Natural way to define a free action of a finite abelian group Let $G$ be a finite abelian group. Then $G \simeq \mathbb{Z}_{u_1} \oplus \cdots \oplus \mathbb{Z}_{u_m}$, where $u_{i}$ is a power of some prime number. Without loss of generality I will consider $G = \mathbb{Z}_{u_1} \oplus \cdots \oplus \mathbb{Z}_{u_m}$. Let $Y$ be an infinite set. I introduce a set
$$
  \mathcal{Y} = Y^{u_1 \times\cdots \times u_m} \setminus \mathcal{Y}_0,
$$
where $\mathcal{Y}_0$ is the diagonal of $Y^{u_1 \times \ldots \times u_m}$:
$$
\mathcal{Y}_0 = \left\{ \left\{ y_{i_1,\ldots,i_m} \right\} \in Y^{u_1 \times \cdots \times u_m} \mid y_{k_1,\ldots,k_m}=y_{j_1,\ldots,j_m} \, \forall k_1,\ldots,k_m, j_1,\ldots,j_m   \right\}.
$$
I define an action of $G$ on $\mathcal{Y}$ by the rule
$$
   (l_1,\ldots,l_m) \cdot \left\{ y_{i_1,\ldots,i_m} \right\} = \left\{y_{i_1 + l_1\pmod{u_1}, \ldots, i_m + l_m\pmod{u_m}} \right\}.
$$
It is just a circular shift of multidimensional matrix $\left\{ y_{i_1,\ldots,i_m} \right\}$ in each dimension by appropriate number of positions. 
If $m=1$ then this action is free if and only if $u_1$ is prime. For $m>1$ this action is never free. For example for $G = \mathbb{Z}_2 \oplus \mathbb{Z}_2$ we have
$$
    (1,0) \cdot \begin{bmatrix} 1  & 2 \\ 1 & 2   \end{bmatrix} = \begin{bmatrix} 1  & 2 \\ 1 & 2   \end{bmatrix}.
$$
My question is if there is a natural way to generalize the above construction $\mathcal{Y}$ so that an analogous (in some sense) action of $G$ will be free for a more general class of finite groups? For example for $\mathbb{Z}_{p^k}$ or $\mathbb{Z}_{p} \oplus \mathbb{Z}_{q}$? 
 A: Notation first: For an element $x$ in an abelian group $H$, we denote $(x)$ the subgroup generated by $x$.
Let us see the first example $G=\mathbb{Z}/p^n$.


*

*We want $+a$ free, we should exclude $\{y\,|\,y_i=y_j, i-j\in (a)\}=\{y:y_i=y_{i+a} \forall i\}$. 

*$+p^{n-1}$ free: cut the set $\{y:y_0=y_{p^{n-1}}=y_{2p^{n-1}}=\cdots, y_1=y_{1+p^{n-1}}=\cdots,\ldots\}$. This is the set $Y_0$ you want.

*$+ap^{k}$ free with $(a,p)=1$, will give the same set of $+p^k$: $i-j\in (ap^k)=(p^k)$.

*And the set of $+p^k$ with $0\leq k\leq n-1$ is a subset of $+p^{n-1}$: since $(p^{k-1})\supseteq (p^{n-1})$. 


OK, We are done for the case for $G=\mathbb{Z}/p^n$.
For $G=\mathbb{Z}/p_1^{n_1}\times \mathbb{Z}/p_2^{n_2} \times\cdots\times \mathbb{Z}/p_m^{n_m}$.


*

*We want $+(a_1,\ldots,a_m)$ free, we should cut $\{y:y_{i_1i_2\ldots i_m}=y_{j_1j_2\ldots j_m}\; , \;(i_1,\ldots,i_m)-(j_1,\ldots,j_m)\in ((a_1,a_2,\ldots,a_m))\}$.

*The minimal subgroups of the form $0\neq ((a_1,\ldots,a_m))$ are that we need to figure out.  So cut the union of these corresponding sets.



Those such minimal subgroups are cyclic, so they must be of prime orders. So what we need to do is to find the elements of prime orders. 
For two concrete examples: 


*

*$G=\mathbb{Z}/4\times \mathbb{Z}/4$. Then $((2,0)), ((0,2)),((2,2))$ are required.

*$G=\mathbb{Z}/4\times \mathbb{Z}/4\times \mathbb{Z}/25$. Then $((2,0,0)),((2,2,0)),((0,2,0)),((0,0,1))$ are required.
