Finding the relationship between the algebraic structure of rings using properties of graphs associated to them became an interesting topic in the last years, for example, see [1-3]. In this paper, let ${{\mathbb{Z}}_{n}}=\{\bar{0}, \bar{1}, \cdots, \overline{n-1}\}$ be the ring of integers modulo $n$, and $\mathbb{Z}_n[\mathbf{i}]=\{\overline a+\overline b\mathbf{i}\, |\, \overline a, \overline b\in \mathbb{Z}_n\}$ be the ring of Gaussian integers modulo $n$. We investigate some properties of the square mapping graphs $\Gamma(n)$, whose vertex set is all the elements of $\mathbb{Z}_n[\mathbf{i}]$, and for which there is a directed edge from $\alpha\in \mathbb{Z}_n[\mathbf{i}]$ to $\beta\in \mathbb{Z}_n[\mathbf{i}]$ if and only if $\alpha^2=\beta$. In [1,4,5], some properties of the square mapping graphs of $\mathbb{Z}_n$ were investigated, and the cubic mapping graphs of $\mathbb{Z}_n[\mathbf{i}]$ were studied in [2].

Let $R$ be a commutative ring, U$(R)$ denotes the unit group of $R$, D$(R)$ the zero-divisor set of $R$. For $\alpha\in$ U$(R)$, $o(\alpha)$ denotes the multiplicative order of $\alpha$ in $R$. If $R=\mathbb{Z}_n$, then we write ${\rm ord}_n\alpha$ instead of $o(\alpha)$. We specify two particular subdigraphs $\Gamma_1(n)$ and $\Gamma_2(n)$ of $\Gamma(n)$, i.e., $\Gamma_1(n)$ is induced by all the vertices of U$(\mathbb{Z}_n[\mathbf{i}])$, and $\Gamma_2(n)$ is induced by all the vertices of D$(\mathbb{Z}_n[\bf{i}])$.

In $\Gamma(n)$, if $\alpha_1, \cdots, \alpha_t$ are pairwise distinct vertices and $\alpha_1^2=\alpha_2, \cdots$, $\alpha_{t-1}^2=\alpha_t$, $\alpha_t^2=\alpha_1$, then the elements $\alpha_1, \alpha_2, \cdots, \alpha_t$ constitute a $t$-cycle. It is obvious that $\alpha$ is a vertex of a $t$-cycle if and only if $t$ is the least positive integer such that $\alpha^{2^t}=\alpha$. For $\alpha\in\mathbb{Z}_n[\mathbf{i}]$, the in-degree indeg$(\alpha)$ of $\alpha$, denotes the number of directed edges coming into $\alpha$.

Example 1.1  The square mapping graph of $\mathbb{Z}_{5}[\mathbf{i}]$ is as follows.

The square mapping graph of $\mathbb{Z}_{5}[\mathbf{i}]$}

Lemma 1.2  (see [6]) Let $n>1$.

$\mathrm{(1)}$ The element $\overline a+\overline b\mathbf{i}$ is a unit of $\mathbb{Z}_n[\mathbf{i}]$ if and only if $\ \overline a^2+\overline b^2$ is a unit of $\mathbb{Z}_n$.

$\mathrm{(2)}$ If $\ n=\prod\limits_{j=1}^sp_j^{k_j}$ is the prime power decomposition of $n$, then $\mathbb{Z}_n[\mathbf{i}]\cong\oplus_{j=1}^s\mathbb{Z}_{p_j^{k_j}}[\mathbf{i}]$.

$\mathrm{(3)}$ $\mathbb{Z}_n[\mathbf{i}]$ is a local ring if and only if $n=p^t$, where $p=2$ or $p$ is a prime congruent to 3 modulo 4, $t≥1$.

$\mathrm{(4)}$ $\mathbb{Z}_n[\mathbf{i}]$ is a field if and only if $n$ is a prime congruent to 3 modulo 4.

Lemma 1.3(see [7])

$\mathrm{(1)}$ $|{\rm U}(\mathbb{Z}_{2^t}[\mathbf{i}])|=2^{2t-1}$, $|{\rm D}(\mathbb{Z}_{2^t}[\mathbf{i}])|=2^{2t-1}$.

$\mathrm{(2)}$ Let $q$ be a prime congruent to 3 modulo 4. Then $|{\rm U}(\mathbb{Z}_{q^t}[\mathbf{i}])|=q^{2t}-q^{2t-2}$, $|{\rm D}(\mathbb{Z}_{q^t}[\mathbf{i}])|=q^{2t-2}$.

$\mathrm{(3)}$ Let $p$ be a prime congruent to 1 modulo 4. Then $|{\rm U}(\mathbb{Z}_{p^t}[\mathbf{i}])|=(p^t-p^{t-1})^2$, $|{\rm D}(\mathbb{Z}_{p^t}[\mathbf{i}])|=2p^{2t-1}-p^{2t-2}$.

By Lemma 1.2 (2), we have the following lemma concerning the in-degree of an arbitrary vertex in $\Gamma(n)$.

Lemma 1.4 Suppose $\alpha=\overline a+\overline b\mathbf{i}\in \mathbb{Z}_n[\mathbf{i}]$, and $n=\prod\limits_{j=1}^sp_j^{k_j}$ is the prime power decomposition of $n$. Then ${\rm indeg}(\alpha)={\rm indeg}(\alpha_1)\times\cdots\times {\rm indeg}(\alpha_s)$, where $\alpha_j=(a\ {\rm mod}\ p_j^{k_j})+ (b\ {\rm mod}\ p_j^{k_j})\mathbf{i}$ and ${\rm indeg}(\alpha_j)$ is the in-degree of $\alpha_j$ in $\Gamma(p_j^{k_j})$, $j=1, \cdots, s$.

By Lemma 1.4, in order to obtain the in-degree of a vertex in $\Gamma(n)$, it suffices to consider the cases of $n$ being a power of a prime.

Theorem 2.5  $\mathrm{(1)}$ Let $n=2^k$, $k≥1$. Then indeg$(\overline0)=2^k$.

$\mathrm{(2)}$ Let $n=p^k$, where $p$ is an odd prime, $k≥1$. Then indeg$(\overline0)=p^k$ if $k$ is even, while indeg$(\overline0)=p^{k-1}$ if $k$ is odd.

Proof  (1) Let $n=2^k$. By inspection, we have indeg$(\overline0)=2^k$ for $k=1, 2$. Now suppose $k≥3$. Assume that $\alpha=\overline a+\overline b\mathbf{i}\in\mathbb{Z}_{2^k}[\mathbf{i}]$ with $\alpha^2=\overline0$. Clearly $2|a$ and $2|b$. Let $a=2^ua_1$, $b=2^vb_1$, where $u, v$ are positive integers, both $a_1$ and $b_1$ are odd. Set $\lambda={\rm min}\{u, v\}$. Then $\alpha=2^\lambda\beta$, where $\beta=2^{u-\lambda}\overline{a_1}+2^{v-\lambda}\overline{b_1}\mathbf{i}$.

Suppose $k$ is even. Clearly $\alpha^2=\overline0$ when $\lambda≥\frac{k}{2}$, and $\alpha^2\neq\overline0$ when $\lambda≤\frac{k}{2}-1$. Hence, $\alpha^2=\overline0$ if and only if $\alpha=2^{k/2}\, \overline a_0+2^{k/2}\, \overline b_0\mathbf{i}$ with $a_0, b_0\in\{0, 1, 2, \cdots, 2^{k/2}-1\}$. Thus indeg$(\overline0)=2^{k/2}\times2^{k/2}=2^k$.

Suppose $k$ is odd. First, if $\lambda≥\frac{k+1}{2}$, then clearly $\alpha^2=\overline0$. Second, if $\lambda=\frac{k-1}{2}$, then $\beta\in {\rm U}(\mathbb{Z}_{2^k}[\mathbf{i}])$ when $u\neq v$. Hence, $\alpha^2=2^{2\lambda}\beta^2\neq\overline0$. Otherwise, $\alpha^2=2^{2u+1}(\frac{\overline {a_1}^2-\overline {b_1}^2}{2}+\overline{a_1}\overline{b_1}\mathbf{i})=\overline0$ when $u=v=\lambda$. Third, if $\lambda≤\frac{k-3}{2}$, then clearly $\alpha^2\neq0$. Therefore, in the case of $k$ odd, $\alpha^2=\overline0$ if and only if $\alpha=2^{(k+1)/2}\, \overline a_0+2^{(k+1)/2}\, \overline b_0\mathbf{i}$ with $a_0, b_0\in\{0, 1, 2, \cdots, 2^{(k-1)/2}-1\}$, or $\alpha=2^{(k-1)/2}\, \overline a_0+2^{(k-1)/2}\, \overline b_0\mathbf{i}$ with $a_0, b_0\in\{1, 3, 5, \cdots, 2^{(k+1)/2}-1\}$. Thus indeg$(\overline0)=2^{(k-1)/2}\times2^{(k-1)/2}+2^{(k-1)/2}\times2^{(k-1)/2}=2^k$.

(2) Let $n=p^k$, where $p$ is an odd prime, $k≥1$. Suppose $k$ is even, then by an argument similar to (1) above, $\alpha^2=\overline0$ if and only if $\alpha=p^{k/2}\, \overline a_0+p^{k/2}\, \overline b_0\mathbf{i}$ with $a_0, b_0\in\{0, 1, 2, \cdots, p^{k/2}-1\}$. Thus indeg$(\overline0)=p^{k/2}\times p^{k/2}=p^k$.

Suppose $k$ is odd. If $\lambda≥\frac{k+1}{2}$, then clearly $\alpha^2=\overline0$. If $\lambda≤\frac{k-1}{2}$, then clearly $\alpha^2\neq\overline0$. Therefore, in the case of $k$ odd, $\alpha^2=\overline0$ if and only if $\alpha=p^{(k+1)/2}\, \overline a_0+p^{(k+1)/2}\, \overline b_0\mathbf{i}$ with $a_0, b_0\in\{0, 1, 2, \cdots, p^{(k-1)/2}-1\}$. Hence, indeg$(\overline0)=p^{(k-1)/2}\times p^{(k-1)/2}=p^k$.

Theorem 2.6  $\mathrm{(1)}$ Let $n=2^k$, $k≥1$. Then indeg$(\overline1)=2^k$ for $k=1, 2$, while indeg$(\overline1)=8$ for $k≥3$.

$\mathrm{(2)}$ Let $n=p^k$, where $p$ is an odd prime, $k≥1$. Then indeg$(\overline1)=2$ if $p\equiv3\ ({\rm mod}\ 4)$, while indeg$(\overline1)=4$ if $p\equiv1\ ({\rm mod}\ 4)$.

Proof  (1) Let $n=2^k$. By inspection, we have indeg$(\overline1)=2^k$ for $k=1, 2$. Now suppose $k≥3$. Assume that $\alpha=\overline a+\overline b\mathbf{i}\in\mathbb{Z}_{2^k}[\mathbf{i}]$ with $\alpha^2=(\overline a^2-\overline b^2)+2\overline a \overline b\mathbf{i}=\overline1$. Clearly the parity of $a$ and $b$ is different. If $a$ is even while $b=2t+1$ is odd, then $2^k|2ab$ if and only if $a=0$ or $2^{k-1}$. However, $a^2-b^2-1\equiv -4t^2-4t-2\not \equiv0\ ({\rm mod}\ 2^k)$, which contradicts to the fact that $\alpha^2=\overline1$. So we must have $a$ is odd and $b$ is even. Then $2^k|2ab$ if and only if $b=0$ or $2^{k-1}$. Hence $a^2-b^2\equiv1\ ({\rm mod}\ 2^k)$ if and only if $a^2\equiv1\ ({\rm mod}\ 2^k)$. The number of solutions of $a^2\equiv1\ ({\rm mod}\ 2^k)$ is 4 for $k≥3$. So we can conclude that indeg$(\overline1)=8$ for $k≥3$.

(2) Let $n=p^k$, where $p$ is an odd prime, $k≥1$. Assume that $\alpha=\overline a+\overline b\mathbf{i}\in\mathbb{Z}_{p^k}[\mathbf{i}]$ with $\alpha^2=(\overline a^2-\overline b^2)+2\overline a \overline b\mathbf{i}=\overline1$. By Lemma 1.2(1), gcd$(p, a^2+b^2)=1$. So gcd$(p, a)=1$ or gcd$(p, b)=1$. Therefore by $p^k|2ab$, we derive that $a=0$ or $b=0$. If $b=0$, then by $a^2-b^2\equiv1\ ({\rm mod}\ p^k)$, we have $a^2\equiv1\ ({\rm mod}\ p^k)$, which has exactly two solutions. If $a=0$, then by $a^2-b^2\equiv1\ ({\rm mod}\ p^k)$, we have $b^2\equiv-1\ ({\rm mod}\ p^k)$, which has exactly two solutions when $p\equiv1\ ({\rm mod}\ 4)$, while no solutions when $p\equiv3\ ({\rm mod}\ 4)$, as claimed.

We call a digraph semiregular if there exists a positive integer $d$ such that the in-degree of each vertex in this digraph is either $d$ or 0. In Example 1.1, we see that $\Gamma_1(5)$ is semiregular. In fact, $\Gamma_1(n)$ is semiregular for $n>1$, by an argument similar to paper [8]. But $\Gamma_2(n)$ is not semiregular for some $n>1$. For example, in $\Gamma_2(5)$, indeg$(\overline0)=1$ while indeg$(\overline3+\mathbf{i})=2$.

Theorem 2.7  (1) $\Gamma_2(2^k)$ is semiregular if and only if $k=$1,2,3,4.

(2) Suppose $p$ is a prime congruent to 1 modulo 4. Then $\Gamma_2(p^k)$ is not semiregular for $k≥1$.

(3) Suppose $p$ is a prime congruent to 3 modulo 4. Then $\Gamma_2(p^k)$ is semiregular if and only if $k=1, 2$.

Proof  (1) By inspection, we readily show that $\Gamma_2(2^k)$ is semiregular for $k=1, 2, 3, 4$. Now suppose $k≥5$. Let $\beta=(\overline1+\mathbf{i})^2=\overline2\mathbf{i}$. Then indeg$(\beta)>0$. Let $\alpha=\overline a+\overline b\mathbf{i}$ such that $\alpha^2=\beta$. Then $a^2-b^2\equiv0\ ({\rm mod}\ 2^k)$ and $2ab\equiv2\ ({\rm mod}\ 2^k)$. By $2ab\equiv2\ ({\rm mod}\ 2^k)$, we have $ab\equiv1\ ({\rm mod}\ 2^{k-1})$ and hence $a^2b^2\equiv1\ ({\rm mod}\ 2^{k-1})$. Moreover, since $a^2-b^2\equiv0\ ({\rm mod}\ 2^k)$, clearly $a^2\equiv b^2\ ({\rm mod}\ 2^{k-1})$. So $b^4\equiv1\ ({\rm mod}\ 2^{k-1})$, which has exactly 4 solutions, since $k≥5$. Hence, $b=b_j+m2^{k-1}$, where $j\in\{1, 2, 3, 4\}$, $m\in\{0, 1\}$ and $b_j^4\equiv1\ ({\rm mod}\ 2^{k-1})$ for $j=1, 2, 3, 4$. For a fixed odd integer $b$, the congruence equation $ab\equiv1\ ({\rm mod}\ 2^{k-1})$ in $a$ has exactly one solution. Therefore $a=a_0+m2^{k-1}$, where $m\in\{0, 1\}$ and $a_0b\equiv1\ ({\rm mod}\ 2^{k-1})$. So we can conclude that indeg$(\beta)=16$. However, by Theorem 2.5, indeg$(\overline0)=2^k>16$ for $k≥5$. So $\Gamma_2(2^k)$ is not semiregular for $k≥5$.

(2) First, by Theorem 2.5, indeg$(\overline0)=1$ in $\Gamma(p)$. However, the in-degree of $\beta=(\overline x+\overline y\mathbf{i})^2\in {\rm D}(\mathbb{Z}_{p}[\mathbf{i}])$ is greater than 1 where $p=x^2+y^2$, since $(\pm \beta)^2=\beta$. Hence $\Gamma_2(p)$ is not semiregular. Second, let $A=\{d^2(\overline x+\overline y \mathbf{i})^2:\ d\in{\rm U}(\mathbb{Z}_{p^2})\ {\rm or}\ d=0\}$. Then indeg$(\gamma)>0$ for $\gamma\in A$. Moreover, since $(\pm d)^2=d^2$, one can derive that $|A|=\frac{1}{2}|{\rm U}(\mathbb{Z}_{p^2})|+1=\frac{1}{2}p^2-\frac{1}{2}p+1$. If $\Gamma_2(p^2)$ is semiregular, then for $\gamma\in A$, indeg$(\gamma)={\rm indeg}(\overline0)=p^2$ by Theorem 2.5. But one can easily check that $p^2|A|>|{\rm D}(\mathbb{Z}_{p^2}[\mathbf{i}])|$, which is impossible. So $\Gamma_2(p^2)$ is not semiregular.

Now, suppose $k≥3$. Let $\beta=\overline p^2\in {\rm D}(\mathbb{Z}_{p^k}[\mathbf{i}])$. Then indeg$(\beta)>0$. Assume that $\alpha= \overline a+\overline b\mathbf{i}$ such that $\alpha^2=\beta$. Then $a^2-b^2\equiv p^2\ ({\rm mod}\ p^k)$ and $2ab\equiv 0\ ({\rm mod}\ p^k)$. It is clear that $p|a$ and $p|b$. Moreover, since $a^2-b^2\equiv p^2\ ({\rm mod}\ p^k)$, one can derive that $p\parallel a$ or $p\parallel b$. If $p\parallel a$, then by $2ab\equiv 0\ ({\rm mod}\ p^k)$, we have $b\equiv0\ ({\rm mod}\ p^{k-1})$. Hence $b=p^{k-1}b_1$ with $b_1=0, 1, \cdots, p-1$. Furthermore, since $a^2-b^2\equiv p^2\ ({\rm mod}\ p^k)$, we derive that $a^2\equiv\ p^2\ ({\rm mod}\ p^k)$. Therefore, $a=p(mp^{k-2}\pm1)$ with $m=0, 1, \cdots, p-1$.

On the other hand, if $p\parallel b$, by an argument similar to above, we have $a=p^{k-1}a_1$ with $a_1=0, 1, \cdots, p-1$ and $b=p(mp^{k-2}\pm1)$ with $m=0, 1, \cdots, p-1$. Therefore, indeg$(\beta)=2p^2+2p^2=4p^2$. However, by Theorem 2.5, the in-degree of $\overline0$ in $\Gamma(p^k)$ is not equal to $4p^2$. So $\Gamma_2(p^k)$ is not semiregular for $k≥3$.

(3) First, by Lemma 1.2 (4), $\mathbb{Z}_{p}[\mathbf{i}]$ is a field when $p\equiv 3\ ({\rm mod}\ 4)$. So $\Gamma_2(p)$ is a 1-cycle and hence is semiregular. Second, by Lemma 1.3 (2), $|{\rm D}(\mathbb{Z}_{p^2}[\mathbf{i}])|=p^2={\rm indeg}(\overline0)$, which implies that $\alpha^2=\overline0$ for $\alpha\in {\rm D}(\mathbb{Z}_{p^2}[\mathbf{i}])$. So $\Gamma_2(p^2)$ is semiregular.

Now, suppose $k≥3$. Let $\beta=\overline p^2\in {\rm D}(\mathbb{Z}_{p^k}[\mathbf{i}])$. Then indeg$(\beta)>0$. Assume that $\alpha= \overline a+\overline b\mathbf{i}$ such that $\alpha^2=\beta$. Similarly to (2) above, we have $p|a$ and $p|b$, and furthermore, $p\parallel a$ or $p\parallel b$. If $p\parallel b$, then $p^2|a$. Let $a=p^ta_1$, while $b=pb_1$, where $t≥2$ and $p\nmid b_1$. Then by $\alpha^2=\beta$, we derive $a^2-b^2\equiv p^2\ ({\rm mod}\ p^k)$. Hence, $2t-2≥2$ and $p^{2t-2}a_1^2\equiv b_1^2+1\ ({\rm mod}\ p^{k-2})$, which contradicts to the fact that $b_1^2+1\not\equiv 0\ ({\rm mod}\ p)$ for any integer $b_1$, since $p\equiv 3\ ({\rm mod}\ 4)$. So we must have $p\parallel a$ and hence, by an argument similar to (2) above, we can conclude that indeg$(\beta)=2p^2\neq{\rm indeg}(\overline0)$. Therefore, $\Gamma_2(p^k)$ is not semiregular for $k≥3$.

We have observed that $\alpha$ is a vertex of a $t$-cycle if and only if $t$ is the least positive integer such that $\alpha^{2^t}=\alpha$. So it is easy to derive the following

Lemma 2.8  (1) $\alpha\in{\rm U}(\mathbb{Z}_{n}[\mathbf{i}])$ is a cycle vertex in $\Gamma_1(n)$ if and only if $2\nmid o(\alpha)$.

(2) $\alpha\in{\rm U}(\mathbb{Z}_{n}[\mathbf{i}])$ is a vertex of a $t$-cycle in $\Gamma_1(n)$ if and only if $t={\rm ord}_{o(\alpha)}2$.

Let $\alpha=\overline a+\overline b{\mathbf{i}}\in \mathbb{Z}_n[\mathbf{i}]$, the norm $N(\alpha)$ of $\alpha$ is defined by $1≤ N(\alpha)≤ n$ and $N(\alpha)\equiv a^2+b^2\ ({\rm mod}\ n)$. It is easy to check that $N(\alpha\beta)\equiv N(\alpha)N(\beta)\ ({\rm mod}\ n)$. If $\alpha$ is a vertex of a $t$-cycle, then $\alpha^{2^t}=\alpha$. So $N(\alpha)^{2^t}\equiv N(\alpha^{2^t})\equiv N(\alpha)\ ({\rm mod}\ n)$, i.e., $N(\alpha)(N(\alpha)^{2^t-1})\equiv0\ ({\rm mod}\ n)$. Since gcd$(N(\alpha), N(\alpha)^{2^t-1})=1$, if $p|N(\alpha)$ with $p^t\parallel n$, clearly $p^t|N(\alpha)$. So we have proved the following lemma.

Lemma 2.9 Let $\ n=\prod\limits_{j=1}^sp_j^{k_j}$ be the prime power decomposition of $n$. If $\alpha$ is a vertex of a $t$-cycle, then $p_j^{k_j}\ |\ N(\alpha)$ whenever $p_j\ |\ N(\alpha)$.

By Lemma 1.2 (3), $\mathbb{Z}_n[\mathbf{i}]$ is a local ring if $n=p^t$, where $p=2$ or $p$ is a prime congruent to 3 modulo 4, $t≥1$. It is easy to show that $\Gamma_2(n)$ has a unique component containing the 1-cycle with $\overline0$ as its only vertex if $\mathbb{Z}_n[\mathbf{i}]$ is a local ring. For the cycle vertices in $\Gamma_2(p^k)$ with $p$ is a prime congruent to 1 modulo 4, we have the following theorem.

Theorem 2.10 Let $p$ be a prime congruent to 1 modulo 4. Then $\alpha=\overline a+\overline b\mathbf{i}\neq\overline0$ lies on a $t$-cycle of $\Gamma_2(p^k)$ if and only if $p^k|N(\alpha)$ and $\overline {2a}$ lies on a $t$-cycle of $\Gamma_1(p^k)$.

Proof Suppose that $\alpha$ is a vertex of a $t$-cycle in $\Gamma_2(p^k)$, then $p|N(\alpha)$. By Lemma 2.9, $p^k|N(\alpha)$. Moreover, since $\alpha\neq\overline0$, it is easy to check that $p\nmid a$ and $p\nmid b$. So by $\alpha^2=(\overline a^2-\overline b^2)+2\overline a\overline b\mathbf{i}$ and $-b^2\equiv a^2\ ({\rm mod}\ p^k)$, we have $\alpha^2=\overline{2a}(\overline a+\overline b\mathbf{i})$. Therefore we can conclude that $\alpha^{2^t}=\overline{2a}^{2^t-1}(\overline a+\overline b\mathbf{i})$. Hence $t$ is the least positive integer such that $(2a)^{2^t-1}\equiv1\ ({\rm mod}\ p^k)$. Thus due to Lemma 2.8, $\overline {2a}$ lies on a $t$-cycle of $\Gamma_1(p^k)$.

Conversely, if $p^k|N(\alpha)$ and $\overline {2a}$ lies on a $t$-cycle of $\Gamma_1(p^k)$, then $\alpha^2=(\overline a^2-\overline b^2)+2\overline a\overline b\mathbf{i}=\overline{2a}(\overline a+\overline b\mathbf{i})$. Hence $\alpha^{2^t}=\overline{2a}^{2^t-1}(\overline a+\overline b\mathbf{i})$. Furthermore, since $t$ is the least positive integer such that $(2a)^{2^t-1}\equiv1\ ({\rm mod}\ p^k)$, we can claim that $t$ is the least positive integer such that $\alpha^{2^t}=\alpha$, which implies that $\alpha$ is a vertex of a $t$-cycle in $\Gamma_2(p^k)$.

For instance, $\overline3+\mathbf{i}$ lies on a 1-cycle of $\Gamma_2(5)$ (see Example 1.1), $2\times 3\equiv 1\ ({\rm mod}\ 5)$ and $\overline1$ lies on a 1-cycle of $\Gamma_1(5)$. If $n=5^2$, one can check that $\alpha=\overline8+\overline6\mathbf{i}$ lies on a 4-cycle of $\Gamma_2(5^2)$, i.e., the cycle ${\rm{\bar 8 + \bar 6}}{\bf{i}} \to {\rm{\bar 3 + }}\overline {{\rm{21}}} {\bf{i}} \to \overline {{\rm{18}}} {\rm{ + }}{\bf{i}} \to \overline {{\rm{23}}} {\rm{ + }}\overline {{\rm{11}}} {\bf{i}} \to {\rm{\bar 8 + \bar 6}}{\bf{i}}.$${\text{While}}\;\;{\mkern 1mu} \overline {{\text{16}}\;\;} {\mkern 1mu} {\text{lies}}\;\;{\mkern 1mu} {\text{on}}\;\;\;{\mkern 1mu} {\text{a}}\;{\text{4 - cycle}}\;{\text{of}}\;\;{\Gamma _{\text{1}}}({{\text{5}}^{\text{2}}}),{\text{i}}.{\text{e}}.,{\text{the}}\;\;{\text{cycle}}\;\;\overline {{\text{16}}} \to {\rm{\bar 6}} \to \overline {{\text{11}}} \to \overline {{\text{21}}} \to \overline {{\text{16}}} .$

Finally, we investigate the height of an arbitrary vertex of $\Gamma_2(p^k)$ for any prime $p$. We say a vertex $\alpha$ in $\Gamma(n)$ is of height $m$ if $m$ is the least nonnegative integer such that $\alpha^{2^m}$ is a vertex of a cycle, and we denote $h_\alpha=m$. Clearly, $h_\alpha=0$ if and only if $\alpha$ is a vertex of a cycle.

Theorem 2.11 Suppose $\alpha=\overline a+\overline b\mathbf{i}\in {\rm D}(\mathbb{Z}_{2^k}[\mathbf{i}])$, $k≥1$. Then the height $h_\alpha$ of $\alpha$ is

${{h}_{\alpha }}=\left\{ \begin{align} & \left\lceil {{\log }_{2}}\frac{k}{\mathit{\lambda }} \right\rceil, {{2}^{x}}\parallel a, \ {{2}^{y}}\parallel b, \ x\ne y, \ \lambda =\rm{min}\{x, y\}\ge 1, \\ & \left\lceil {{\log }_{2}}\frac{2k}{2\lambda +1} \right\rceil, \ {{2}^{\lambda }}\parallel a, \ {{2}^{\lambda }}\parallel b, \ \lambda 0. \\ \end{align} \right.$

Proof Suppose that $2^x\parallel a$, $2^y\parallel b$, $\lambda={\rm min}\{x, y\}$. Then $\alpha=2^\lambda\beta$, where $\beta=\overline {a_1}+\overline {b_1}\mathbf{i}$ with $2\nmid{\rm gcd}(a_1, b_1)$.

If $x\neq y$, then $\lambda≥1$, and $\beta^{2^j}\in {\rm U}(\mathbb{Z}_{2^k}[\mathbf{i}])$ for $j≥0$. Hence $\alpha^{2^j}=(2^\lambda)^{2^j}\beta^{2^j}=\overline0$ if and only if $2^j\lambda≥ k$, if and only if $j≥ \log_{_2}\frac{k}{\lambda}$. So $h_\alpha=\lceil \log_{_2}\frac{k}{\lambda}\rceil$.

If $x=y=\lambda≥0$, then $\alpha=2^\lambda\beta$ with both $a_1$ and $b_1$ are odd. Thus $\beta\in{\rm D}(\mathbb{Z}_{2^k}[\mathbf{i}])$. Let $\beta^2=2\gamma$ where $\gamma=\frac{1}{2}(\overline {a_1}^2-\overline {b_1}^2)+\overline {a_1}\overline{b_1}\mathbf{i}$. Then clearly $\gamma\in {\rm U}(\mathbb{Z}_{2^k}[\mathbf{i}])$ since $4|a_1^2-b_1^2$. Hence, $\alpha^{2^j}=(2^\lambda)^{2^j}\beta^{2^j}=2^{2^j\lambda}(2\gamma)^{2^{j-1}}=2^{2^j\lambda+2^{j-1}}\gamma^{2^{j-1}}$. So $\alpha^{2^j}=\overline0$ if and only if $2^j\lambda+2^{j-1}≥ k$, if and only if $j≥ \log_{_2}\frac{2k}{2\lambda+1}$. So $h_\alpha=\lceil \log_{_2}\frac{2k}{2\lambda+1}\rceil$.

Theorem 2.12 Suppose $\alpha=\overline a+\overline b\mathbf{i}\in {\rm D}(\mathbb{Z}_{p^k}[\mathbf{i}])$, where $p$ is a prime congruent to 3 modulo 4, $k≥1$. Then the height $h_\alpha$ of $\alpha$ is $h_\alpha=\lceil \log_{_2}\frac{k}{\lambda}\rceil$, where $p^x\parallel a$, $p^y\parallel b$ and $\lambda={\rm min}\{x, y\}≥1$.

Proof Since $p\equiv3\ ({\rm mod}\ 4)$, $\alpha\in{\rm D}(\mathbb{Z}_{p^k}[\mathbf{i}])$ if and only if $p|a$ and $p|b$. Let $p^x\parallel a$, $p^y\parallel b$ and $\lambda={\rm min}\{x, y\}≥1$. Then $\alpha=p^\lambda\beta$, where $\beta=\overline a_1+\overline b_1\mathbf{i}$ and $p\nmid{\rm gcd}(a_1, b_1)$. Hence $\beta\in{\rm U}(\mathbb{Z}_{p^k}[\mathbf{i}])$. So $\alpha^{2^j}=(p^\lambda)^{2^j}\beta^{2^j}=\overline0$ if and only if $2^j\lambda≥ k$, if and only if $j≥ \log_{_2}\frac{k}{\lambda}$. So $h_\alpha=\lceil \log_{_2}\frac{k}{\lambda}\rceil$.

Theorem 2.13 Suppose $\alpha=\overline a+\overline b\mathbf{i}\in {\rm D}(\mathbb{Z}_{p^k}[\mathbf{i}])$, where $p$ is a prime congruent to 1 modulo 4, $k≥1$. Then the height $h_\alpha$ of $\alpha$ is

${{h}_{\alpha }}=\left\{ \begin{align} & \left\lceil {{\log }_{2}}\frac{k}{\lambda } \right\rceil, {{p}^{x}}\parallel a, \ {{p}^{y}}\parallel b, \ k=\rm{min}\{\mathit{x}, \mathit{y}\}\ge 1, \\ & j, \ \quad \quad \mathit{p}\nmid a, p\nmid b, \ \rm{and}\ \mathit{j}\ \rm{is}\ \rm{the}\ \rm{least}\ \rm{nonnegative}\ \rm{integer} \\ & \quad \quad \quad \rm{such}\ \rm{that}\ \rm{both}\ {{\mathit{p}}^{k}}\ |\ {{(\mathit{N}(\alpha ))}^{{{2}^{\mathit{j}}}}}\ \rm{and}\ 2\nmid o(2\rm{Re}({{\alpha }^{{{2}^{\mathit{j}}}}})), \\ \end{align} \right.$

where ${\rm Re}(\gamma)=\overline c$ if $\gamma=\overline c+\overline d\mathbf{i}$.

Proof Since $p\equiv1\ ({\rm mod}\ 4)$, $\alpha=\overline a+\overline b\mathbf{i}\in{\rm D}(\mathbb{Z}_{p^k}[\mathbf{i}])$ if and only if $p|a^2+b^2$.

First, suppose $p|{\rm gcd}(a, b)$. Let $p^x\parallel a$, $p^y\parallel b$, where $x≥1$ and $y≥1$. Let $\lambda={\rm min}\{x, y\}$. Then $\alpha=p^{\lambda}\beta$, where $\beta=\overline a_0+\overline b_0\mathbf{i}$ with $p\nmid{\rm gcd}(a_0, b_0)$. Hence, $\alpha^{2^j}=\overline0$ for some $j≥1$. Now, suppose that $\alpha^2=p^{2\lambda}(\overline a_1+\overline b_1\mathbf{i})$, where $\overline{a_1}=\overline{a_0}^2-\overline{b_0}^2$ and $\overline{b_1}=2\overline{a_0}\overline{b_0}$. Then, clearly $p\nmid{\rm gcd}(a_1, b_1)$ since $p\nmid{\rm gcd}(a_0, b_0)$. So we can conclude that $\alpha^{2^j}=p^{2^j\lambda}(\overline a_j+\overline b_j\mathbf{i})$ with $p\nmid{\rm gcd}(a_j, b_j)$. Therefore $\alpha^{2^j}=\overline0$ if and only if $2^j\lambda≥ k$, if and only if $j≥ \log_{_2}\frac{k}{\lambda}$. So $h_\alpha=\lceil \log_{_2}\frac{k}{\lambda}\rceil$.

Second, suppose $p|a^2+b^2$ but $p\nmid{\rm gcd}(a, b)$. Then $\alpha^{2^j}\neq\overline0$ for any $j≥0$. It is easy to show that if $\alpha^{2^j}=\overline c+\overline d\mathbf{i}$, then $p\nmid{\rm gcd}(c, d)$. Moreover, by Theorem 2.10 and Lemma 2.8, $\alpha^{2^j}$ lies on a $t$-cycle of $\Gamma_2(p^k)$ if and only if $p^k|N(\alpha)^{2^j}$ and $\overline{2c}$ lies on a $t$-cycle of $\Gamma_1(p^k)$, if and only if $j$ is the least nonnegative integer such that both $p^k|N(\alpha)^{2^j}$ and ${\rm ord}_{ o(\overline{2c})}2=t$, if and only if $j$ is the least nonnegative integer such that both $p^k|N(\alpha)^{2^j}$ and $2\nmid o(\overline{2c})$. Hence the result follows.