prove : a finite abelian group G has a subgroup of order d for all d dividing the order of G

Use Cauchy’s Theorem and induction to prove that a finite abelian group $G$ has a subgroup of order $d$ for all $d$ dividing the order of $G$.

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@Asaf Karagila , I don't know how to put the right place ! the fourm dont allow me to do this !!! –  Maths Lover Jan 16 '13 at 15:19
@Arthur Fischer , thanks –  Maths Lover Jan 16 '13 at 15:19
@MathsLover Please don't cross-post on MO. –  user38268 Jan 17 '13 at 0:49

Hints:

1) From Cauchy's Theorem for any prime dividing the group's order there's an element $\,x_p\,$ of order $\,p\,$

2) The claim follows at once for $\,|G|=1,2,3,4\,$ and, in fact, for every prime.

3) Use (1) above and the fact that every subgroup of an abelian group is normal and look at the group $\,G/\langle x_p\rangle\,$

4) Apply the inductive hypothesis to the group in (3) and end the argument.

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it's the first time i face a question need induction in group theory , it will be great if you add the complete solution because i don't know any thing of this kind !! specially , you said , " The claim follows at once for |G|=1,2,3,4 and, in fact, for every prime." why it follows ?!!! –  Maths Lover Jan 16 '13 at 15:28
Well, you can work your own way for the groups of order 1,2,3,4,5 and perhaps a little more. A group of order a prime only has two subgroups: the trivial one $\,\{1\}\,$ and itself, and every non-identity element has order that prime, so it obviously fulfills the condition. The quotient $\,G/\langle x_p\rangle\,$ has, by the ind. hyp., a sbgp. of any order dividing $\,|G|/p\,$ , so... –  DonAntonio Jan 16 '13 at 18:04