Exercise 3.4.4 (Converse of Lagrange's Theorem for finite abelian groups)

Proof. Consider for every positive integer $n$ the proposition

$𝒫(n)⟺$ “every abelian group $G$ of order $|G|<n$ has a subgroup of order $d$ for each positive divisor $d$ of $|G|$”.

If $n=1$, then $G=\{1\}$ and the only positive divisor $d$ of $1$ is $d=1$. There is a subgroup of $G$ of order $1$, which is $G$ itself, so $𝒫(1)$ is true.

Suppose that $𝒫(n)$ is true. Consider now an abelian group $G$ of order $n$, et let $d$ be a positive divisor of $n$.

If $d=1$, the subgroup trivial $\{1\}$ has order $1$. We may assume now that $d>1$. Then there is some prime number $p$ such that $p\mid d$. By Cauchy’s Theorem there exists $x\in G$ of order $p$. Consider the subgroup $N=⟨x⟩$, which is normal in $G$ since $G$ is abelian. Then $|G∕N|=|G|∕|N|=n∕p<n$, so we may apply the induction hypothesis to $G∕N$ and the divisor $\delta =d∕p$ of $|G∕N|=n∕p$: there exists a subgroup $\bar{H}\le G∕N$ of order $|\bar{H}|=\delta =d∕p$. By the Lattice isomorphism theorem, there is a subgroup $H$ of $G$ such that $\bar{H}=H∕N$, and $|H|=|\bar{H}|\cdot |N|=(d∕p)\cdot p=d$. This shows that every abelian group $G$ of order $|G|<n+1$ has a subgroup of order $d$ for each positive divisor $d$ of $|G|$, so $𝒫(n+1)$ is true.

The induction is done, which proves that $𝒫(n)$ is true for every positive integer $n$.

So a finite abelian group has a subgroup of order $d$ for each positive divisor $d$ of its order. □