They are not isomorphic.
In what follows, $\bigoplus$ and $\prod$ will denote direct sums or
products of copies of groups indexed by $\mathbb{N}$, and $e_{n}$ will
denote the element of $\prod\mathbb{Z}$ whose $n$th coordinate is $1$,
with all other coordinates zero.
I will show that there is no surjective homomorphism $\bigoplus\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$.
An abelian group $G$ is called slender if, for every homomorphism
$\alpha:\prod\mathbb{Z}\to G$, $\alpha(e_{n})=0$ for all but finitely
many $n$. The following properties of slender groups are standard
(see, for example, Section 13.2 of Fuchs's 2015 book Abelian
Groups).
$\mathbb{Z}$ is slender (this is just the well-known theorem of
Specker).
(Infinite) direct sums of slender groups are slender. So in particular
$\bigoplus\mathbb{Z}$ is slender.
The image of any homomorphism from $\prod\mathbb{Z}$ to a slender
group is a finite rank free abelian group.
So for any homomorphism
$\alpha:\prod\mathbb{Z}\to\bigoplus\mathbb{Z}$, there is some $n$ such
that the image of $\alpha$ contains only elements of
$\bigoplus\mathbb{Z}$ whose $m$th coordinates are zero for $m\geq
n$. Call the least such $n$ the type of $\alpha$.
A homomorphism $\beta:\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$ consists
of a sequence of homomorphisms
$\beta_{i}:\prod\mathbb{Z}\to\bigoplus\mathbb{Z}$. If $\beta_{i}$ has
type $n_{i}$, we'll say that $\beta$ has type $(n_{0},n_{1},n_{2},\dots)$.
A homomorphism
$\gamma:\bigoplus\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$ consists
of a sequence of homomorphisms
$\gamma_{i}:\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$.
Suppose $\gamma_{i}$ has type $(n_{i,0},n_{i,1},n_{i,2},\dots)$. Then
every element of the image of $\gamma$, considered as an
$\mathbb{N}\times\mathbb{N}$ matrix with finitely supported rows, has
the following property:
- There is some $k\in\mathbb{N}$ such that, for every $j$, the $m$th
coordinates of the $j$th row are zero for
$$m\geq\max(n_{0,j},n_{1,j},\dots,n_{k,j}).$$
But one can easily construct elements of $\prod\bigoplus\mathbb{Z}$
that do not have this property, and so $\gamma$ cannot be
surjective, and in particular cannot be an isomorphism. For example, take an $\mathbb{N}\times\mathbb{N}$ matrix
where the $j$th row has a single nonzero entry, in the $N_{j}$th
column, where $N_{j}=\max(n_{0,j},n_{1,j},\dots,n_{j,j})$.
In fact, something more general is true. Contained in Corollary 3 of
Zimmermann-Huisgen, Birge, On Fuchs’ problem 76, J. Reine Angew. Math. 309, 86-91 (1979). ZBL0408.20038.
is the fact that the groups
$\bigoplus\mathbb{Z},\;\prod\mathbb{Z},\;\bigoplus\prod\mathbb{Z},\;\prod\bigoplus\mathbb{Z},\;\bigoplus\prod\bigoplus\mathbb{Z},\;\prod\bigoplus\prod\mathbb{Z},\dots$ are pairwise nonisomorphic.
Best Answer
Here's a low-tech approach. Start with a simple case: suppose $a$ and $b$ are powers of some prime $p$, $a = p^i$ and $b = p^j$. Then $m = p^{\min(i, j)}$ and $n = p^{\max(i, j)}$. So $m$ and $n$ are the same as $a$ and $b$ (perhaps swapped), and there's clearly an isomorphism as required.
Now suppose there are two different primes involved, $p_0$ and $p_1$: $a = {p_0}^{i_0} {p_1}^{i_1}$, $b = {p_0}^{j_0} {p_1}^{j_1}$. Intuitively, because $p_0$ and $p_1$ are relatively prime, they operate independently. So $\Bbb{Z}/(a) \simeq \Bbb{Z}/({p_0}^{i_0}) \times \Bbb{Z}/({p_1}^{i_1})$ and $\Bbb{Z}/(b) \simeq \Bbb{Z}/({p_0}^{j_0}) \times \Bbb{Z}/({p_1}^{j_1})$. Also, $m = {p_0}^{\min(i_0, j_0)} {p_1}^{\min(i_1, j_1)}$ and $n = {p_0}^{\max(i_0, j_0)} {p_1}^{\max(i_1, j_1)}$, so $\Bbb{Z}/(m) \simeq \Bbb{Z}/({p_0}^{\min(i_0, j_0)}) \times \Bbb{Z}/({p_1}^{\min(i_1, j_1)})$ and $\Bbb{Z}/(n) \simeq \Bbb{Z}/({p_0}^{\max(i_0, j_0)}) \times \Bbb{Z}/({p_1}^{\max(i_1, j_1)})$. This shows that $\Bbb{Z}/(a) \times \Bbb{Z}/(b)$ and $\Bbb{Z}/(m) \times \Bbb{Z}/(n)$ are (up to isomorphism) just rearrangements of the same factors, so again we're done.
Finally we extend this argument to any number of primes $p_k$.