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

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 $\frac{m}{p}$. This subgroup is of the form $H/\langle x \rangle$ for some $H \le G$. Since $|H/\langle x \rangle| = \frac{m}{p}$, 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.

Answer

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. $\blacksquare$

Attribution
Source : Link , Question Author : Alex Petzke , Answer Author : Yadati Kiran

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