Prove that if $G$ is an abelian group of prime-power order $p^n$ and let $a$ be an element of maximum order in $G$, then $G$ can be written in the form $\langle a\rangle \times K.$
Our teacher told us to prove this lemma in this way:
Assume $|a| = p^{k_1}$ for some $k_1 ≤ n$. If $k_1 = n$ the proof is completed by choosing $K = ⟨e⟩$. Suppose $k_1 < n$ and pick $b' ∈ G$ such that $\bar{b'}$ has maximum order $p^{k_2}$ in $G/\langle a\rangle$. Note that $1 ≤ k_2 ≤ k_1$ by the maximum order of a, and $(b′)^{p^{k_2}} = a ^i ∈ ⟨a⟩$ for some $0 ≤ i < p^{k_1}$. Since $a^{ip^{k_1−k_2}} = (a^i)^{p^{k_1−k_2}} = (b'^{p^{k_2}})^{p^{k_1−k_2}} = (b′)^{p^{k_1}} = e$, $i$ is a multiple of $p^{k_2}$. Let $b = b′a^{−i/p^{k_2}}$. Note that $\bar{b}$ has the same order $p^{k_2}$ of $\bar{b′}$ in $G/⟨a⟩$, which means $b^i \not\in ⟨a⟩$ if $i < p^{k_2}$. Note that $b^{p^{k_2}} = b'^{k_2}a^{−i} = e$. This proves $⟨a⟩ ∩ ⟨b⟩$ = {e} and $|⟨a⟩ × ⟨b⟩| = p^{k_1+k_2}$. If $k_1 + k_2 = n$ then $G = ⟨a⟩ × ⟨b⟩$ and the proof is completed by choosing $K = ⟨b⟩$. Suppose $k_1 + k_2 < n$, and pick $c′ \in G$ such that $\bar{\bar{c′}}$ has maximum order $p^{k_3}$ in $G/(⟨a⟩ × ⟨b⟩)$. Note that $1 ≤ k_3 ≤ k_2$ since $p^{k_3} = |\bar{\bar{c′}}| ≤ |\bar{c′}| ≤ |\bar{b}| = p^{k_2}$...
But I don't know how to continue. Can someone help me? Thanks!