# 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.

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 $$k. Let $$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. 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. $$\blacksquare$$