Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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've solved the following problem in what seems to me to be a very inelegant way:

Let $G$ be an abelian group containing elements $a$ and $b$ of orders $m$ and $n$, respectively, where $m$ and $n$ are not necessarily relatively prime. Show that $G$ must contain an element $c$ whose order $\mathcal{O}(c)$ equals the least common multiple of $m$ and $n$.

My solution is the following:

Let $d=\gcd(m,n)$ and note that $m=dk$, $n=dl$ where $k$ and $l$ are relatively prime. Now, define $$ r=\prod_{p\mid d, p\mid k}p^i, $$ where $p^i\mid d$ but $p^{i+1}\nmid d$. Now, by construction, since $k$ and $l$ are relatively prime, $r$ and $l$ are relatively prime. Likewise, since $r$ takes all of the primes in $d$ that are in common with $k$, we know $r$ and $\frac{d}{r}$ are relatively prime. Now we have \begin{align} &m=dk=r\left(\frac{d}{r}\right)k\\ &n=dl=r\left(\frac{d}{r}\right)l \end{align} Clearly the element $a^{d/r}$ has order $rk$ and the element $b^{r}$ has order $\frac{dl}{r}$. Now the orders are relatively prime, so from a previous problem (where we showed that elements with rel. prime orders has the property that their product has the order of the product of their individual orders), we know the order of $a^{d/r}b^{r}$ is $dkl$, which by definition is the least common multiple of $m$ and $n$.

This doesn't strike me as being an elegant proof, so I'm wondering if there are any other ways to prove the statement?

share|cite|improve this question
It seems fine to me, just a bit on the long side (the general idea is certainly also the first one that comes to my mind, finding suitable powers of $a$ and $b$ whose orders are coprime and whose product is the lcm of their orders). – Tobias Kildetoft May 30 '13 at 15:34
@TobiasKildetoft: I was pretty sure it was correct, I was just hoping for a solution that seemed cleaner, rather than just running numbers all over the place. Thanks! – Clayton May 30 '13 at 15:41
Well, an alternative idea is to mod out the subgroup generated by one of those elements, take an element of a suitable order in the quotient and pull that back to get the sought for element (the trick here is to use the other element to get that element of a suitable order). – Tobias Kildetoft May 30 '13 at 15:49
Actually, thinking a bit more about it, that other method seems like it will take an even longer argument. – Tobias Kildetoft May 30 '13 at 15:55
I was just thinking about it, but I'm not sure it would work (for example, if $b$ were in the group generated by $a$, then modding out by $\langle a\rangle$ would kill $b$). I could be on the wrong track; I'm just thinking about it. – Clayton May 30 '13 at 15:57
up vote 0 down vote accepted

Let $m=dm'$ and $n=dn'$ with $gcd(m',n')=1$.


$$f: Z \times Z \to G \,;\, f(k,l)=a^{kd}b^{ld} \,.$$

We claim $\ker(f)=m'Z \times n'Z$. Since $\frac{Z \times Z}{m'Z \times n'Z} \sim Z_{m' n'}$ pick a generator $a^{id}b^{jd}$ for $Im(f)$ and then look to $a^ib^j$.

We prove now the claim

$$f(k,l)=0 \Rightarrow a^{kd}b^{ld}=e \,.$$

Thus, since $m',n'$ are relatively prime

$$b^{n'dl}=e \Rightarrow dm'|dln' \Rightarrow m'|ln' \Rightarrow m'|l \,,$$ $$a^{m'dk}=e \Rightarrow dn'|dkm' \Rightarrow n'|km' \Rightarrow n'|k \,,$$

share|cite|improve this answer
To make it simpler to follow, what I do is replace $a,b$ by $a^d,b^d$ and then you only need to prove the result in the case $m,n$ relatively prime. This implies that $a^kb^l=e$ only when $a^k=e$ and $b^l=e$. – N. S. May 30 '13 at 16:26
I'm not sure I'm following. What happens in the case $a^d=b^d=e$ (as an example, see my comment under Alex Ravsky's solution). – Clayton May 30 '13 at 16:56
@Clayton If the element is $a'^ib'^j$ the desired element is $a^ib^j$. In your case the element you seek in $<a',b'>$ is $a'^1$ then $our element is $a^1$. – N. S. May 30 '13 at 17:08

An alternative way is to do the prime fiddling inductively as follows.

LEMMA $\ $ A finite abelian group $\rm\:G\:$ has an lcm-closed order set, i.e.

$$\rm X,Y \in G\ \ \Rightarrow\ \ \exists\ Z \in G:\ o(Z) = lcm(o(X),o(Y)),\,\ \ o(X) =\: order\ of\ X$$

Proof $\ \ $ By induction on $\rm\, o(X)\ o(Y).\ $ If it is $\,1\,$ then $\rm\:Z = 1\, $ works. $\ $ Otherwise

write $\rm\ o(X)\ =\ AP,\: \ \ o(Y) = BP',\ \ \ P'\mid P = p^m > 1,\ \ $ prime $\rm\: p\:$ coprime to $\rm\: A,B.$

Then $\rm\: o(X^P) = A,\ \ o(Y^{P'}) = B.\ $ By induction there is a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$

so $\rm\ \ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y)).\ \ $ QED

share|cite|improve this answer

Order of $ab$ need not be equal to the lcm of $|a|$ and $|b|$ if the gcd is not relatively prime. For example consider $G=\langle x\rangle$ with $a=x$ and $b=x^3$. Here $|a|=4, |b|=4$ but $|ab|=1$.

Unless $\gcd(|a|, |b|)=1$ is given we can't say that order of $ab$ is same as lcm$(|a|, |b|)$

share|cite|improve this answer
The question is to prove the existence of an element; in fact, in some previously-deleted answers, you can see some of my own counterexamples. – Clayton Dec 4 '13 at 1:25

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.