{Z}_{n_i}$ with $n_i | n_{i+1}$, consider the element $(0,0, cdots ,1)$ with order $n_k$. It is clear that $exp G = n_k$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah okay..Thanks!
    $endgroup$
    – Infinity
    Jan 7 at 4:52
















1












$begingroup$

If $G = bigoplus_{1 leq i leq k} mathbb{Z}_{n_i}$ with $n_i | n_{i+1}$, consider the element $(0,0, cdots ,1)$ with order $n_k$. It is clear that $exp G = n_k$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ah okay..Thanks!
    $endgroup$
    – Infinity
    Jan 7 at 4:52














1












1








1





$begingroup$

If $G = bigoplus_{1 leq i leq k} mathbb{Z}_{n_i}$ with $n_i | n_{i+1}$, consider the element $(0,0, cdots ,1)$ with order $n_k$. It is clear that $exp G = n_k$.






share|cite|improve this answer









$endgroup$



If $G = bigoplus_{1 leq i leq k} mathbb{Z}_{n_i}$ with $n_i | n_{i+1}$, consider the element $(0,0, cdots ,1)$ with order $n_k$. It is clear that $exp G = n_k$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 4:48









MathematicsStudent1122MathematicsStudent1122

8,62122467




8,62122467












  • $begingroup$
    Ah okay..Thanks!
    $endgroup$
    – Infinity
    Jan 7 at 4:52


















  • $begingroup$
    Ah okay..Thanks!
    $endgroup$
    – Infinity
    Jan 7 at 4:52
















$begingroup$
Ah okay..Thanks!
$endgroup$
– Infinity
Jan 7 at 4:52




$begingroup$
Ah okay..Thanks!
$endgroup$
– Infinity
Jan 7 at 4:52


















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