Does there exist a group with two elements of order 9: $o(h) = o(g) = 9$, where their product is of order 3: $o(gh) = 3$?
I know, for a non-abelian group anything can happen, but I can't find any example or counterexample for this.
Order of product of elements of a group
3
$\begingroup$
abstract-algebra
group-theory
2 Answers
3
Well, in the group $\mathbb Z_9,+$, both $1$ and $2$ have order $9$, and their 'product' $1+2=3$ has order $3$. There are more elements with order $9$ in that group, though, but that's true for every group with elements of order $9$.
-
0Thanks! But why is it true for every group with element of order 9? Is there any general rule for that? – 2017-01-29
-
0If $o(h) = 9$, $o(h^2) = 9$ as well because 2 and 9 are relatively prime. The same holds for $h^4$, $h^5$, $h^7$ and $h^8$. – 2017-01-29
1
Consider the group $G=S_9 \times \mathbb{Z}_3$, then the elements $\alpha = ((123456789), 1)$ is of order $9$ and likewise $\beta = ((123456789)^{-1}, 1)$ but $\alpha\beta=(e,2)$ is of order $3$.