Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $G$.

share|improve this question

migrated from meta.math.stackexchange.com Jan 16 '13 at 14:35

This question came from our discussion, support, and feature requests site for people studying math at any level and professionals in related fields.

    
@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
add comment

1 Answer

up vote 6 down vote accepted

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.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.