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I have been trying to do this exercise and found two answers:

Let $G$ be an abelian group of order $m$. If $n$ divides $m$, prove that $G$ has a subgroup of order $n$.

Showing that a finite abelian group has a subgroup of order m for each divisor m of n

I think I found a clearer way of doing that, and I would like to share it and have it "peer-reviewed".


Lemma: Let $G$ be a finite abelian group of order $m$. Then $\forall g\in G$, $g^m=1$.

Let $G$ be a finite abelian group of order $m$, such that $n\mid m$. Let $g\in G$ be an element of $G$ with (finite) order $O(g)=r$.

Since $g\in G$, then $\langle g\rangle\le G$. By Lagrange's Theorem, then $r\mid m$; then let $m=kr$ for a certain $k$. Thus:

$$g^m=g^{kr}=(g^r)^k=(g^k)^r=1^r=1$$


Proposition: Let $G$ be a finite abelian group of order $m$. If $n\mid m$, then $G$ has a subgroup of order $n$.

If $n\mid m$, let $m=kn$ for a certain $k$. Then for every $g\in G$ (with $g \ne e$, $e$ the identity element of $G$): $$g^m=g^{kn}=1\implies (g^k)^n=1$$

Thus with $h=g^k$, we have $h^n=1$; so the subgroup generated by $h=g^k$ has order $n$.


What do people think about this proof? Does it contain any flaws or holes?

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All you've shown is that the subgroup generated by $h$ has an order that divides $n$, not that it is $n$. For instance, if you had (by mistake) picked the identity element $e$ as $g$ in your proposition, you've have correctly shown that $e^n = e$, but that doesn't mean that $e$ has order $n$; in fact, it has order $1$.

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  • $\begingroup$ What if we excluded $e$? Maybe then I could reformulate it to give an inductive argument? $\endgroup$ – AspiringMathematician Apr 21 '17 at 21:35
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    $\begingroup$ That doesn't work either. For in $Z/32Z$, for instance, you might have $n= 16$, and pick, accidentally, the element $24$. You'd then know that $16 \cdot 24= 0$, but that doesn't make $24$ have order $16$ -- it in fact has order $4$. You've done something good here, but it'll take more than this to produce a proof. Bandaids like "I didn't mean the identity" won't help. And that's (part of) why the usual proof doesn't look like the one you've come up with. $\endgroup$ – John Hughes Apr 21 '17 at 21:37
  • $\begingroup$ I meant about inducting the argument on the order of the subgroup generated by $h$, but I can see then that my proof would become about the same as the other ones (maybe rephrased, but still). Thanks for your response! $\endgroup$ – AspiringMathematician Apr 21 '17 at 21:44

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