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This is a question from Aluffi's Algebra Chapter 0, which I am self studying. Specifically this is from Chapter 2, Page 69, Question 4.10

Let $p\neq q$ be odd prime integers; show that $(\mathbb{Z}/ pq\mathbb{Z})^*$ is not cylcic.

The hint we have been given is to find the order of $(\mathbb{Z}/ pq\mathbb{Z})^*$, and show that no element of this order exists.

It is clear that the order of $(\mathbb{Z}/ pq\mathbb{Z})^*$ is $(p-1)(q-1)$, however I have trouble seeing why no element $[m]_{pq}\in\mathbb{Z}/ pq\mathbb{Z}^*$ exists such that $m^{(p-1)(q-1)}\equiv 1 (\mod pq)$.

Note that there have been similar questions which have been solved using either the Chinese Remainder Theorem or the Fundamental Theorem of finitely generated Abelian Groups. However Aluffi has introduced none of these until now, and so I would appreciate any form of help which does not make use of them.

  • It's not just that you want $m^{(p-1)(q-1)} \equiv 1$, you also want $m^k \not\equiv 1$ for $1\leq k < (p-1)(q-1)$. Most likely, you want to use Fermat's Little Theorem here. – Andrew Dudzik May 22 '16 at 09:27
  • If I show that, won't I be proving that such an element necessarily has order $(p-1)(q-1)$? –  May 22 '16 at 09:33

1 Answers1

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Let $n=(p-1)(q-1)/2$ and $m\in (\mathbb{Z}/pq\mathbb{Z})^\times$. Then, $m\in \mathbb{Z}/p\mathbb{Z}$ also.

So, $m^{|(\mathbb{Z}/p\mathbb{Z})^\times|}\equiv m^{p-1}\equiv 1\pmod{p}$. Similarly, $m^{|(\mathbb{Z}/q\mathbb{Z})^\times|}\equiv m^{q-1}\equiv 1\pmod{q}$.

As $p,q$ are odd $p-1$ and $q-1$ divides $n$.Thus, $p|m^{n}-1$ and $q|m^{n}-1$. As $p,q$ are relatively prime, $pq|m^{n}-1$. So, $m^n\equiv 1\pmod{pq}$. Thus, there is no element of order $(p-1)(q-1)$ in $(\mathbb{Z}/pq\mathbb{Z})^\times$.

user5826
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Emre
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  • This works, however Aluffi does not even introduce Fermat's Little Theorem until 34 pages later. I wonder if there is a method to do this even more directly, without using Fermat's Little Theorem? –  May 22 '16 at 10:04
  • What about using $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic? – Emre May 22 '16 at 10:08
  • We prove it just after this exercise. –  May 22 '16 at 10:26
  • I edited the solution accordingly, by plugging the proof of Fermat's little theorem. This is a cheap trick, but I can't think of anything else to remove Fermat's little theorem completely, as the question is basically asking for it. – Emre May 22 '16 at 10:33
  • Could you explain how you got from $p-1|n, q-1|n$ to $p|m^n-1$? –  May 22 '16 at 10:44
  • $x-1|x^k-1$ for every $k\in\mathbb{N}$. So, $m^{p-1}-1|(m^{p-1})^{\frac{n}{p-1}}-1$ – Emre May 22 '16 at 10:52
  • I'm also reading Aluffi (with little background in group theory). There is a simple way to "avoid" Fermat's theorem in the nice proof given by Emre. We may prove without previous knowledge Lagrange's theorem (or just believe into it since it's given further in Aluffi's book). The Fermat's theorem now follows from the consequence that the order of $m$ divides $p-1$. – Bertrand Haskell Mar 11 '20 at 16:47