I think there is a corollary for Cauchy's theorem such that a finite group $G$ contains a subgroup of order $n$ for each divisor $n$ of $|G|$ with a certain condition (I can't remember what it was). If you know this corollary, could you give me a proof for it or any relevant link? I think induction with Cauchy's theorem can be used, but I have no idea how to do.
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I am not sure if this is what you are looking for: cyclic groups have this property. So if $G$ is cyclic then you will have this statement to be true. – Anurag A Dec 12 '14 at 20:20
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Maybe you are looking for Sylow's theorems? Link: http://en.wikipedia.org/wiki/Sylow_theorems – Lukas Geyer Dec 12 '14 at 20:26
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I think some Abelian group satisfies this property, and the condition is not as strong as being cyclic or Sylow p-group. I found it on D&F's book. – Math.StackExchange Dec 12 '14 at 20:41
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There is such a result that $G$ has a subgroup of order $p^{a}$ whenever $p$ is a prime, $a$ is a positive integer, and $p^{a}$ divides $|G|$. – Geoff Robinson Dec 12 '14 at 21:06
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@GeoffRobinson yea, Sylow's theorems – Dec 12 '14 at 21:09
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I just found the proof, and this was the question 3.4.4 of D&F. It was about finite Abelian group. Thanks for your answer! https://crazyproject.wordpress.com/2010/04/11/finite-abelian-groups-have-subgroups-of-every-order-dividing-the-order-of-the-group/ – Math.StackExchange Dec 12 '14 at 21:14
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No problem, I'm happy I could help. – Asinomás Dec 12 '14 at 21:16
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@forgetfulfunctor: Indeed. – Geoff Robinson Dec 12 '14 at 21:38
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A group $G$ satisfying it has a subgroup of order $n$ of $|G|$ for all divisor $n$ is called a clt group (converse lagrange theorem) .If the group is supersolvable then it is clt. Also all clt groups are solvable.
You can prove for solvable groups that the group has a subgroup of order $n$ for all $n$ such that $\frac{|G|}{n}$ and $n$ are relatively prime.
Something which is true for all groups is there is a subgroup of size $p^\alpha$ for all such values dividing $G$ (you can prove this by first proving it for $p$-groups and using Sylow's theorem).See a proof here.
Added: since abelian groups are supersolvable clearly abelian groups are clt. For a direct proof abelian groups are clt see here (although there is a shorter proof by the fundamental theorem of finite abelian groups).
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clt group sounds really interesting! Your post gave me a huge motivation for this kind of topic. – Math.StackExchange Dec 12 '14 at 21:19