## Finite abelian groups have subgroups of every order dividing the order of the group

Use Cauchy’s Theorem and induction to prove that a finite abelian group has a subgroup of order $d$ for all $d$ dividing $|G|$.

First we make a quick definition. Let $G$ be a finite abelian group, say with $|G| = n$. By the Fundamental Theorem of Arithmetic, $n$ factors completely as a product of primes in an essentially unique way; that is, $n = p_1p_2 \ldots p_k$ for some primes $p_i$, where $k$ is unique and the $p_i$ are unique up to a permutation. We call this $k$ the width of $G$. We will proceed by induction on the width of $G$.

If $G$ has width 0, then $|G| = 1$ so that $G = 1$. Now 1 itself is the only (positive) divisor of 1 and $G$ itself is a subgroup of order 1. Thus the conclusion holds in the base case.

Now suppose that every group $G$ of width $n$ has the property that if $d$ divides $|G|$, then there exists a subgroup $H \leq G$ with $|H| = d$. Let $G$ be a group of width $n+1$; say $|G| = p_1 \cdots p_{n+1} = m$. Let $d|m$. If $d = 1$, then $1$ is a subgroup of $G$ of order $d$. Suppose $d > 1$ and without loss of generality, say $p_1$ divides $d$. By Cauchy’s Theorem, there exists an element $x \in G$ of order $p_1$, and $\langle x \rangle$ is normal in $G$ since $G$ is abelian. Now $G/\langle x \rangle$ has width $n$ and order $m/p$. By the induction hypothesis, there is a subgroup $H/\langle x \rangle$ of $G/\langle x \rangle$ of order $d/p$, and so $H \leq G$ has order $d$ as desired.

• Robert Bradford  On January 17, 2011 at 5:35 pm

How does d=m imply G is order d?

• nbloomf  On January 17, 2011 at 7:59 pm

Good question.

I just noticed that this proof is broken for another reason: the line “if $d|pm$ then $d = p$ or $d|m$” is wrong.

I’ll add this to the growing list of broken proofs that will have to be fixed later when I have more time.

Thanks!

• Gobi Ree  On November 29, 2011 at 1:56 am

Why does the image of ‘pm’ seem $r$? lol

• Gobi Ree  On November 29, 2011 at 2:01 am

And it may be an error that $G \leq G$ is a subgroup of order $d$