If $G$ is abelian, it has a subgroup in every order of $|G|'s$ divisors?

Question

Assume that $G$ is an abelian group, I read somewhere that it can be derived from Lagrange's theorem that it has a number of subgroups that is equal to the number of $G$'s divisors.

Why does it hold? I'm confused...

Answer 1

Yes, the fundamental theorem of finite abelian groups trivializes it. We also prove it without this theorem using strong induction. The base case when the order is $n$ and $n=1$ is trivial.

Inductive step: Let $|G|=n$ and $d|n$ we need to prove there is a subgroup of order $d$. If $d=1$ we are done. Else let $p$ be a prime dividing $d$. By Cauchy's theorem there is a subgroup $P$ of $G$ of order $p$. Since $G$ is abelian, $P$ is normal. Hence we can consider the homomorphism $\varphi:G\rightarrow \frac{G}{P}$. The group $\frac{G}{P}$ has order $\frac{n}{p}$. By the inductive hypothesis there is a subgroup $C$ of $\frac{G}{P}$ of order $\frac{d}{p}$.

Apply the first isomorphism theorem:

$\frac{\varphi^{-1}(C)}{P}\cong C$. This tells us $|\varphi^{-1}(C)|=|P||\frac{G}{P}|=p\frac{d}{p}=d$. So $\varphi^{-1}(C)$ is the group we were looking for.

Answer 2

Perhaps what you mean is that if $G$ is a finite abelian group of order $n$, then $G$ has at least one subgroup of order $d$ for every divisor $d\ge 1$ of $n$. This claim follows easily from the fundamental theorem of finite abelian groups. Note that it does not state that there is a unique subgroup of every divisor. In fact, a finite group $G$ is cyclic if, and only if, it has at most one subgroup of every possible divisor. In any case, the general theorem does not state how many subgroups of a particular order there are. Only that there is at least one of every possible divisor.

Obviously, there is precisely one subgroup of order $1$ and precisely one subgroup of order $n$. Each of the cyclic groups $\mathbb Z_n$ has precisely one subgroup of every possible order. The Klein group $\mathbb Z_2 \times \mathbb Z_2$ has three subgroups of order $2$.