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This question is inspired by $\mathbb{Z}_a\oplus\mathbb{Z}_b\cong \mathbb{Z}_c\oplus\mathbb{Z}_d$ question.

We change the additive structure to multiplicative:

Problem 1: If $a|b$, $c|d$ and $\mathbb{Z}^*_a \times \mathbb{Z}^*_b \cong \mathbb{Z}^*_c \times \mathbb{Z}^*_d$. Does this imply $(a,b)=(c,d)$?

The counter-example to Problem 1 is: $$\mathbb{Z}^*_3 \times \mathbb{Z}^*_{12} \cong \mathbb{Z}^*_4 \times \mathbb{Z}^*_{12}$$

And so I slightly modified it.

Problem 2: If $a|b$, $c|d$, $\mathbf{ab=cd}$ and $\mathbb{Z}^*_a \times \mathbb{Z}^*_b \cong \mathbb{Z}^*_c \times \mathbb{Z}^*_d$. Does this imply $(a,b)=(c,d)$?

Is there a counter-example now?

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nice question.+1 for the problem statment – dato datuashvili Feb 26 '12 at 11:26
up vote 4 down vote accepted

Isn't $(a,b,c,d)=(2,8,4,4,)$ a counterexample?

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$Z^*_8$ is not $Z^*_4 \times Z^*_4$, because $Z^*_8 \cong C_4$, $Z^*_4 \times Z^*_4 \cong C_2 \times C_2$. One group has an element of order 4, all proper elements are of order 2 in the other. – Tom Artiom Fiodorov Feb 26 '12 at 11:56
@Artiom: Actually, $C_4\not\cong(\mathbb{Z}/2^3\mathbb{Z})^\times\cong C_2\times C_2$. (Wikipedia) – anon Feb 26 '12 at 12:04

Initially I tried to come up with a sufficient condition to conclude that $(a,b)=(c,d)$, and $ab=cd$ is actually suffice ignoring the snag above.

So restrict ourselves to $n$ just that $Z^*_n \cong C_{\phi(n)}$, i.e. $4\nmid n$.

Hence the problem 3:

$$ C_{\phi(a)}\times C_{\phi(b)} \cong C_{\phi(c)}\times C_{\phi(d)}$$ $a|b,c|d,ab=cd. 4\nmid b,d.$ Is $(a,b)=(c,d)$?

Then by considering the highest order in each product group, we conclude that $$\phi(b)=\phi(d)$$ Quoting out $C_{\phi(b)}$ yields $\phi(a)=\phi(c).$

From $ab=cd$ and $c|d$ it's possible to show that $a|d$ and $c|b$.

Put $b=\operatorname{LCM}(a,c)l_1$ and $d=\operatorname{LCM}(a,c)l_2$, and from $ab=cd$ deduce that $l_1=\frac{c}{\operatorname{GCD(a,c)}}j$, $l_2=\frac{a}{\operatorname{GCD(a,c)}}j$


$$ \phi\big(\frac{\operatorname{LCM}(a,c)}{\operatorname{GCD(a,c)}}cj\big)=\phi\big(\frac{\operatorname{LCM}(a,c)}{\operatorname{GCD(a,c)}}aj\big) $$

Now by cancelling out factors of $j$ which are coprime with $\operatorname{LCM}(a,c)$ we can assume that all prime factors of $j$ are present in the factorisation of $\operatorname{LCM}(a,c)$. But now RHS and LHS have the same primes, and so the powers must be the same!

This concludes that $a=c$.

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