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Suppose $a,b,c,d$ are positive integers such that $b$ is an integer multiple of $a$ and $d$ is an integer multiple of $c$. Prove that if the direct products $$\mathbb{Z}_a\oplus\mathbb{Z}_b \text{ and } \mathbb{Z}_c\oplus\mathbb{Z}_d$$ are isomorphic, then $a=c$ and $b=d$.

My attempt: We can see that because these two products are isomorphic, $a*b=c*d$, so $a^2k=c^2l$ for some integers $k,l$. Hence,

$a = c\sqrt{\frac{l}{k}}$, so since $a$ is an integer, $k$ must divide $l$.

I got this far, but now I'm stuck.

Any help appreciated!

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    It's not clear why $\sqrt{\frac{l}{k}}$ is an integer - can't $a=c\cdot \frac{1}{2}$ with $a,c$ integers? – 2017-02-24

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You aren't going to get there purely by counting.

For example, $a=2,b=18$ and $c=3,d=12$ will give the same counts. Why are they not isomorphic? (Note that in this case, $k=9,l=4$ and $l\not\mid k$.)

Hint: What is the maximum order of an element of the left side, given that $a\mid b$?

What is the maximum order of an element on the right side, given that $c\mid d$?

If the two are isomorphic then...?

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    This helps! So the maximum order of $\mathbb{Z}_a\oplus\mathbb{Z}_b$ must be $b$, and the maximum order of $\mathbb{Z}_c\oplus\mathbb{Z}_d$ must be $d$. Is it then true that those two elements are isomorphic, i.e. $c=d$? – 2017-02-24
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    @Jess - If the groups are isomorphic, then any element of maximum order in one must be sent to *some* element of maximum order in the other, so yes, the maximum orders match. (But it would take more work to show that any specific pair of elements of maximum order, one from each group, were paired up by some isomorphism.) Btw, I think $c=d$ was a typo? – 2017-02-26