Subgroup of free abelian group

Question

Let $G$ be a free abelian group of finite rank. Let $H$ be a subgroup of $G$ such that $$ G = H + pG$$ for some prime number $p$. How can I show that $[G:H]<\infty$?

My proof is like this;

$G \cong \mathbb{Z}^{n}$ for some $n \in \mathbb{N}$, thus $pG = \{ (p{i_1},p{i_2}, \cdots, p{i_n}) : i_{1}, \cdots i_{n} \in \mathbb{N} \}.$ Thus, by Euclidean algorithm, for any $g= (a_{1}, \cdots, a_n) \in G$, $\exists (p{i_1},p{i_2}, \cdots, p{i_n}) \in pG$ such that $$g-(p{i_1},p{i_2}, \cdots, p{i_n}) = (r_{1}, \cdots, r_{n})$$ where $0 <= r_i < p$ for any $0 <= i <=n$. By the condition $G =H+pG$, $(r_1 , \cdots, r_{n}) \in H$. Thus $$H = <\{ (r_{1},\cdots, r_{n}): 0<= r_{i}<p \}>.$$ Therefore, H=G implies $[G:H]=1$

Is this proof right? Or could you give me some idea?

Answer 1

Use additive notation, which is implied by the question.

Consider $$ p(G/H)=(pG+H)/H=G/H $$ Hence $K=G/H$ is a finitely generated abelian group with the property that $pK=K$.

Write $K=F\oplus C$, where $F$ is free and $C$ is finite, by the fundamental theorem on finitely generated abelian groups.