# $\operatorname{ord}(u)=r,\,\operatorname{ord}(v)=s$ $\,\Rightarrow\,\operatorname{ord}(uv)=rs\,$ if $\,r,s\,$ (co)primes

Let u and v be elements of a commutative group, and suppose that their orders are r and s, respectively. Show that if r and s are distinct primes then the order of uv is rs.

I will use the follwing theorem to proof this.

Theorem: Let x be an element of order m in a finite group G. Then $$x^s=1$$ in G if and only if s i a multiple of m.

My attempt

We can here see that $$(uv)^{rs}=[commutative]=u^{rs}v^{rs}=1^s1^r=1$$ By the theorem, this tells us that this is true only if only rs is a multiple of the order of $uv$. Let the order be $ord(uv)=k$. We know that every positive integer can be written as $$a=bm+l, 0<=l<b$$ and since $r$ and $s$ is primes (positive integers), $rs$ can be written as $$rs=km+l, 0<=l<k$$ This gives ut that $$1=(uv)^{rs}=(uv)^{km+l}=1^muv^l=(uv)^l$$ If l>0 then the equation $(uv)^l=1$ contradicts that the positive integer k is the least positive integer for which $(uv)^k=1$. Therefore l=0 and $rs=km$ . We can here see that $rs$ is a multiple of k (and this proof also the theorem). $$rs=km => k | rs => k|r ORk|s$$ But we know that $r$ and $s$ are distinct primes, so k must be equal to one of the primes.

$$k=r => (uv)^k=(uv)^r=1v^r=v^r=1$$ And since we know that $ord(v)=s$ the theorem tells us that $r$ must be a multiple of $s$. But since both of them is primes, we get that $r=s$ but this is a contradiction (r and s are distincts as the exercise tells us).

Now, we try with $k=s$. Same as above, we get that $s=r$ which is a contradiction. Neither k=r nor k=s is right since by the theorem we get that $k=r=s$ and we have that $r$ and $s$ are distincts. This tells us that k cannot be equal to r nor s, since this gives us that $r=s$. And since we know (and as i did proof)that $$rs=km$$, this tells us, as i wrote before, that $$k|rs$$ Since k is neither equal to $r$ or $s$, this tells us that k must be equal to $rs$. Therefore $$ord(uv)=k=rs$$.

Edit: i know that this can be prooved by l.c.m(x,y) but we have not used it so far. That's why i did not proof it with l.c.m(x,y)

• Your argument seems good. The double use of the letter $r$ is confusing though, please use another name for the remainder. – Servaes May 20 '14 at 18:32
• oh sorry. i did not think of that :D – Bonnie May 20 '14 at 18:32
• There is a gap: the prime product $\,rs\,$ has three smaller divisors $\,r,s,\color{#c00}1.\,$ You've ruled out the possibilities $\,r,s\,$ for $\,k,\,$ but not the possibility $\,k = \color{#c00}1.\$ What is your question? If the proof is correct? If there are other proofs? – Bill Dubuque May 20 '14 at 19:25
• @BillDubuque: Yes, if the proof was correct. – Bonnie May 20 '14 at 19:50
• @Bonnie I added an answer. If you need some other way besides Euclid's Lemma to avoid lcm properties let me know what you already know (e.g. Bezout's Identity) and I can show you how to do it. – Bill Dubuque May 20 '14 at 19:53

There is a gap: the prime product $$\,rs\,$$ has three smaller divisors $$\,r,s,\color{#c00}1.\,$$ You've ruled out the possibilities $$\, k = r,s\,$$ but not $$\,k = \color{#c00}1.\$$ Below is a quicker way to proceed, where we prove it more generally for any $$r,s$$ that are coprime, i.e. their gcd $$(r,s) = 1.\,$$ Let $$\,o(x) := {\rm ord}\,x$$

$$(uv)^k\! = 1\,\Rightarrow\, u^k\! = v^{-k} =\color{#c00}t \in \langle u\rangle\cap\langle v\rangle\Rightarrow \bigg\lbrace\begin{eqnarray} && {o(t)\mid\ r,s = o(u),\,o(v)},{\rm\ by\ Lagrange}\\ \Rightarrow && o(t)\mid(r,s)\!=\!1\, \Rightarrow\, \color{#c00}{t = 1}\end{eqnarray}$$

So $$\ (uv)^k\! = 1\iff u^k =\color{#c00}{1} = v^k\iff r,s\mid k\underset{\large (r,s)\,=\,1}\iff rs\mid k,\,$$ so $$\ rs = o(uv)\$$ by here.

Remark  If Lagrange in unknown, instead: $$\,t^{\large r}\! = (u^{\large k})^{\large r}\! = (u^{\large r})^{\large k}\! = 1^{\large k}\! = 1$$ thus $$\,o(t)\mid r$$

To prove that (co)primes $$\,r,s\mid k\,\Rightarrow\,rs\mid k\,$$ without knowledge of lcm, you can either use the Fundamental Theorem of Arithmetic, or equivalent properties such as Euclid's Lemma, e.g. $$\, s\mid k\,\Rightarrow\, k = sn,\,$$ so $$\,(r,s)=1,\,r\mid sn\,\Rightarrow\, r\mid n,\,$$ so $$\,n = rm,\,$$ so $$\,k = sn = s(rm),\,$$ so $$\,rs\mid k.$$ Alternatively, using Bezout's identity, since $$\,s,r\,$$ are coprime there are integers $$\,j,k\,$$ with $$\,js\!+\!kr = 1\,$$ so $$\,r\mid sn,rn\,\Rightarrow\,r\mid j(sn)+k(rn) = (js\!+\!kr)n = n,\,$$ so the result follows as above. See also this answer for further proofs and comparisons.

• Thanks for the answer. Yes i know Bezout's Identity. If you have time, then sure :) – Bonnie May 20 '14 at 19:55
• @Bonnie Ok, I added a proof of Euclid's Lemma using Bezout. – Bill Dubuque May 20 '14 at 20:02
• +1 for the interesting trick of considering $u^k=v^{-k}$. I've seen this technique in another form in the lessons linear algebra when proving the linear independence of bases of disjoint subspaces. Not that these are such astonishing results but it's interesting how the same manipulation is used in different contexts. It looks like extracting more information from a single equality. – punctured dusk May 20 '14 at 20:12
• I appreciate the time you took to help me. thanks again :) – Bonnie May 20 '14 at 21:00

One can use the same (correct) idea, streamline things a bit, and relax the primality assumptions.

Suppose that $u$ has order $r$ and $v$ has order $s$, where $r$ and $s$ are relatively prime. We show that $uv$ has order $rs$. Since $(uv)^{rs}=1$, it is enough to show that if $k$ is the order of $uv$, then $rs$ divides $k$.

Note that $u^{ks}=u^{ks}v^{ks}=1$. So the order $r$ of $u$ divides $ks$, and since $r$ and $s$ are relatively prime, we conclude that $r$ divides $k$.

Similarly, $s$ divides $k$.

Since $r$ and $s$ are relatively prime, it follows that $rs$ divides $k$.

• You are welcome. – André Nicolas May 20 '14 at 21:03

Using this answer by setting $t=1$ you'll get what you wanted and even more general.