Converse Lagrange's Theorem for abelian groups

Question

I just came up with a very simple proof of the converse Lagrange's Theorem for abelian groups. Here is a sketch:

First, denote by $\pi_p$ the quotiening out by (some) element of order $p$. Now, if $A$ is an abelian group of order $n$ and $d\mid n$, with $d =p_1 p_2 ...p_k$ (not necessarily distinct primes), then we can construct a chain of homomorphisms \begin{array}{lclcl} A \stackrel{\pi_{p_1}} \longrightarrow A_1 \stackrel{\pi_{p_2}} \longrightarrow A_2 \stackrel{\pi_{p_3}} \longrightarrow ... \stackrel{\pi_{p_k}} \longrightarrow A_k \end{array} Such homomorphisms exist by the Cauchy's theorem. So, if $\pi=\pi_{p_k}\circ\pi_{p_{k-1}}\circ ...\circ\pi_{p_1}$, then $\mid\operatorname{im}\pi|=n/d$, therefore $\mid\ker\pi\mid$=d.

It looks simpler than what I have seen before, so I am wondering if I am fooling myself here.

Answer 1

This sort of argument suggests a proof by induction. We'll prove that for all natural numbers $n$, if $G$ is a group of order $n$, then for any divisor $d$ of $n$, $G$ has a subgroup of order $d$. The base case is trivial. Suppose now that $|G|=n=dk$. We want to find a subgroup $H\subset G$ of order $d$. Let $p$ be a prime dividing $d$. By Cauchy's theorem, $G$ has an element of order $p$, say $x$. Consider the quotient group $G/\langle x\rangle.$ By the induction hypothesis, the latter group has a subgroup of order $d/p$, and by the Lattice Isomorphism Theorem, this subgroup is of the form $H/\langle x \rangle.$ What is the order of $H$? It's $|H|=|H/\langle x \rangle|\cdot |\langle x\rangle| = (d/p)\cdot p = d.$ So $H$ is the desired group. I hadn't thought of this proof until reading lanskey's post, so thanks.

Remark: Another argument for finite abelian groups using the induction hypothesis on a quotient group is used in Prop. 21 of Ch. 3 in Dummit and Foote to prove Cauchy's theorem for abelian groups. The author's comment: "The philosophy behind this method of proof is that if we have a sufficient amount of information about some normal subgroup, N, of a group G and sufficient information on $G/N$, then somehow we can piece this information together to force $G$ itself to have some desired property."