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Prove that $U_{pq}$ is not cyclic if $p$ and $q$ are distinct odd primes.

I am a self taught person. I just learned this and tried this on my own and came up with this.

$x \equiv 1 \pmod{p}$ and $x \equiv -1 \pmod{q}$ has a solution $[a]_{pq}$, since $p$ and $q$ are relatively prime. Because $q$ is an odd prime, $[-1]_{pq}$ is not a solution, so $[a]_{pq}\neq [-1]_{pq}$. But $a^2 \equiv 1 \pmod{p}$ and $a^2 \equiv 1 \pmod{q}$, so $a^2 \equiv 1\pmod{pq}$ since $p$ and $q$ are relatively prime, and thus $[a]_{pq}$ has order 2.

Can someone please tell me if this proof this correct? Please help with proof as I learned it just now as a self taught person.

Zev Chonoles
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9959
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  • Why is that a contradiction? You need to finish off your argument. – Zev Chonoles Apr 08 '13 at 04:43
  • This is the most I could come up with. Can yoiu please help me with details? – 9959 Apr 08 '13 at 04:51
  • The idea is good. You have identified an element of order $2$ which is not congruent to $-1$ modulo $pq$. There is then a need to show that a cyclic group has a unique element of order $2$. – André Nicolas Apr 08 '13 at 04:51
  • That is the part which I need help with. I have no idea how to do this. Is it possible you can show me. I did this part of the proof myself. Thanks for telling me it is good. – 9959 Apr 08 '13 at 05:20
  • @9959 Let $C$ be a cyclic group of order $n$. If $d|n$, then there is a unique subgroup $D\subset C$ of order $d$. Now consider the fact that $2$ divides the order of $U_{pq}$. – Karl Kroningfeld Apr 08 '13 at 05:38
  • @9959 Your proof is correct and it'd be perfect, imo, if you'd mention what user1 wrote: a finite cyclic group has one unique subgroup of any order dividing the group's order, and in particular, if the group order is even, it has one unique element of order two. Way to go. – DonAntonio Apr 08 '13 at 10:17

2 Answers2

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This proof is indeed correct. You have produced an element of order $2$ which is not $-1$--this proves the result as Andre Nicolas said.

Phrasing what you said differently, note that the Chinese Remainder Theorem gives you a ring isomorphism $\def\Z{\mathbb{Z}}$

$$\mathbb{Z}/pq\Z\cong (\Z/p\Z)\times(\Z/q\Z)$$

This gives rise to an isomorphism of unit groups:

$$(\Z/pq\Z)^\times\cong (\Z/p\Z)^\times\times(\Z/q\Z)^\times$$

Now, $(\Z/p\Z)^\times$ and $(\Z/q\Z)^\times$ are both abelian groups of order $p-1$ and $q-1$. From here you can observe (just as you did!) that we have produced two distinct elements of order $2$, or you can note that the product of two finite abelian groups of non-coprime orders is never cyclic by FTFGAG.

Alex Youcis
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  • Thanks for telling me I did good on it. I just learned it too. I suppose this would be the entire proof because the product of two finite abelian groups of non coprime orders are never cyclic. – 9959 Apr 08 '13 at 05:29
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We prove the lemma needed to finish your argument.

Suppose that $A$ is a cyclic group of order $n$, where $n$ is even. Let $g$ be a generator of $A$. Then $g^{n/2}$ is the only element of order $2$.

For suppose that $g^k$ has order $2$, where $1\le k\le n-1$. Then $g^{2k}$ is the identity. It follows that $n$ divides $2k$, which forces $k=n/2$.

Remark: In a similar way, one can prove the following useful result. Let $A$ be an abelian group, and suppose that $a$ has finite order $d$. Then $a^k$ has order equal to $\frac{d}{\gcd(k,d)}$.

André Nicolas
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  • Alternatively, you could note that if $A$ had two elements of order $2$, say $x,y$ then $\langle x,y\rangle$ is obviously isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$ which is well-known to not be cyclic. Since every subgroup of a cyclic group is cyclic, this implies the theorem. – Alex Youcis Apr 08 '13 at 06:26