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I am working through Aluffi's Algebra Chapter $0$ and I'm not sure how the author intended us to use the conclusion from exercise $4.9$ in $4.10$

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Since $p$ and $q$ are distinct prime numbers, they are relatively prime. So using $4.9$, $C_p \times C_q \cong C_{pq} \cong (\mathbb{Z} / pq \mathbb{Z})^*$. However $|(\mathbb{Z} / pq \mathbb{Z})^*| \ne pq$. The order of $(\mathbb{Z} / pq \mathbb{Z})^*$ is $pq-p-q+1 = (p-1)(q-1)$. Since it is assumed that $p$ and $q$ are odd, $gcd(p-1,q-1) \geq 2$ and we can no longer utilize the result of $4.9$

Aydin Ozbek
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  • Isn't saying $C_{pq}\cong (\mathbb Z/pq\mathbb Z)^$ assuming that $(\mathbb Z/pq\mathbb Z)^$ is cyclic? – Mark Schultz-Wu Mar 31 '17 at 18:39
  • It surprises me that you have to use 4.9. I think the idea is to say ${\mathbb Z}_{pq} \cong {\mathbb Z}_p \times {\mathbb Z}_q$ and take units on both sides as in the anser by Dietrich Burde. However, for this you need the isomorphism as rings and 4.9 only gives you the isomorphism as additive groups. – Magdiragdag Mar 31 '17 at 18:56

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The mistake here is your statement $C_{pq}\cong (\mathbb{Z} / pq \mathbb{Z})^*$, which is not true, because, as you said, the order of these two groups is different, namely $pq\neq (p-1)(q-1)$. Instead you should conclude $$ (\mathbb{Z}/p)^*\times (\mathbb{Z}/q)^*\cong (\mathbb{Z}/(pq))^* $$ from $\mathbb{Z}/p\times \mathbb{Z}/q\cong \mathbb{Z}/(pq)$. Then you have shown that $N=(p-1)(q-1)$.

Dietrich Burde
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  • So the operation of taking the units of both groups preserves the isomorphism? – Aydin Ozbek Mar 31 '17 at 19:51
  • Yes, but it is better to conclude this from the isomorphism of rings (see the answer of Alex Youcis at the duplicate). – Dietrich Burde Apr 01 '17 at 08:50
  • @Aydin Ozbek, taking units preserves isomorphisms, but this property isn't preserving isomorphisms, it is commuting with products. Preserving isomorphisms would be $(\Bbb Z/p \times \Bbb Z/q)^\cong (\Bbb Z/(pq) )^$. – Ennar Apr 01 '17 at 09:10