Converse of Lagrange for finite abelian groups: every divisor occurs as a subgroup order

Given: The objects and hypotheses in the statement.

Let $d,a\in \mathbb{Z}$ (def-integers). We say $d$ divides $a$, and write $d\mid a$, when $a=dq\text{for some }q\in \mathbb{Z},$ the product being that of def-int-operations. We write $d\nmid a$ when this fails. In this situation $d$ is called a divisor, or a factor, of $a$, and $a$ is called a multiple of $d$. This is the relation the library already has, not a second one. The published thm-division-algorithm-in-z introduces it in its own Statement, in these words: "We say $b$ divides $a$, written $b\mid a$, when $a=qb$ for some $q\in \mathbb{Z}$." Since multiplication on $\mathbb{Z}$ is commutative (thm-int-comm-ring), $a=qd$ and $a=dq$ are the same condition, so the definition above is that relation verbatim and the two usages agree everywhere. The theorem defined it for use on its own page and left the systematic theory to a later page; this is that page, and this item records the agreement rather than introducing a rival notion. The remainder test. For $b>0$ the same Statement records that $b\mid a$ holds exactly when the remainder $r$ in $a=qb+r$, $0\le r<b$, is $0$. Boundary values. Each is one line from the ring axioms, and each is used below, so all three are recorded here rather than assumed: - $d\mid 0$ for every integer $d$, including $d=0$, since $0=d\cdot 0$; - $0\mid a$ only for $a=0$, since $a=0\cdot q$ forces $a=0$; - $1\mid a$ and $a\mid a$ for every $a$, since $a=1\cdot a$ and $a=a\cdot 1$. (Divisibility in $\mathbb{Z}$: $d\mid a$ when $a=dq$ for some integer $q$).

The order of a finite group. Let $G$ be a group (def-group) whose underlying set is finite (def-countable), so that $G\approx n$ for some $n\in \mathbb{N}$ (def-equinumerous). That natural number is unique: if $G\approx n$ and $G\approx n^{\mathrm{\prime }}$ then $n\approx n^{\mathrm{\prime }}$, since $\approx$ is symmetric and transitive, and then $n=n^{\mathrm{\prime }}$ by claim 3 of lem-pigeonhole. The order of $G$ is that unique natural number, written $\mathrm{\mid }G\mathrm{\mid }$. A group is infinite when its underlying set is not finite, and $\mathrm{\mid }G\mathrm{\mid }$ is then not defined. The order of an element. Let $G$ be any group and $g\in G$, with natural powers as in def-group-power. Put $S_g:=\{k\in \mathbb{N}:k\ge 1\text{ and }{g}^k=e\}\subseteq \mathbb{N}\mathrm{.}$ - If $S_g\mathrm{\ne }\mathrm{\varnothing }$, the order of $g$ is its least element, $\mathrm{ord}(g):=\min S_g\in \mathbb{N},$ which exists by the well-ordering principle (thm-well-ordering-principle): every nonempty subset of $\mathbb{N}$ has a least element, and that element is unique, being $\le$ every element of $S_g$ and a member of it. We then say $g$ has finite order. - If $S_g=\mathrm{\varnothing }$ we say $g$ has infinite order and write $\mathrm{ord}(g)=\mathrm{\infty }$, where $\mathrm{\infty }$ is a symbol reserved for this case and is not a natural number. No arithmetic is performed with it here. By construction $\mathrm{ord}(g)\ge 1$ whenever it is finite, and $\mathrm{ord}(g)=1$ exactly when $g=e$, since $g^1=g$. Every element of a finite group has finite order. If $G$ is finite then $S_g\mathrm{\ne }\mathrm{\varnothing }$ for every $g\in G$, by lem-order-of-element-exists, so $\mathrm{ord}(g)$ is a natural number. (The order $\mathrm{\mid }G\mathrm{\mid }$ of a finite group and the order $\mathrm{ord}(g)$ of an element, with $\mathrm{ord}(g)=\mathrm{\infty }$ when no positive power of $g$ is the identity).

Let $P$ be a property of naturals such that for every $n\in \mathbb{N}$, if $P(m)$ holds for all $m<n$ then $P(n)$. Then $P(n)$ holds for all $n\in \mathbb{N}$. (At $n=0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

Let $n\in \mathbb{Z}$ with $n>1$, and put $S:=\{d\in \mathbb{Z}:d\mid n\text{ and }d>1\}$ (def-divides-in-z). Then $S$ is nonempty and has a least element $q$, and $q$ is prime (def-prime). In particular every integer greater than $1$ has a prime divisor. (Every integer $n>1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime).

Let $G$ be a finite abelian group and let $p$ be a prime dividing $\mathrm{\mid }G\mathrm{\mid }$. Then $G$ contains an element, and hence a subgroup, of order $p$. (Cauchy's theorem for finite abelian groups).

Let $G$ be a group and let $N⊴G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G\mathrm{/}N$ has the left cosets $G\mathrm{/}N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN\mathrm{.}$ Independence of the chosen representatives is proved in thm-coset-multiplication-well-defined-iff-normal, and the group axioms are proved in thm-quotient-group-laws. (The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$).

If $G$ is abelian and $N⊴G$, then $G\mathrm{/}N$ is abelian. (Every quotient group of an abelian group is abelian).

Let $N⊴G$. If $[G:N]$ is finite, then the quotient group $G\mathrm{/}N$ is finite and $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]\mathrm{.}$ In particular, if $G$ is finite, then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=\frac{\mathrm{\mid }G\mathrm{\mid }}{\mathrm{\mid }N\mathrm{\mid }}\mathrm{.}$ (If $[G:N]$ is finite then $\mathrm{\mid }G\mathrm{/}N\mathrm{\mid }=[G:N]$; for finite $G$ this equals $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$).

Correspondence theorem: subgroups of $G\mathrm{/}N$ correspond to subgroups of $G$ containing $N$, with normality preserved. For $N⊴G$, the maps $H\mapsto H\mathrm{/}N$ and $K\mapsto {\pi }^{-1}(K)$ are inverse inclusion-preserving bijections between subgroups $H$ with $N\le H\le G$ and subgroups $K\le G\mathrm{/}N$; they preserve normality. (Correspondence theorem: subgroups of $G\mathrm{/}N$ correspond to subgroups of $G$ containing $N$, with normality preserved).