If $\mathbb{Z}_{pq} \cong \mathbb{Z}_{p} \times\mathbb{Z}_{q}$, then $(\mathbb{Z}_{pq})^* \cong (\mathbb{Z}_{p})^* \times (\mathbb{Z}_{q})^*$

Question

I'm reading Aluffi's Algebra Chapter 0, and one of the problems asks to compute the order of $(\mathbb{Z}_{pq})^*$ (the group of integers relatively prime to $pq$ under multiplication) when $p, q$ are prime and $p \neq q$

The previous problem asks to prove that $\mathbb{Z}_{mn} \cong \mathbb{Z}_m \times \mathbb{Z}_n$ whenever $m,n$ are relatively prime. So, naively I thought, with $p$ and $q$, $(\mathbb{Z}_{pq})^* \cong (\mathbb{Z}_p)^* \times (\mathbb{Z}_q)^*$, yielding the correct result that the the number of relatively prime integers under $pq$ is $(p-1)(q-1)$. However, I can't seem to prove this.

Help appreciated :-)

Do you know that $Aut(G\times H) = Aut(G)\times Aut(H)$ when $|G|$ and $|H|$ are coprime? — Giorgos Giapitzakis, Sep 02 '21 at 12:35
The group of units of the ring $R=\Bbb Z_n$ is $U(R)=\Bbb Z_n^*$. So the duplicate exactly solves your question. Actually the direct sum is the direct product. If $R=R_1\times R_2$, then $U(R)=U(R_1)\times U(R_2)$. — Dietrich Burde, Sep 02 '21 at 12:38
@DietrichBurde we haven't hit a discussion of rings yet in the book. The concepts we hit so far are only categories, groups and homomorphisms — Timothy Samson, Sep 02 '21 at 12:44
You don't need rings in this argument really. Just consider it as a group in the proof there. On the other hand, this duplicate proves it directly, if you prefer this. It is precisely for those not knowing rings or Chinese remainder theorem. — Dietrich Burde, Sep 02 '21 at 12:54
Also this duplicate gives a proof without using rings. They write $R_m$ for $\Bbb Z_m^*$. — Dietrich Burde, Sep 02 '21 at 13:02

If $\mathbb{Z}_{pq} \cong \mathbb{Z}_{p} \times\mathbb{Z}_{q}$, then $(\mathbb{Z}_{pq})^* \cong (\mathbb{Z}_{p})^* \times (\mathbb{Z}_{q})^*$

0 Answers0