A subgroup in which every element has order a power of a fixed prime $p$ is
called a $p$-(sub)group.
A finite abelian $p$-group (Exercise 7) is generated by its elements of maximal
order.
Each subgroup / element has order power of some prime and if $|G| = p^n$, there exist an element of each order $p$ such that $p^i$ for $0\leq i<n$.
Now, I have to prove that it is generated by an element of order $p^n$ , hope I am correct! But I am at loss of ideas.
Do you mind giving some hints?
Best Answer
We will prove that a finite abelian p-group is generated by its elements of maximal order.
Let $G$ be an Abelian group of order $p^n$ and let $M$ be the set of all elements of maximal order $p^m$. (Note in parentheses that generally speaking $m\leq n$.) Let $H=\langle M\rangle$.
We will prove that $H=G$.
Let $g\in G$ and $\operatorname{ord}(g)=p^k<p^m$. Let us choose $h\in H$ such that $\operatorname{ord}(h)=p^m$.
Then $(gh)^{p^k}=g^{p^k}h^{p^k}=h^{p^k}$. It follows that $\operatorname{ord}(gh)=p^m$ (think about why) and hence $gh\in H$. Since $g=(gh)h^{-1}$ therefore $g\in H$.