# If ord$(a)=m$, ord$(b)=n$ then does there exist $c$ such that ord $(c)=lcm(m,n)$? [duplicate]

Possible Duplicate:
Order of elements in abelian groups

Let $G$ be an abelian group and suppose that $G$ has elements of orders $m$ and $n$, respectively. Prove that $G$ has an element whose order is the least common multiple of $m$ and $n$.

I've attempted this problem for quite some time, but didn't seem to get anywhere.

First, let $a$ and $b$ be the elements whose orders are $m$ and $n$, respectively. I guessed that we can find the element of order $lcm(m,n)$ explicitly, instead of simply proving its existence. Furthermore, I also guessed that the element can be expressed in the form $a^kb^l$, because the statement must also hold when $G$ is generated by $a$ and $b$.

Then I let $k$ be the smallest positive integer such that $a^k$ is a power of $b$, say $a^k=b^l$. Then I proved that $l$ is also the smallest positive integer such that $b^l$ is a power of $a$, and that $ml=nk$. I'm not sure whether it's correct though.

Then I tried to find the order of ab. I can prove that the order is divisible by $\frac{lcm(m,n)}{\gcd(m,n)}$, but I can't prove whether it is equal to $lcm(m,n)$. Apparently, taking any $a^ib^j$ won't be any better. And now, I'm at wits end.

Please tell me whether I'm on the right path. If not, please give me some adequate hints so I can work on it.

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Please do not use titles entirely composed of $\TeX$. – Mariano Suárez-Alvarez Nov 3 '11 at 11:13
Sorry. Is that inappropriate? – adjwhrwe Nov 3 '11 at 11:18
It's often a good idea, when you've failed to prove something, to try to disprove it. For example, taking $ab$ doesn't work in general, since we could have $b=a^{-1} \neq id$. $a$ itself (which is of the form $a^i b^j$) works in this case. – Chris Eagle Nov 3 '11 at 11:34
@adjwhrwe, it is quite inconvenient. – Mariano Suárez-Alvarez Nov 3 '11 at 11:55
@adjwhrwe yours first guesses were correct, infact there is an element of order $lcm(n,m)$ of the form $a^kb^l$. Because you're so close to the solution I'll give you something I hope will help you: "Try to do first the case of $gcd(n,m)=1$". – ineff Nov 3 '11 at 12:23

## marked as duplicate by Gerry Myerson, lhf, Mike Spivey, Henning Makholm, Arturo MagidinNov 3 '11 at 16:52

First we consider the case where $(m,n)=1$. Because $(m,n)=1$ the least common multiple of $m$ and $n$ is $mn$. Consider the element $ab$. Because $G$ is Abelian, $$(ab)^{mn} = \overbrace{ab\ ab\ ab \ldots ab}^{mn} = \overbrace{aaa\ldots a}^{mn} \overbrace{bbb\ldots b}^{mn} = a^{mn}b^{mn} = ee =e$$ so the order of $ab$ is at most $mn$. Say the order of $ab$ is $k$, then we just showed that $k \leq mn$. We see that $$e= (ab)^k = (ab)^{km} = a^{km}b^{km} = eb^{km} = b^{km}$$ so $n$ divides $km$. But since $(m,n)=1$ it follows that $n$ divides $k$. Similarly $$e=(ab)^k = (ab)^{kn} = a^{kn}b^{kn} = a^{kn}e = a^{kn}$$ so $m$ divides $kn$, and so $m$ divides $k$. Thus $mn$ divides $k$. And since $k \leq mn$ it follows that $k = mn$, and so $ab$ has order $mn$, the lowest common multiple of $m$ and $n$. We need to take from this the fact that the order of a product of elements that have relatively prime orders is the product of the orders of those elements.
Now we consider the case where $m$ and $n$ are not relatively prime. Call $L$ the least common multiple of $m$ and $n$. We write $L=p^{r_1}_1\cdots p^{r_s}_s$ for distinct primes $p_i$ and strictly positive powers $r_i$. If we could find an element of $G$ with order $p^{r_i}_i$ for every $i$, then by the first half of this proof, the product of these elements would have order $L$ because prime powers are all relatively prime to prime powers of different primes. Let $i$ with $1 \leq i \leq s$ be given. We note that $p^{r_i}_i$ divides either $m$ or $n$. Thus $a^{m/p^{r_i}_i}$ or $b^{n/p^{r_i}_i}$ (whichever one divides evenly) has order $p^{r_i}_i$. Therefore we have shown that an element of order $L$, the least common multiple of $m$ and $n$, is in $G$.
The statement is not necessarily true if $G$ is not Abelian. Consider $G=S_3$. This group contains a cycle of order 3, and a transposition (of order 2), but no element of order 6.