Order of a given element in an abelian group.

Question

Question. Let $G$ be an abelian group and $f,g\in G$. Suppose the order of $f$ is $3$ and the order of $g$ is $2$.

What is the order of $fg$?

Attempt. We need to find a non-negative integer $x$ such that $(fg)^x=1$. Since the order of $f$ is $3$ we have $f^3=1$, or equivalently $f^6=1$. Similarly, $g^2=1$ gives $g^6=1$. So: $x$ must equal $6$.

Is this correct? If so, is the answer concrete enough? If not, where have I gone wrong?

Thanks in advance.

Answer 1

You just need slightly more detail. $(fg)^x=1$ does not always imply $f^x=1$. It does in this case because $(fg)^x=1$ implies $f^x=(g^x)^{-1}$, and powers of $f$ have order dividing $3$ whilst powers of $g$ have order dividing $2$, and the only number dividing both $2$ and $3$ is $1$, so $f^x$ has order 1, i.e., $f^x=1$ and similarly for $g$. Now you just check that the smallest positive $x$ with $f^x=g^x=1$ is $6$ since $6$ is the least common multiple of $2$ and $3$.

To see why your argument doesn't work in general, consider $G$ the group of rotations about an axis. $G$ is abelian, and is $f$ is counterclockwise rotation by $45$ degrees and $g$ is counterclockwise rotation by $135$ degrees, then $f$ and $g$ each have order $8$, but $fg$ has order $2$

Answer 2

That's good, except your haven't stated why the order of $fg$ is $6$, as opposed to say, given $f^{12} = 1$, and $g^{12} = 1$, so that $12$ is the order of $12$. Of course, $|fg| = 6$, because it is the least common multiple of $2$ and $3$.

So, it is better to say that "we need to find the least non-negative integer $x$ such that $(fg)^x=1,$ given that $f^3 =1,$ and $g^2 = 1$. And indeed, $x=6$ is the least possible $x$ such that $(fg)^x = 1$.

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An example that might be helpful to consider would be the group $\mathbb Z_12$ under addition modulo $12$. It's a cyclic group, and hence it is abelian. Now, for example, the element $4$ has order 3, because $3(4) = 12 = 0\pmod {12}.$ The element 6 has order 2, since $2(6) = 12 = 0\pmod{12}$.

So what can we say about the order of element $10 = 4+6$? It has the order of $\operatorname{lcm}(3, 2) = 6$. Indeed, $10^6 = 6(10)$ (given the group operation is additive,) and $6(10) = 60 = 0 \pmod {12}$, and there is no lesser positive integer n, such that $n(10) = 0\pmod {12}$.