This is Exercise 4.3.9(b) of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
This question is referred to later on in the text.
The previous part is here:
Questions of mine on similar topics include:
- For $H\le G$, abelian $G$, show that $r_0(H)+r_0(G/H)=r_0(G)$, where $r_0(K)$ is the torsionfree rank of $K$
- If $H\le G$ for an abelian group $G$, then $r_p(H)+r_p(G/H)\ge r_p(G)$, where $r_p(K)$ is the $p$-rank of $K$.
The Details:
On page 98 to 99, ibid., we have
Let $G$ be an abelian group and let $S$ be a nonempty subset of $G$. Then $S$ is called linearly independent, or simply independent, if $0\notin S$ and, given distinct elements $s_1,\dots, s_r$ of $S$ and integers $m_1,\dots, m_r$, the relation $m_1s_1+\dots+m_rs_r=0$ implies $m_is_i=0$ for all $i$.
. . . and . . .
If $p$ is prime and $G$ is an abelian group, the $p$-rank of $G$
$$r_p(G)$$
is defined as the cardinality of a maximal independent subset of elements of $p$-power order. Similarly the $0$-rank or torsion-free rank
$$r_0(G)$$
is the cardinality of a maximal independent subset of elements of infinite order.
Note that
$$nG=\{ng\mid g\in G\}$$
is the set of all $n$th multiples of elements of an abelian group $G$.
The Question:
Let $G$ be an abelian group such that $r_p(G)<\infty$ if $p=0$ or a prime.
(b) Prove that $G/nG$ is finite for all $n>0$.
Thoughts:
Let $G$ be as defined above.
The Fundamental Theorem of Finitely Generated Abelian Groups is not discussed in Robinson's book yet. However, I get the impression that the case when $G$ is finitely generated is much easier.
Since $G$ is abelian, $nG\unlhd G$, so the question makes sense.
Proof: We have $e=ne\in nG$, so $nG\neq \varnothing$. All multiples of any element of $G$ are in $G$ by closure of $G$, so $nG\subseteq G$. Let $x,y\in nG$. Then there exist $g,h\in G$ such that $x=ng, y=nh$. Thus
$$\begin{align} x-y&=ng-(nh)\\ &=n(g-h), \end{align}$$
for $g-h\in G$, so $x-y\in nG$. Hence $nG\le G$ by the one-step subgroup test. Moreover, for $k\in G$, we have $k+x-k=x\in nG$, so $nG\unlhd G$.$\square$
Indeed, all subgroups of an abelian group are normal in that group.
If $G$ is finite, then so is $G/nG$, since
$$\lvert G\rvert=\lvert G/nG\rvert\lvert nG\rvert.$$
This is rather easy to see.
I'm not sure how to incorporate the $p$-rank and $0$-rank of $G$ nor of $G/nG$, only each of $G/nG$ is at most that of $G$, and it is necessary to have/show that $r_0(G/nG)=0$.
The question is easy if $G=\Bbb Z$.
The kind of answer I'm hoping for is a full solution, please.
I think I could answer this myself if I had more time. As with the previous part of this question, though, I am trying to maintain a reasonable pace in the book, and after a week, I think it's ripe for asking here.
Please help :)