The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements

Abelian Group problems and solutions

Problem 497

Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively.
Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.

Also determine whether the statement is true if $G$ is a non-abelian group.

 
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Hint.

First, consider the case when $m$ and $n$ are relatively prime.

Proof.

When $m$ and $n$ are relatively prime

Recall that if the orders $m, n$ of elements $a, b$ of an abelian group are relatively prime, then the order of the product $ab$ is $mn$.
(For a proof, see the post “Order of the Product of Two Elements in an Abelian Group“.)

So if $m, n$ are relatively prime, then we can take $c=ab\in G$ and $c$ has order $mn$, which is the least common multiple.

The general Case

Now we consider the general case.
Let $d$ be the greatest common divisor of the orders $m$ and $n$”
\[d=\gcd(m,n).\]

Then the least common multiple $l$ of $m$ and $n$ is given by
\[l=\frac{mn}{d}.\]

Consider the element $a^d$.
We claim that the order of $a^d$ is $n/d$.

Let $k$ be the order of the element $a^d$.
Then we have
\begin{align*}
a^{dk}=(a^d)^k=e,
\end{align*}
where $e$ is the identity element in the group $G$.
This yields that $m$ divides $dk$ since $m$ is the order of $a$.
It follows that $m/d$ divides $k$.

On the other hand, we have
\begin{align*}
(a^d)^{\frac{m}{d}}=a^m=e,
\end{align*}
and hence $k$ divides $m/d$ since $k$ is the order of the element $a^d$.

As a result, we have $k=m/d$.
So the order of $a^d$ is $m/d$.

Since $d=\gcd(m,n)$, we know that $m/d$ and $n$ are relatively prime.
Thus, the orders of elements $a^d$ and $b$ are $m/d$ and $n$, and they are relatively prime.
Hence we can apply the first case and we conclude that the element $a^db$ has order
\[\frac{m}{d}\cdot n=l.\] Thus, we can take $c=a^db$.

The Case When $G$ is a Non-Abelian Group

Next, we show that if $G$ is a non-abelian group then the statement does not hold.

For example, consider the symmetric group $S_3$ with three letters.
Let
\[a=(1\,2\,3) \text{ and } b=(1 \,2).\]

Then the order of $a$ is $3$ and the order of $b$ is $2$.
The least common multiple of $2$ and $3$ is $6$.
However, the symmetric group $S_3$ have no elements of order $6$.

Hence the statement of the problem does not hold for non-abelian groups.

Related Question.

Problem. Let $G$ be a group. Let $a$ and $b$ be elements of $G$.
If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.

For a solution of this problem, see the post “Order of Product of Two Elements in a Group“.


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  1. 06/30/2017

    […] A proof of this problem is given in the post “The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements“. […]

  2. 06/30/2017

    […] A proof of this problem is given in the post “The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements“. […]

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