Proof that a finite abelian group has a subgroup of order $n$ for every $n$ which divides $|G|$

Question

As per the title, I'm trying to prove that a finite abelian group has a subgroup of order $n$ for every $n$ which divides $|G|$, and I think I have a solution, but I would like someone to verify this that it is correct.

Proof: Take a group $G$ that is finite and abelian. Suppose that for our desired result holds for all abelian groups with order less than $|G|$. Take any prime $p$ such that $|G| = p^{\alpha}m$ where $p$ does not divide $m$, so that by Theorem 12, $G$ has a subgroup of order $p^{\alpha}$, which we denote $H$. By our inductive hypothesis, $H$ has subgroups of order $p^{\alpha - 1}, p^{\alpha - 2}, \ldots, p^2, p$. Then, take the prime factorization of any integer $n$ for which $|G|$ divides $n$, say $n = x_1^{y_1}x_2^{y_2}\ldots x_z^{y_z}$ where $y_1,\ldots,y_z$ are primes which divide $|G|$.

For any of these factors, say $x_a^{y_a}$ and $x_b^{y_b}$, we recall that there are subgroups in $G$ with order equal to each of these terms, say $A$ and $B$. By Lagrange's Theorem, all non-identity elements in $A$ have order divisible by $x_a$, and the same may be said for elements in $B$. Since $x_a \neq x_b$ and both are prime, we see that $A \cap B = \{1\}$. By Proposition 13, we have $|AB| = \frac{|A||B|}{|A \cap B|} = \frac{x_a^{y_a}x_b^{y_b}}{1} = x_a^{y_a}x_b^{y_b}$, and since $G$ is an abelian group, Proposition 14 states that $AB$ is a subgroup.

Applying the previous result to the entire prime factorization of $n$, we therefore construct a subgroup of $G$ that has order $n$, and we are done. We never did use Cauchy's Theorem either, did we?

Theorem 12 is by Sylow and says that if $G$ is a finite group of order $p^{\alpha}m$ where $p$ is a prime and $p$ does not divide $m$, then $G$ has a subgroup of order $p^{\alpha}$.

Proposition 13 says that if $H$ and $K$ are finite subgroups of a group then $|HK| = \frac{|H||K|}{|H \cap K|}$.

Proposition 14 says that if $H$ and $K$ are subgroups of a group, $HK$ is a subgroup if and only if $HK = KH$.

My reasons for posting this are twofold; firstly, I haven't found a proof of this sort on the internet, so I am afraid that I made a mistake somewhere. The second reason is that Dummit and Foote request that Cauchy's Theorem be used in the proof, which I have not, and I would greatly appreciate it if you could point me to a proof that does use Cauchy's Theorem. Thanks so much!

Answer 1

There is a gap in your argument. You have $|G|=p^am$, with $\gcd(p,m)=1$, and you take $H$ to be a subgroup of order $p^a$. Fine. You then just assume that $|H|\lt |G|$ so that you can apply the induction hypothesis.

However, you don’t know for sure that this happens! What if $G$ is a $p$-group, so that $|G|=p^a$, and so $H=G$? You still need to deal with the case of a $p$-group: you have to show that if $G$ is an abelian group of order $p^n$, $p$ prime, and $n\geq 1$, then it contains a subgroup $H$ of order $p^a$ for all $a$, $0\leq a\leq n$.

And I expect you’ll need to use Cauchy’s Theorem to do that (if you want to do induction on $n$, say...).

Answer 2

This seems fine. It's basically a divide-and-conquer strategy very common in number theory. You basically show a lemma:

If a finite abelian group $G$ contains subgroups of order $n$ and $m$, with $n$ and $m$ coprime, then $G$ contains a subgroup of order $nm$.

This and Theorem 12 show your result basically immediately.

I don't see off the top of my head how to prove this using Cauchy's theorem.