False statement: every divisor of the order of a finite group occurs as a subgroup order

Given: The hypotheses and objects in the false claim.

Let $G$ be finite, let $p$ be prime, and write $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then $G$ has a subgroup of order $p^a$, hence a Sylow $p$-subgroup (def-sylow-p-subgroup). (Sylow I: every finite group has a Sylow $p$-subgroup).

Let $G$ be finite abelian and let $d$ be a positive divisor of $\mathrm{\mid }G\mathrm{\mid }$. Then $G$ has a subgroup of order $d$. (Converse of Lagrange for finite abelian groups: every divisor occurs as a subgroup order).

For $n\in \mathbb{N}$, the alternating group is the kernel of the sign homomorphism, $A_n:=\ker (\mathrm{sgn}:S_n\to \{+1,-1\})=\{\sigma \in S_n:\mathrm{sgn}(\sigma )=1\}\mathrm{.}$ Thus $A_n$ consists exactly of the even permutations. The subgroup and normality assertions implicit in the word “group” follow from thm-image-subgroup-and-kernel-normal. (The alternating group $A_n=\ker (\mathrm{sgn})$ of even permutations).

Let $G$ be a finite group and $H\le G$. Then $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }\mathrm{.}$ Consequently, under the canonical embedding $\iota :\mathbb{N}\to \mathbb{Z}$, $\mathrm{\mid }H\mathrm{\mid }$ divides $\mathrm{\mid }G\mathrm{\mid }$. (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).