Putnam 2009, Problem A-5

Is there a finite abelian group $G$ such that the product of the orders of all its elements is $2^{2009}$?

Solution:
No, there is no such group. By the structure theorem for finitely generated abelian groups, \( G \) can be written as a product of cyclic groups. If any of these factors has odd order, then \( G \) has an element of odd order, so the product of the orders of all of its elements cannot be a power of 2. We may thus consider only abelian 2-groups hereafter. For such a group \( G \), the product of the orders of all of its elements has the form \( 2^{k(G)} \) for some nonnegative integer \( G \), and we must show that it is impossible to achieve \( k(G) = 2009 \). Again by the structure theorem, we may write
\[
G \cong \prod_{i=1}^\infty (\mathbb{Z}/2^i\mathbb{Z})^{e_i}
\]
for some nonnegative integers \( e_1, e_2, \ldots \), all but finitely many of which are 0. For any nonnegative integer \( m \), the elements of \( G \) of order at most \( 2^m \) form a subgroup isomorphic to
\[
\prod_{i=1}^\infty (\mathbb{Z}/2^{\min\{i,m\}}\mathbb{Z})^{e_i},
\]
which has \( 2^{s_m} \) elements for \( s_m = \sum_{i=1}^\infty \min\{i,m\}e_i \). Hence
\[
k(G) = \sum_{i=1}^\infty i(2^{s_i} - 2^{s_{i-1}}).
\]
Since \( s_1 \leq s_2 \leq \cdots \), \( k(G) + 1 \) is always divisible by \( 2^{s_1} \). In particular, \( k(G) = 2009 \) forces \( s_1 \leq 1 \). However, the only cases where \( s_1 \leq 1 \) are where all of the \( e_i \) are 0, in which case \( k(G) = 0 \), or where \( e_i = 1 \) for some \( i \) and \( e_j = 0 \) for \( j \neq i \), in which case \( k(G) = (i - 1)2^i + 1 \). The right side is a strictly increasing function of \( i \) which equals 1793 for \( i = 8 \) and 4097 for \( i = 9 \), so it can never equal 2009. This proves the claim.

**Remark.** One can also arrive at the key congruence by dividing \( G \) into equivalence classes, by declaring two elements to be equivalent if they generate the same cyclic subgroup of \( G \). For \( h > 0 \), an element of order \( 2^h \) belongs to an equivalence class of size \( 2^{h-1} \), so the products of the orders of the elements of this equivalence class is \( 2^j \) for \( j = h2^{h-1} \). This quantity is divisible by 4 as long as \( h > 1 \); thus to have \( k(G) \equiv 1 \pmod{4} \), the number of elements of \( G \) of order 2 must be congruent to 1 modulo 4. However, there are exactly \( 2^e - 1 \) such elements, for \( e \) the number of cyclic factors of \( G \). Hence \( e = 1 \), and one concludes as in the given solution.