Order of a Product in a Finite Abelian Group

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Problem

If $G$ is a finite abelian group and $a, b \in G$, then show that the order of the element $ab$, denoted $o(ab)$, is a divisor of the least common multiple of the orders of $a$ and $b$.

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Definitions Used

1. Abelian Group: A group in which the operation is commutative, i.e., $ab = ba$ for all elements $a, b$.
2. Order of an Element, $o(x)$: The smallest positive integer $n$ such that $x^n = e$, where $e$ is the identity.
3. Least Common Multiple, $\text{l.c.m.}(m, n)$: The smallest positive integer that is a multiple of both $m$ and $n$. If $l = \text{l.c.m.}(m, n)$, then $m \mid l$ and $n \mid l$.
4. Key Property of Order: If $o(x) = k$ and $x^l = e$ for some integer $l$, then $k$ must divide $l$.

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Proof

Let $o(a) = n$ and $o(b) = m$. Let the order of the product be $o(ab) = k$.

Let $l$ be the least common multiple of the orders of $a$ and $b$:

$$ l = \text{l.c.m.}(n, m) $$

By the definition of the least common multiple, we know that $n$ divides $l$ and $m$ divides $l$. This means we can write $l$ as an integer multiple of both $n$ and $m$:

$$ l = nr_1 \quad \text{and} \quad l = mr_2 \quad \text{for some integers } r_1, r_2. $$

Now, let’s consider the element $(ab)$ raised to the power of $l$:

$$ (ab)^l $$

Since the group $G$ is abelian, we can distribute the exponent:

$$ (ab)^l = a^l b^l $$

Now, we substitute the expressions for $l$:

$$ a^l b^l = a^{nr_1} b^{mr_2} $$

Using the rules of exponents, this can be rewritten as:

$$ (a^n)^{r_1} (b^m)^{r_2} $$

By the definition of order, we know that $a^n = e$ and $b^m = e$. Substituting these values in, we get:

$$ (e)^{r_1} (e)^{r_2} = e \cdot e = e $$

So, we have shown that $(ab)^l = e$.

We know that the order of $(ab)$ is $k$. Using the key property of orders, if $(ab)^l = e$, then the order of $(ab)$ must divide $l$. Therefore:

$$ k \mid l $$

Conclusion

We have proven that $o(ab)$ divides $\text{l.c.m.}(o(a), o(b))$.