I have an abstract algebra proof I can not complete. The proposition goes as follows: Consider a group $G$ of order $m$. If $n$ divides $m$, prove that $G$ has a subgroup of order $n$. This question is based off the fundamental theorem of finitely generated abelian groups.

Proof: My idea was to let $m = P_{1}^{\alpha_{1}}...P_{k}^{\alpha_{k}}$ and $n = P_{1}^{\beta_{1}}...P_{l}^{\beta_{l}}$ where $\beta_{i} \leq \alpha_{i}$. Now after this, I am a little unsure what to do. Applying the theorem can let me write this as a direct product, but I am not sure how that leads to show $G$ has a subgroup of order $n$.

I am open to answers since this is something I am independently studying in my spare time.


NOTE: I am not looking to prove this by induction. I want to use the fundamental theorem of finitely generated abelian groups.


marked as duplicate by Martin Brandenburg, Jared, user61527, Thomas Andrews, Adriano Sep 6 '13 at 0:47

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First, you must assume $G$ is abelian, otherwise, it is not the case that if $n$ divides $|G|$, then $G$ contains a subgroup of order $n$ (e.g., $A_4$ has no subgroup of order $6$).

Now, hints for the proof:

Use the fundamental theorem of finite abelian groups to deduce $G$ is isomorphic to a product of cyclic groups. Now, decompose $n$ as a product of primes. Necessarily these are related nicely to the decomposition of $G$ as a product of cyclic groups. Namely, you can write $n$ as a product of prime-powers, with each prime power dividing the order of precisely one of the cyclic subgroups composing $G$.

Now, recall that cyclic subgroups enjoy the property that their subgroups correspond precisely to divisors, so you get a subgroup in each component of $G$, corresponding to the prime-powers decomposition of $n$. Show you can simply take the product of these subgroups to obtain a subgroup of order $n$ inside the isomorphic copy of $G$ written as a product of cyclic groups. Don't forget to mention that the isomorphism thus gives you the desired subgroup in $G$, and you are done.


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