Proving that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where $l=\mathrm{lcm}(m,n)$ and $d=\gcd(m,n)$ 
How would one go about proving that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l $ as groups, where  $l=\mathrm{lcm}(m,n)$ and $d=\gcd(m,n)$? 

I am attempting to use the fundamental theorem of finitely generated abelian groups but am struggling.
In the interest of honesty, this is a past exam question that I am attempting for which the solutions are not available.
EDIT: As the question was asked in the paper, the Chinese Remainder Theorem would not be permitted since this is only proved in the follow-up course.
 A: Write $m=dm', n=dn', d=mu+nv$. Then $l=m'n=mn'$.
These row and columns operations prove that $\mathbb{Z}_m\oplus \mathbb{Z}_n \cong \mathbb{Z}_d\oplus \mathbb{Z}_l$:
$$
A=\pmatrix{ m & 0 \\ 0 & n}
\to \pmatrix{ m & mu \\ 0 & n}
\to \pmatrix{ m & mu+nv \\ 0 & n}
= \pmatrix{ m & d \\ 0 & n}\\
\to \pmatrix{ 0 & d \\ -m'n & n}
= \pmatrix{ 0 & d \\ -l & n}
\to \pmatrix{ 0 & d \\ -l & 0}
\to \pmatrix{ d & 0 \\ 0 & l}=B
$$
An explicit isomorphism can be written by collecting the row and columns operations into two matrices $P,Q$ so that $B=PAQ$:
$$
P = 
\pmatrix{ 1 & 0 \\ -n' & 1}
\pmatrix{ 1 & v \\ 0 & 1}
=\pmatrix{1 & v \\ -n' & 1 - v n'}
\\
Q =
\pmatrix{ 1 & u \\ 0 & 1}
\pmatrix{ 1 & 0 \\ -m' & 1}
\pmatrix{ 0 & -1 \\ 1 & 0}
= \pmatrix{u & -1 + u m' \\ 1 & m'}
$$
If $e_1, e_2$ is the canonical basis for $\mathbb Z^2$, then the basis $f_1, f_2$ given by $F=Q^{-1}E$ is such that this diagram commutes:
$$
\matrix { \mathbb Z^2 , \{ e_1, e_2\} & \to & \mathbb Z^2, \{ f_1, f_2\} \\
\downarrow & & \downarrow \\
\mathbb{Z}_m\oplus \mathbb{Z}_n & \to & \mathbb{Z}_d\oplus \mathbb{Z}_l
}
$$
This isomorphism does not use prime factorizations nor explicitly the Chinese Remainder Theorem.
A: Let $p_1, \dots, p_n$ be all prime numbers that divide either $m$ or $n$.
Suppose $m = p_1^{i_1}\dots p_n^{i_n}$ and $n = p_1^{j_1} \dots p_n^{j_n}$.
Now $d = p_1^{\min(i_1,j_1)} \dots p_n^{\min(i_n,j_n)}$ and $l = p_1^{\max(i_1,j_1)} \dots p_n^{\max(i_n,j_n)}$
Now we apply the Chinese remainder theorem, using that powers of distinct primes are coprime $\mathbb{Z}_m \oplus \mathbb{Z}_n \cong \mathbb{Z}_{p_1^{i_1}} \oplus \dots \oplus \mathbb{Z}_{p_n^{i_n}} \oplus \mathbb{Z}_{p_1^{j_1}} \oplus \dots \oplus \mathbb{Z}_{p_n^{j_n}}$
Now we simply rearrange depending on whether $i_k$ or $j_k$ is bigger:
$\cong \mathbb{Z}_{p_1^{\min(i_1,j_1)}} \oplus \dots \oplus \mathbb{Z}_{p_n^{\min(i_n,j_n)}} \oplus \mathbb{Z}_{p_1^{\max(i_1,j_1)}} \oplus \dots \oplus \mathbb{Z}_{p_n^{\max(i_n,j_n)}} \cong \mathbb{Z}_d \oplus \mathbb{Z}_l$
In the last step we used the Chinese remainder theorem again.
