TITLE: Subgroup of elements of order at most $2^{m}$
QUESTION [8 upvotes]: The problem A5 in Putnam 2009 reads as follows:

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

The answer is No. I am reading the official solution here. The solution starts by observing that if such group existed, it would be a 2-group. By structure theorem for finitely generated abelian groups, we may write
$$G\cong\prod_{i=1}^{\infty} (\mathbb{Z}/2^{i}\mathbb{Z})^{e_i}$$
for some nonnegative integers $e_1, e_2, ...$ all but finitely many of which are $0$. I am having trouble understanding the step that immediately follows: 

For any nonnegative integer $m$, the elements of $G$ of order at most
  $2^m$ form a subgroup isomorphic to  $$\prod_{i=1}^{\infty}
 (\mathbb{Z}/2^{\min(i, m)}\mathbb{Z})$$

I can see that the elements of order at most $2^{m}$ forms a subgroup. But I don't see why this subgroup must be isomorphic to the group $\prod_{i=1}^{\infty}
 (\mathbb{Z}/2^{\min(i, m)}\mathbb{Z})$. Can someone shed light on this matter? Perhaps even show an explicit isomorphism if possible? 
I appreciate any input.

REPLY [3 votes]: Just look at each factor individually.
The subgroup of $\Bbb{Z}/2^i\Bbb{Z}$ if elements of order at most $2^m$ is isomorphic to $\Bbb{Z}/2^{\min(i, m)}\Bbb{Z}$, because, as a subgroup of a cyclic group, this must be cyclic, and if $ i < m$ then all $2^i$ elements have order at most $2^m$, otherwise only $2^m$ of them do because $\Bbb{Z}/2^i\Bbb{Z}$ is cyclic, so the subgroup is generated by a single element $g$, and, because $g$ has order $2^m$, the subgroup generated by $g$ must have $2^m$ elements (if $g$ had order less that $2^m$, then all of the elements of the subgroup would also have order less than $2^m$, but clearly, if $i\ge m$, $\Bbb{Z}/2^i\Bbb{Z}$ has an element of order exactly $2^m$).
Since the order of an element of a product of groups is the LCM of the orders of it's projections onto the factors, an element of order at most $2^m$ in $G$ is given by simply choosing an element of order at most $2^m$ from each factor, which is exactly the desired isomorphism.