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

## 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.

Add to solve later

Sponsored Links

Contents

## 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 $p_i$ be the prime factors of either $m$ or $n$.

Then write prime factorizations of $m$ and $n$ as

\[m=\prod_{i}p_i^{\alpha_i} \text{ and } n=\prod_{i} p_i^{\beta_i}.\]
Here $\alpha_i$ and $\beta_i$ are nonzero integers (could be zero).

Define

\[m’=\prod_{i: \alpha_i \geq \beta_i}p_i^{\alpha_i} \text{ and } n’=\prod_{i: \beta_i> \alpha_i} p_i^{\beta_i}.\]

(For example, if $m=2^3\cdot 3^2\cdot 5$ and $n=3^2\cdot 7$ then $m’=2^3\cdot 3^2\cdot 5$ and $n’=7$.)

Note that $m’\mid m$ and $n’\mid n$, and also $m’$ and $n’$ are relatively prime. The least common multiple $l$ of $m$ and $n$ is given by

\[l=m’n’\]

Consider the element $a’:=a^{m/m’}$. We claim that the order of $a’$ is $m’$.

Let $k$ be the order of the element $a’$. Then we have

\begin{align*}

e=(a’)^k=(a^{\frac{m}{m’}})^k=a^{mk/m’},

\end{align*}

where $e$ is the identity element in the group $G$.

This yields that $m$ divides $mk/m’$ since $m$ is the order of $a$.

It follows that $m’$ divides $k$.

On the other hand, we have

\begin{align*}

(a^{m/m’})^{m’}=a^m=e,

\end{align*}

and hence $k$ divides $m’$ since $k$ is the order of the element $a^{m/m’}$.

As a result, we have $k=m’$.

So the order of $a’$ is $m’$.

Similarly, the order of $b’:=b^{n/n’}$ is $n’$.

The orders of elements $a’$ and $b$ are $m’$ and $n’$, and they are relatively prime.

Hence we can apply the first case and we conclude that the element $a’b’$ has order

\[m’n’=l.\]
Thus, we can take $c=a’b’$.

### 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“.

Add to solve later

Sponsored Links

“Since d=gcd(m,n), we know that m/d and n are relatively prime.” This is a WRONG claim!

Consider, m=125 and n=175.

Dear Soumya,

You are absolutely right. I fixed the error and rewrite the proof.

Thank you for pointing out the mistake.

Yes, it is perfect now.