Let $G_k = \Bbb Z_3 × · · · × \Bbb Z_3$. Let$ \,\,\alpha(z_1, . . . , z_{k−1}, z_k )=(−z_1, . . . ,−z_{k−1}, z_k ) \text{ where} \,\,z_i \in\Bbb Z_3$ for $i = 1, 2, . . . ,k$. Then $α ∈ \operatorname{Aut}(G_k )$ and $A =< \alpha>$ is a subgroup of $\operatorname{Aut}(G_k )$ of order 2. Why is 2 the order $A$?
Now let $A$ act on $ G_k$, so $\operatorname{orbit}((z_1, . . . , z_k )) = \{(z_1, . . . , z_{k−1}, z_k ), (−z_1, . . . ,−z_{k−1}, z_k)\}$, where $z_i ∈ \Bbb Z_3$ for $i = 1, 2, . . . ,k$. Finally let $X_k$ = $G_k/A$. $|X_k|=(3^k+3)/2$ why? How to find the order of $ X_k$?
Thanks for your answers.
Because $\alpha^2 = \mathrm{id}$. (Notice that $\alpha$ is not just some arbitrary thing acting by automorphisms on $G_k$, it is an automorphism. So if it acts as the identity everywhere, it is the identity.)
$G_k$ has size $3^k$. The orbits in $G_k$ under the action of $A$ almost all have size $2$, except a few that have size $1$. In particular, an element of the form $(z_1, \dots, z_{k-1}, z_k)$ have orbit of size $1$ if and only if $z_1 = \dots = z_{k-1} = 0$ (why?), and there are only $3$ of these elements (why?), so there are $3^k - 3$ elements with order $2$. So $G_k/A$ has size $\dfrac{3^k - 3}{2} + \dfrac{3}{1} = \dfrac{3^k + 3}{2}$, as required.