Standard texts prove that $\textrm{ord}_n(ab)=\textrm{ord}_n(a)\,\textrm{ord}_n(b)$ when $\textrm{gcd}(\textrm{ord}_n(a),\textrm{ord}_n(b))=1$. What if they are not relatively prime? Here $\textrm{ord}_n(b)$ is the multiplicative order of $b$, i.e. the smallest positive integer $m$ such that $b^m=1 (\textrm{mod}\,n)$.

The naive guess that $\textrm{ord}_n(ab)$ is $\textrm{lcm}(\textrm{ord}_n(a),\textrm{ord}_n(b))$ can not be right since it gives $\textrm{ord}_n(b^k)=\textrm{ord}_n(b)$, whereas in fact $\textrm{ord}_n(b^k)=\frac{\textrm{ord}_n(b)}{\textrm{gcd}(\textrm{ord}_n(b),k)}$. So the order of the product must depend on more than just orders of the factors, perhaps on something in their prime factorizations?

What does it depend on exactly? Is there a formula for the general case? If not, are there formulas for some special cases, e.g. when $a,b$ are relatively prime, square-free, prime powers, when $n$ is prime, etc.? Are there useful bounds for it? Since the product of orders is always an upper bound a lower bound would be interesting. References appreciated.


This is a rather long answer, so feel free to skip over to the formula. $\def\ord{\text{ord}}\def\div{\text{ divides }}\def\v#1#2{v_{#1}(#2)}$

Let $G$ be an abelian group with two elements $a,b$. Suppose $a$ has order $m$ and $b$ has order $n$. Denote $l=\text{lcm}(m,n)$ and $g=\text{gcd}(m,n)$. Also define $H=\langle a\rangle,K=\langle b\rangle$ and $s=|H\cap K|$.

Our goal is to determine the order of $ab$.

Since $H\cap K$ is a subgroup of $H$ and $H$ is cyclic, we know $H\cap K$ must be generated by $a^{m/s}$. Similarly $H\cap K$ is generated by $b^{n/s}$. Hence there is some $k_0$ which is relatively prime to $s$ such that $a^{m/s}=b^{nk_0/s}$.

Lemma. $$\frac ls\div\ord(ab)\div l.$$


Let $x=\ord(ab)$. Since $(ab)^x=e$, we see $a^x\in H\cap K=\langle a^{m/s}\rangle$, so $m/s\mid x$, i.e. $m\mid xs$. Similarly, $n\mid xs$. Hence $l\mid xs$, namely, $l/s\mid\ord(ab)$.

Also, $(ab)^l=e$, so $\ord(ab)\mid l$.

Theorem. $$\ord(ab)=\frac l{\gcd(s,\frac{m+nk_0}g)}.$$


From the above lemma we know $\ord(ab)=lk/s$ for some $k$.

Now we consider $(ab)^{lk/s}=e$. $$ \eqalign{ (ab)^{lk/s}&=(a^{m/s})^{lk/m}b^{lk/s}\cr &=(b^{nk_0/s})^{lk/m}b^{lk/s}\cr &=b^{lk/s(1+nk_0/m)}\cr &=b^{lk(m+nk_0)/ms} } $$ So $(ab)^{lk/s}=e$ is equivalent with $n\mid lk(m+nk_0)/ms$, i.e. $mns\mid lk(m+nk_0)$.

Note that $mns=lgs$, so it is further equivalent with $s$ dividing $k(m+nk_0)/g$.

Let $g_1=\gcd(s,\frac{m+nk_0}g)$. Then it is equivalent with $s/g_1$ dividing $k\frac{m+nk_0}{g}/g_1$. But we know $s/g_1$ and $\frac{m+nk_0}{g}/g_1$ are relatively prime, so this is equivalent with $s/g_1$ dividing $k$, that is to say, the least such $k$ is $s/g_1$.

Therefore $\ord(ab)=lk/s=\frac{l(s/g_1)}s=\frac l{g_1}$ as desired.

From this formula we can derive a necessary and sufficient condition for $\ord(ab)=l$: $\gcd(s,\frac{m+nk_0}g)=1$. A non-trivial special case is when, for every prime divisor $p$ of $s$, $p$ does not divide $\frac{m+nk_0}g$, e.g. when $p$ divides $m$ and $n$ to different degrees.

Hope this helps.

  • $\begingroup$ Thank you, this is a lot work. Are there references where I can read more about these kinds of calculations? $\endgroup$ – Conifold Aug 30 '18 at 21:07
  • $\begingroup$ I did these calculations exactly because I couldn't find a good reference in the past. XD It is my pleasure to help. :) $\endgroup$ – awllower Aug 31 '18 at 1:07
  • $\begingroup$ @Conifold I have simplified the proof and the notation, and provided a not-so-trivial application. Hope you find this of some value. ;-) $\endgroup$ – awllower Aug 31 '18 at 12:55
  • $\begingroup$ Yes, thank you, it looks much prettier now and suggests that $m+nk_0$ has some invariant significance symmetric in $a$ and $b$. I found it hard to find $s$ directly from $a,b$ and $n$, but I guess this is the heart of the difficulty, if $a$ is a primitive root then $s=n$ and primitive roots are hard to detect. Is there a "strategy" for choosing $a,b$ to keep $s$ as small as possible? I wanted to build a number that has "high" orders modulo "large" primes. $\endgroup$ – Conifold Sep 1 '18 at 3:40

Note the extreme case $b \equiv 1/a \mod n$, where $\text{ord}_n(a) = \text{ord}_n(b)$ but $\text{ord}_n(ab) = 1$.

What is true is that $\text{ord}_n(ab) \mid \text{lcm}(\text{ord}_n(a), \text{ord}_n(b))$, since if $\text{ord}_n(a) | m$ and $\text{ord}_n(b) | m$ then $(ab)^n = a^n b^n \equiv 1 \mod n$.

Moreover, using $a = b^{-1} (ab)$ we get $\text{ord}_n(a) \mid \text{lcm}(\text{ord}_n(b), \text{ord}_n(ab))$, and similarly $\text{ord}_n(b) \mid \text{lcm}(\text{ord}_n(a), \text{ord}_n(ab))$. Thus if for some prime $p$ and positive integer $k$, $p^k$ divides $\text{ord}_n(a)$ but not $\text{ord}_n(b)$, we must have $p^k | \text{ord}_n(ab)$.

  • $\begingroup$ Are there conditions on $n,a,b$ that would bound $\textrm{ord}_n(ab)$ from below? I am curious as to how $\min\{\textrm{ord}_p|p>k\}$ behaves for large $k$, say for primes. $\endgroup$ – Conifold Aug 30 '18 at 17:41

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