Step in proof showing $p$-groups are product of cyclic groups

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I am trying to follow a proof that shows that an abelian group of order $p^n $ for some $p$ prime and $n\geq 1$ then $G$ is a product of cyclic groups.

We have chosen $a \in G$ to have maximal order of $p^m $ in $G$.

Confusion here

Supposing $G \neq \langle a \rangle $, choose $b \in G \setminus \langle a \rangle $ such that $c \in G $ with $|c| < |b| $ implies $c \in \langle a \rangle $.

Why can this be assumed? I tried to work it out and I came up with something:

Maybe if we let $b \in G\setminus \langle a \rangle $ such that $b$ has minimal order. Note $|b| \neq 1 $ as $b\neq 1 $ since $b \notin \langle a \rangle $.

Then $b $ has this propert described above? If $c \in G $ with $|c|< |b| $ then we can’t have $c \in G \setminus \langle a \rangle $. So $c \in \langle a \rangle $.

Is this ok?