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If the order of a group G is divisible by $p^n$ for some prime p and natural number n, then prove G has a subgroup of order $p^n$.

My try has been by induction on the exponent of divisor $p^n$. I use Cauchy's Theorem for Abelian case which basically is a corollary of Theorem of Finite Abelian Groups. Note that I want to prove this without Sylow's theorems.

$n = 1$ it is clearly true, as Cauchy's Theorem for Abelian Groups says that if order of G is divisible by p, then it contains an element a of order p which clearly generates a cyclic group $$ of order p. Inductive assumption will then of course be that if order of G is divisible by $p^{n-1}$, then G contains a subgroup of this order.

Assume order of G is divisible by $p^n$. Then it certainly also is divisible by $p^{n-1}$, so it contains a nontrivial subgroup of order $p^{n-1}$. Call this subgroup for S. Then, since G is abelian, quotient group $G/S$ certainly exists, and we have relation of orders of groups: $|G| = |G/S||S|$ Since order of G is divisible by $p^n$, then it follows from the relation that p divides $|G/S|$ and by Cauchy's Theorem, G/S contains an element of order p. Could I somehow 'construct' a group of order $p^n$ by direct product of the S and the group generated by element of order P in G/S?

I'm pretty much stuck. Is there a way to go on, or have I hit a wall? If somebody knows, could you point me to a place on the internet where this or similar proof has been written out?

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    Are you allowed to use the classification of finite abelian groups?2011-08-16
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    Look at the correspondence theorem between groups and their quotient groups.2011-08-16
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    Yes, classification is ok.2011-08-16
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    [When you write a comment for a given user, it's better to put an "@x" in your comment, where x is the user's name. (I didn't get the notification; I saw your answer by chance. I don't have to put an "@Barre" because you're automatically notified since it's **your** question.)] You can use the fact that a cyclic group of order $n$ has an order $d$ subgroup iff $d$ divides $n$.2011-08-16
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    By the classification, $G=A\times B$ where $A$ is a $p$-group and the order of $B$ is prime to $p$, and $A=A_1\times\cdots\times A_n$ where each $A_i$ is a cyclic $p$-group. The order of $A$ is $p^k$ with $k\ge n$. Take $H=A_1\times\cdots\times A_i$ such that the order $p^ j$, say, of $H$ divides $p^n$, and $i$ is maximal for this property. Then $A_{i+1}$ has a subgroup $K$ of order $p^ {n-j}$, and $H\times K$ does the job.2011-08-16
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    Variation: $A_i$ is cyclic of order $p^ {f(i)}$. The sum of the $f(i)$ is $\ge n$. For all $i$ choose $0\le g(i)\le f(i)$ such that the sum of the $g(i)$ is $n$. Each $A_i$ has a subgroup of order $p^ {g(i)}$. Take the product of these subgroups.2011-08-16
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    @Pierre-Yves Gaillard , thanks for the tip about notifications, I'm a new user and frankly have not read FAQs/guides yet. Also, especially the second proof is easily understandable for me. Both are something a bit above my mathematical sophistication, so I'd never come up with them on my own.2011-08-16
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    If it looks complicated, it means that I didn’t express myself well. A way to think about the argument is to start with particular cases. Case 1: $G$ is a cyclic $p$-group. Case 2: $G$ is a product of cyclic $p$-groups. Case 3 (general case): $G$ is the product of a product of cyclic $p$-groups by a group of order prime to $p$ (this last factor won’t count). [About the notification system: sometimes it works even without the @, but not always. Also it suffices to write @x where x is the first part of the user's name (for instance, for me, you can write @Pierre-Yves, or even @Pierre).]2011-08-16
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    Another of thinking about this: Consider the class $C$ of all finite groups $G$ such that $G$ has a subgroup of order $d$ whenever $d$ divides its order. The cyclic groups are in $C$. If $G$ and $H$ are in $C$, then so is $G\times H$. Then, by the classification theorem, all finite abelian groups are in $C$.2011-08-16
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    @Pierre Both are pretty clear to me, it just took a while to get 'used' to them (I am not a student of pure mathematics and frankly I have not yet taken an introduction to abstract algebra class). Your argument is actually one that I myself explored previously, it seemed intuitively true but I could not put it down on paper properly. Isn't that so that a similar argument to yours could prove the general converse of Lagrange's Theorem for abelian case? [Your last example: mindblown :)]2011-08-16
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    Exactly! The converse of Lagrange's Theorem holds for abelian groups. (Thanks for your compliments, but don't get too easily impressed!)2011-08-16

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You are almost there. So you have a surjective homomorphism of abelian groups $G\longrightarrow G/S$, which simply sends $g$ to the coset $gS$. The group $G/S$ has a subgroup or order $p$. What can you say about the pre-image of this subgroup under the above homomorphism?

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    I've tried to follow up on your thought. Since $G/S$ contains a subgroup of order $p$, this subgroup is cyclic and generated by some element $a$. Then $(Sa)^p = Sa^p = S$, so we see that $a^p \in S$. Also, $a$'s order in G is a multiple of p. Actually this is true for any member $a^d$ of subgroup generated by $a$, as for example if $d = 2$, $(Sa^2)^p = Sa^{2p} = S$. Am I on the right track, is there some fact I'm missing?2011-08-16
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    @Barre If $\overline{a}=aS$ generates a cyclic subgroup of $G/S$ of order $p$, then the preimage of that subgroup in $G$ is simply the disjoint union $$\bigsqcup_{i=0}^{p-1} a^iS.$$ What is its order? Show that it's a subgroup of $G$ and you are done.2011-08-16