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I have made an attempt to prove that a finite abelian group of order $n$ has a subgroup of order $m$ for every divisor $m$ of $n$.

Specifically, I am asked to use a quotient group-induction argument to show this. I'd appreciate comments on the validity or lack thereof of my attempted proof below.

Let $G$ be a finite abelian group of order $n$ and let $m$ be a divisor of $n$. The proposition is true for $n=1$, so we'll proceed by induction and assume $n \ge 2$. Let $p$ be a prime dividing $m$ and let $x$ be an element of order $p$ in $G$ (which exists by Cauchy's Theorem for Abelian Groups). By the induction hypothesis, $G/\langle x \rangle$ has a subgroup of order $\dfrac{m}{p}$. This subgroup is of the form $H/\langle x \rangle$ for some $H \le G$. Since $|H/\langle x \rangle| = \dfrac{m}{p}$, it follows that $H \le G$ has order $m$.

I chose to use a prime divisor of $m$, but I don't see why it wouldn't work to use any proper divisor of $m$. Am I correct on this point?

Thanks, I appreciate the help.

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  • $\begingroup$ What exactly are you inducting on? $\endgroup$ – vadim123 Aug 27 '13 at 17:05
  • $\begingroup$ @vadim123 He's inducting on the order of the parent group. $\endgroup$ – FireGarden Aug 27 '13 at 17:09
  • $\begingroup$ @AlexPetzke Since you used the Cauchy Theorem I don't think one can use a non-prime divisor. $\endgroup$ – user26857 Nov 30 '16 at 8:04
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I found this question while searching for some hints for the same problem. Here is the way I found to solve this problem.

Let $G$ be a finite abelian group. We induct on the order of $G$. If $\lvert G\rvert=1$, then we know that the only divisor of the order is $1$, and we are done. Next, suppose that $\lvert G\rvert=n$, and that the statement holds for $k<n$. Let $d\in \mathbb{N}$ be a divisor of $n$, $d|n$. We can decompose $d=kp$, for some prime $p$ and $k\in \mathbb{N}$. By Cauchy's Theorem, there exists a subgroup $H\le G$ of order $p$. We can form the quotient group because $G$ is abelian, so $H$ is normal.

Then, $\lvert G/H\rvert$ satisfies the inductive hypothesis, since $\lvert G/H\rvert<n$. So, we have that all of the divisors of $|G/H|$ correspond to a subgroup of $G/H$ with appropriate order. In particular, $k$ divides $\lvert G/H\rvert$. By the inductive hypothesis and the Lattice Isomorphism Theorem, we have a subgroup $H\le K\le G$ such that $K/H\le G/H$ and $K/H$ has order $k$. Since $H$ is finite, this implies that $\lvert K\rvert=k\lvert H\rvert=kp=d.$ So, we have a subgroup of order $d$. This completes the induction. $\blacksquare$

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  • $\begingroup$ Typo: I think you mean $H\le K\le G$ not $K\le H\le G$. $\endgroup$ – almagest Jan 10 '18 at 14:12
  • $\begingroup$ Sure do. Thanks. $\endgroup$ – Antonios-Alexandros Robotis Jan 10 '18 at 14:25
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HINT(S):

1) The property is true for finite cyclic groups.

2) Any abelian finite group is product of finite cyclic groups.

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    $\begingroup$ Is this to say that my attempted proof does not work? $\endgroup$ – Alex Petzke Aug 28 '13 at 17:44
  • $\begingroup$ @AlexPetzke : This is to say what is the standard way to tackle your question. $\endgroup$ – Andrea Mori Aug 28 '13 at 18:59
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    $\begingroup$ Alright, but does it work as it is? That's really what I wanted to know, since I was specifically trying to prove it in the way indicated. $\endgroup$ – Alex Petzke Aug 29 '13 at 13:49
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    $\begingroup$ The OP has asked a very concrete question about a very concrete proof, not for an alternative proof which let me say is "standard" only if one knows the structure theorem of finite (finitely generated) abelian groups. The OP's approach is much more natural for students who just begun to study group theory. $\endgroup$ – user26857 Nov 30 '16 at 7:58

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