Let \(A\) be a finite abelian group, and let \(m\) be the maximum of the orders of the elements of \(A\). Put \[ S=\{a\in A:|a|=m\}. \] Prove that \(A\) is generated by \(S\).
Solution
1. If \(A\) is trivial, the assertion holds.
Then \(m=1\), the identity belongs to \(S\), and \(A=\langle S\rangle\).
2. Suppose \(A\) is nontrivial. By the invariant-factor form of the structure theorem,\[ A\cong C_{d_1}\oplus\cdots\oplus C_{d_r}, \qquad d_1\mid d_2\mid\cdots\mid d_r. \]Then\[ m=d_r. \]
Every element \[ (a_1,\ldots,a_r) \] has order equal to the least common multiple of the orders of its coordinates. The order of the \(i\)th coordinate divides \(d_i\), and \(d_i\mid d_r\), so every element of \(A\) has order dividing \(d_r\).
On the other hand, a generator of the last cyclic factor has order \(d_r\). Hence the maximum element order is exactly \(d_r\).
3. Let \(e_i\) denote a standard generator of the \(i\)th cyclic factor. Then\[ e_r\in S \]and, for every \(i<r\),\[ e_i+e_r\in S. \]
The element \(e_r\) has order \(d_r=m\) by step 2.
For \(i<r\), the last coordinate of \(e_i+e_r\) is \(e_r\). Hence if \[ k(e_i+e_r)=0, \] then in particular \(ke_r=0\), so \(d_r\mid k\). Conversely, \(d_r\) kills both \(e_i\) and \(e_r\) because \(d_i\mid d_r\). Thus \[ |e_i+e_r|=d_r=m, \] so \(e_i+e_r\in S\).
4. Every standard generator \(e_i\) belongs to \(\langle S\rangle\).
Step 3 gives \(e_r\in S\). For \(i<r\), both \(e_i+e_r\) and \(e_r\) lie in \(S\), so \[ e_i = (e_i+e_r)-e_r \in \langle S\rangle. \] Thus all standard generators lie in \(\langle S\rangle\).
5. Therefore\[ \boxed{A=\langle S\rangle}. \]
The standard generators \(e_1,\ldots,e_r\) generate \(A\), and step 4 places all of them in \(\langle S\rangle\). Hence \[ A\subseteq\langle S\rangle. \] The reverse inclusion is automatic because \(S\subseteq A\).
6. Q.E.D.
Steps 1 and 5 cover the trivial and nontrivial cases.