I'm trying to prove that a finite abelian group $G$ is generated by elements of maximal order.
I can sort of see why that happens in a vague instinctive kind of way but no real hard logic. So far, I have tried to use this lemma:
Let $G$ be a finite abelian group and a be an element of maximal order in $G$. Then any element $b$ is such that $|b|$ divides $|a|$.
Then each cyclic group generated by the maximal order element contains exactly 1 element of each possible order for an element in the group. I have tried expanding from this idea in a few directions but I don't think it's the correct way to go since it's not going anywhere.
Any help appreciated. Thanks guys.
Best Answer
Let $G$ be a finite abelian group, then $G = \mathbb{Z}_{n_1}\oplus\cdots\oplus \mathbb{Z}_{n_k}$ for some $n_1|\cdots|n_k$. Then clearly these elements $$\{(1,\cdots,1),(0,1,\cdots,1),(0,0,1,\cdots,1),\cdots,(0,\cdots,0,1)\}$$ are of maximal order $n_k$ and they generate $G$.