Basis theorem for finite abelian groups question (Pinter's "A book of Abstract Algebra")

Question

I am stuck at the exercise Q5 of Chapter 16. Here is a summary of the question:

Let $G$ be a finite Abelian $p$-group where $p$ is a prime. Let $a \in G$ be an element of highest possible order in $G$. Let $H$ be the subgroup generated by $a$. Now $G/H$ = [$Hb_1$, $Hb_2$, ... $Hb_n$] and we can choose $b_i$ such that $ord(b_i) = ord(Hb_i)$ (this is proven in another exercise O).

The question is to prove that for any intergers $l_0, l_1, l_2,..., l_n$, if $a^{l_0}b_1^{l_1}b_2^{l_2}...b_n^{l_n} = e$, then $a^{l_0}=b_1^{l_1}=b_2^{l_2}=...=b_n^{l_n} = e$.

What I have tried: it looks like I need to use induction. So I started with simple cases.

If $n=0$ this is true.

If $n=1$, $a^{l_0}b_1^{l_1}=e$ means $(Ha^{l_0})(Hb_1^{l_1})=H$, which in turn gives $(Hb_1)^{l_1}=H$. So $l_1$ is a multiple of $ord(Hb_1)$, which is also a multiple of $ord(b_1)$. This indicates $b_1^{l_1}=1$.

If $n=2$, assume $a^{l_0}b_1^{l_1}b_2^{l_2}=e$. There are two cases, if $b_2^{l_2}=1$, by the $n=1$ case it's done. If, however, if $b_2^{l_2} \neq 1$, I am stuck. Trying the logic $(Hb_1^{l_1})(Hb_2^{l_2})=H$ and I cannot use the information that $ord(b_i) = ord(Hb_i)$.

I also notice that I haven't used the information that $G$ is a $p$-group, but I have struggled a few days without any progress. I've tried searching for proof and found a question on Q2, which unfortunately does not seem to help in my case. Could anyone shed some light on how to prove this? Appreciate your help!

Answer 1

Pinter gives a specific meaning to the notation $[\dotsc]$, namely, he writes $G = [a_1,\dotsc,a_n]$ if

1. Every $x \in G$ can be written as $x = a_1^{k_1} \dotsm a_n^{k_n}$ for some integers $k_n$,
2. if $a_1^{l_1} \dotsm a_n^{l_n} = e$ then $a_1^{l_1}=\dotsb=a_n^{l_n}=e$.

For the exercise you're asking about, you'll have to use that $G/H = [Hb_1,\dotsc,Hb_n]$ in this sense, plus $\operatorname{ord}(Hb_i) = \operatorname{ord}(b_i)$.