# Converse Lagrange's Theorem for abelian groups

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.

• You are not fooling yourself, there is a proof along the lines that you suggest. Apr 30, 2018 at 22:28

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."