Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm going over some past papers and have been able to show that if $p$, $q$ are distinct odd primes and $\gcd (a, pq)=1$ then $a^{\operatorname{lcm}(p-1,q-1)} \equiv 1 \pmod {pq}$

the next part says taking $g$ to be a primitive root $\mod p$ and $h$ a primitive root $\pmod q$, apply the Chinese Remainder Theorem to specify an integer a whose order $\pmod {pq}$ is exactly $\operatorname{lcm}(p-1,q-1)$ but I'm not sure how to do this.

So far I've written out this but don't see how to solve

$$g^{p-1} \equiv 1 \pmod p$$

$$h^{q-1} \equiv 1 \pmod q$$

$$a^{\operatorname{lcm}(p-1,q-1)} \equiv 1 \pmod p$$

$$a^{\operatorname{lcm}(p-1,q-1)} \equiv 1 \pmod q$$


share|cite|improve this question
Remember that the order of $a$ is $r$ if and only if $a^r \equiv 1$ and $a^{r/d} \not\equiv 1$ for proper divisors $d$ of $r$. – TMM Apr 28 '12 at 19:18
By the Chinese Remainder Theorem, there is an $e$ congruent to $g$ modulo $p$ and congruent to $h$ modulo $q$. – André Nicolas Apr 28 '12 at 19:35
OK so $e \equiv g$ mod p and $e \equiv h$ mod q but then $e=g \cdot m_1 +h \cdot m_2$ where $m_1$ and $m_2$ are inverses such that $qm_1 \equiv 1$ mod p and $pm_2 \equiv 1$ mod q and don't know how to find those? – Mike Davies Apr 28 '12 at 20:11
Unfortunately I used $e$ instead of the $a$ in your statement. Is your query about how to find $a$ which is simultaneously congruent to $g$ mod $p$ and to $h$ mod $q$? Or is your query about showing that this $a$ has the desired order mod $pq$? – André Nicolas Apr 28 '12 at 20:52
It asks how to specify such an $a$ and to show that a has the required order. – Mike Davies Apr 28 '12 at 21:18
up vote 1 down vote accepted

Let $g$ be a primitive root modulo $p$, and $h$ a primitive root modulo $q$.

By the Chinese Remainder Theorem, there exists a unique integer $a$ modulo $pq$ such that $a\equiv g\pmod{p}$ and $a\equiv h\pmod{q}$.

(To "find" this $a$, since $p$ and $q$ are relatively there exist integers $x$ and $y$ such that $xp+yq = 1$. Then $xp\equiv 1\pmod{q}$, and $yq\equiv 1\pmod{p}$. Therefore, $a=gyq+hxp$ satisfies $$\begin{align*} a &= gyq+hxp \equiv gyq \equiv g(yq)\equiv g\pmod{p}\\ a &= gyq+hxp \equiv hxp \equiv h(xp) \equiv h\pmod{q} \end{align*}$$ so $gyq+hxp\bmod pq$ will do.)

By the Chinese Remainder Theorem, $$\begin{align*} a^t\equiv 1\pmod{pq} &\iff a^t\equiv 1\pmod{p}\text{ and }a^t\equiv 1\pmod{q}\\ &\iff g^t\equiv 1\pmod{p} \text{ and }h^t\equiv 1\pmod{q}\\ &\iff t\equiv 0\pmod{p-1} \text{ and }t\equiv 0\pmod{q-1}\\ &\qquad\qquad\text{(since }g\text{ is a primitive root mod }p,\ h\text{ a primitive root mod }q)\\ &\iff p-1|t\text{ and }q-1|t\\ &\iff \mathrm{lcm}(p-1,q-1)|t. \end{align*}$$ So the order of $a$ is exactly $\mathrm{lcm}(p-1,q-1)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.