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The lemma to be proved is:

Let $G$ be a finite abelian $p$-group that is not cyclic. Suppose that $g \in G$ has maximal order. If $h \in G \setminus \langle g \rangle$ has smallest possible order, then $|h| = p$.

And the proof is written as:

Let $g \in G$ be of maximal order in $G$, say $|g| = p^m$ for some $m \le n$. Since $G$ is not cyclic, $G \neq \langle g \rangle$. Choose $h \in G \setminus \langle g \rangle$ where $h$ has smallest possible order, say $|h|=p^l$. Since $e \in \langle g \rangle$, then $h \neq e$ and so $l > 0$. But $|h^p|=p^{l-1}$ and so $|h^p|$ has smaller order than $|h|$, whence $h^p \in \langle g \rangle$.

What I could not understand is the bolded line: how can we say that $h^p \in \langle g \rangle$ because $|h^p|$ has smaller order than $|h|$?

Maths Rahul
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GalaxyY
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    Is $\langle g\rangle$ the cyclic subgroup generated by $g$? And if so, what is $G\langle g\rangle$? Do you mean $G \setminus \langle g\rangle$ (the set of elements of $G$ not in $\langle g\rangle$? – Misha Lavrov Oct 12 '21 at 02:57
  • You are right. Typo is corrected. – GalaxyY Oct 12 '21 at 04:04
  • If order of $h$ is $p$, we are done. If order of $h$ is $p^2$, what it means? $h^{p^2}=1$ but no smaller positive power is $1$. Now, what is order of $h^p$? It is $p$ - why? $(h^p)^p=1$. So, $h^p$ is an element of order $p$, which is less than the smallest possible order of element in $G\setminus \langle g\rangle$; so $h^p$ should lie inside $\langle g\rangle$. – Maths Rahul Oct 12 '21 at 04:15
  • Got it. Thanks for the proof. – GalaxyY Oct 14 '21 at 11:23

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We chose $h$ to be an element of $G \setminus \langle g\rangle$ whose order is as small as possible.

Therefore if $h^p$ has smaller order than $h$, $h^p$ cannot be an element of $G \setminus \langle g\rangle$: otherwise, we would have chosen $h^p$ instead of $h$. This means it must be an element of $\langle g\rangle$.

Misha Lavrov
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