Question: Show that there exists an injective homomorphism
$$r:\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}\to S_{27}$$
whose image is contained in the union of two conjugacy classes of $S_{27}$
My solution I thought:
I know that two permutation are conjugate if and only if they have the same cycle structure.
I thought to take a function $f:\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}\to\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}$ such that if $f(a,b,c) = (f_{1}(a,b,c),f_{2}(a,b,c),f_{3}(a,b,c))$, then for every $(a,b,c)\in\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3}$ and for every $1\le i \le3$ $$f_{i}(a,b,c) = \begin{cases} 0&, (a,b,c) = (0,0,0) \\ \neq 0 & ,otherwise \\ \end{cases}$$ Using this function I thought to take the homomorphism $$r(a,b,c) = (1, 2,3,...,9)^{f_{1}(a,b,c)}(10,11,12,..,18)^{f_{2}(a,b,c)}(19,20,21,...,27)^{f_{3}(a,b,c)}$$
My problem is that I don't know how to find this kind of function,
Any help for finding this kind of function or for alternative solution for this problem will be appreciated.
Thanks.
There is an elementary abelian group $H$ of order $9$ that lives in $S_9$ and is generated by two elements $a$ and $b$ with three fixed points each: $$H:=\langle a, b\rangle\text{ where }a=(1 2 3)(4 5 6),\;b=(4 5 6)(7 8 9)\;$$ Now let $t$ be the three-fold "shift" of the range $(1, \ldots, 9)$: $$t:=(1\;10\;19)(2\;11\;20)(3\;12\;21)(4\;13\;22)(5\;14\;23)(6\;15\;24)(7\;16\;25)(8\;17\;26)(9\;18\;27)$$ Then $a^\ast:=a\cdot a^t\cdot a^{t^2}$, $b^\ast:=b\cdot b^t\cdot b^{t^2}$ and $t$ generate an elementary abelian group of order $27$ (and thus an injective image of $\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_3$) in $S_{27}$ that contains some elements with $9$ fixed points and some with no fixed points, i.e. elements from two conjugacy classes.