%% Antes de processar este arquivo LaTeX (LaTeX2e) deve %% verificar que o arquivo TEMA.cls estah no mesmo %% diretorio. O arquivo TEMA.cls pode ser obtido do %% endereco www.sbmac.org.br/tema. \documentclass{TEMA} %\usepackage[brazil]{babel} % para texto em Português \usepackage[english]{babel} % para texto em Inglês \usepackage[latin1]{inputenc} % para acentuação em Português %\input{P-margin.inf} \usepackage[dvips]{graphics} \usepackage{subfigure} \usepackage{graphicx} \usepackage{epsfig} %\usepackage{graphics} %\usepackage{subfigure} %\usepackage{graphicx} %\usepackage{epsfig} \usepackage[centertags]{amsmath} \usepackage{graphicx,indentfirst,amsmath,amsfonts,amssymb,amsthm,newlfont} \usepackage{longtable} \usepackage{cite} \usepackage[usenames,dvipsnames]{color} \usepackage{algorithm} \usepackage{algorithmic} \newcommand{\B}{{\tt\symbol{92}}} \newcommand{\til}{{\tt\symbol{126}}} \newcommand{\chap}{{\tt\symbol{94}}} \newcommand{\agud}{{\tt\symbol{13}}} \newcommand{\crav}{{\tt\symbol{18}}} \newenvironment{demo}{\noindent{\bf Proof:}}{\\\hspace*{\fill}\bloco} \newcommand{\bloco}{\hfill\rule{2mm}{2mm}} \newcommand{\al}{\alpha} \newcommand{\p}{\psi} \newcommand{\ep}{\varepsilon} \newcommand{\lam}{\lambda} \newcommand{\ka}{\kappa} \newcommand{\LB}{\left|} \newcommand{\RB}{\right|} \newcommand{\rnor}{\vartriangleright} \newcommand{\lnor}{\vartriangleleft} \newcommand{\f}{\phi} \newcommand{\vf}{\varphi} \newcommand{\LL}{\left\langle} \newcommand{\RR}{\right\rangle} \newcommand{\LP}{\left(} \newcommand{\RP}{\right)} \newcommand{\pt}{\forall} \newcommand{\dsp}{\displaystyle} \newcommand{\om}[2]{\omega_{#1}^{#2}} \newcommand{\braket}[2]{\ensuremath{\langle #1 |#2\rangle}} % quantum bracket \newcommand{\brakett}[3]{\ensuremath{\langle #1 |#2| #3 \rangle}} % 3 place quantum bracket \newcommand{\aut}[1]{{\rm{Aut}}\left(#1\right)} \newcommand{\mdc}[1]{{\rm{mdc}}\left(#1\right)} \newcommand{\qbit}{{\it qbit}} \newcommand{\qbits}{{\it qbits}} \def\grp#1{\left\langle #1\right\rangle} % variable < > \def\sVEV#1{\left\langle #1\right\rangle} % variable < > %\def\bra#1{\Big\langle #1\Big|} % < | %\def\ket#1{\Big| #1\Big\rangle} % | > \def\sbra#1{\left\langle #1\right|} % variable < | \def\sket#1{\left| #1\right\rangle} % variable | > \def\R{\mathbb{R}} \def\N{\mathbb{N}} \def\C{\mathbb{C}} \def\Z{\mathbb{Z}} \def\X{\mathcal{X}} \def\G{\mathcal{G}} \def\H{\mathcal{H}} \def\A{\mathcal{A}} \def\ze#1{\mathcal{Z}\left( #1\right)} \def\w{\omega} %\def\cl{\mathcal{c}} \def\up{\textup} \floatname{algorithm}{Algorithm} \begin{document} %******************************************************** \title {An Efficient Quantum Algorithm for the Hidden Subgroup Problem over some Non-Abelian Groups~\thanks{Paper presented to III CMAC-SE, Vitória, 2015.}} \author { % Author A% % \thanks{email@email.br}\,, % Address Brazil % \\ \\ % Author B% % \thanks{email@email.br}\,, % Address Brazil % \\ \\ % Author C% % \thanks{email@email.br}\,, % Address Brazil } \criartitulo %\runningheads {Author A, Author B and Author C}{Instruções para %Preparação de Trabalhos} % \begin{abstract} {\bf Abstract}. The hidden subgroup problem (HSP) plays an important role in quantum computation, because many quantum algorithms that are exponentially faster than classical algorithms are special cases of the HSP. In this paper we show that there exist a new efficient quantum algorithm for the HSP on groups $\Z_{N}\rtimes\Z_{q^s}$ where $N$ is an integer with a special prime factorization, $q$ prime number and $s$ any positive integer. %{\bf Palavras-chave}. Palavra-chave 1, palavra-chave 2, %palavra-chave 3. \noindent {\bf Keywords}. Quantum Algorithms, Hidden Subgroup Problem, Quantum Computational Group Theory \end{abstract} %******************************************************** \newsec{Introduction} The most important problem in group theory in terms of quantum algorithms is called hidden subgroup problem (HSP)~\cite{cris}. The HSP can be described as follows: given a group $G$ and a function $f:G \rightarrow X$ on some set $X$ such that $f(x)=f(y)$ iff $x\cdot H=y\cdot H$ for some subgroup $H$, the problem consists in determining a generating set for $H$ by querying the function $f$. We say that the function $f$ hides the subgroup $H$ in $G$ or that $f$ separates the cosets of $H$ in $G$. A quantum algorithm for the HSP is said to be efficient when the running time is $O(\up{poly}(\log |G|))$. There are many examples of efficient quantum algorithms for the HSP in particular groups~\cite{simon94,shor2}. It is known that for finite abelian groups, the HSP can be solved efficiently on a quantum computer~\cite{cris}. On the other hand, an efficient solution for a generic non-abelian group is not known. Two important groups in this context are the symmetric and the dihedral groups. An efficient algorithm for solving the HSP for the former would imply in an efficient solution for the graph isomorphism problem~\cite{Beals,boneh, hoyer1997, ettinger1999} and for the latter would solve instances of the problem of finding the shortest vector in a lattice, which has applications in cryptography~\cite{Regev1}. One way to design new quantum algorithms for the HSP is to investigate the structures of all subgroups of a given group, and then to find a quantum algorithm applicable to each subgroup structure. Following this, Inui and Le Gall presented an efficient quantum algorithm for the HSP on groups of the form $\mathbb{Z}_{p^r}^m\rtimes \mathbb{Z}_{p}$ with prime $p$ and positive integers $r$ and $m$~\cite{inui2005efficient}. Later, Cosme~\cite{cmagnothesis} presented an efficient quantum algorithm for the HSP in $\Z_{p^r}\rtimes_\phi\Z_{p^s}$ where $p$ is any odd prime number, $r$ and $s$ are positives integers and the homomorphism $\phi$ is given by the root $tp^{r-s+l}+1$ such that $r\geq 2s-l$. Subsequently, Gonçalves et al. presented an efficient quantum algorithm for the HSP in $\Z_{p} \rtimes \Z_{q^s}$, with $p/q = \up{poly}( \log p)$, where $p$, $q$ are distinct odd prime numbers and $s$ is an arbitrary positive integer~\cite{DRC1, demersonthesis}. The case $\Z_{p^r} \rtimes \Z_{q^s}$ with $p,q$ distinct odd prime numbers and $r,s>0$ such that $p^r/q^t= \up{poly}(\log p^r)$ was discussed in~\cite{DNGarXiv:1104.1361}. The parameter $t\in\{0,1,\ldots,s\}$ characterizes the group. The case $t=0$ reduces to the abelian group $\Z_{p^r} \times \Z_{q^s}$ and the case $t=1$ was addressed by the authors, with unsuccessful results. This work established, for the first time, a complete description of the structure of the subgroups of $\Z_{p^r} \rtimes \Z_{q^s}$. Recently, using the algebraic structure of the subgroups of $\Z_{p^r} \rtimes \Z_{q^s}$, van Dam and Dey~\cite{wim14} presented a new quantum algorithm for the HSP over $\Z_{p^r} \rtimes \Z_{q^s}$ for all possible values of $t\in\{0,1,\ldots,s\}$ by imposing a restriction on the parameters $p$ and $q$: the relative sizes of subgroups are bounded by $p^r/q^{t-j}\in O(\mbox{poly}(\log p^r))$, where $j\in\{0,\ldots,t-1\}$. In this article, we describe a new efficient quantum algorithm to solve the HSP in the specific class of non-abelian groups, i.e., the semi-direct product groups of the form $G= \Z_{N}\rtimes\Z_{q^s}$, where, $N$ is factorized as $p_1^{r_1}\ldots p_n^{r_n}$ and there exists a $1\leq k \leq n$ such that $q^t$ ($q$ odd prime) divides $p_k-1$ and $q$ does not divide $p_i-1$ for all $i\neq k$. The parameter $t$ is related to the type of the homomorphism that describes the group, as can be checked in Sec.~\ref{semi-direct} Using a similar approach presented in~\cite{dong}, we define an isomorphism between $ \Z_{N}\rtimes\Z_{q^s}$ and the direct product of $\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}$ with cyclic groups, reducing the HSP in $G$ to similar HSPs the solutions of which are already known. %we show that the HSP in $G$ reduces the problem to similar ones %with known efficient solution. We show that the HSP over $G$ %reduces to the HSP over % %By employing an isomorphism between $ \Z_{N}\rtimes\Z_{q^s}$ and %the direct product of $\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}$ with %cyclic groups we have shown that the HSP can be reduced to similar %HSPs the solutions of which are already known. %Our algorithm was inspired by the work of Chi et. al.~\cite{dong} %that solved the HSP in $\Z_{N}\rtimes\Z_{p}$, where $p$ is a prime %that does not divide $p_i-1$ for any of the prime factors $p_i$ of %$N$. As in~\cite{dong}, we use the homomorphic properties of $G= %\Z_{N}\rtimes\Z_{q^s}$ to reduce the problem to similar ones with %known efficient solutions. As far as we know this is the first %efficient quantum algorithm to solve the HSP in this class of %groups. This work is organized as follows. In Section~\ref{semi-direct}, we give the relevant definitions and results concerning the semi-direct product groups and explain its homomorphisms and their properties. In Section~\ref{sec3}, we present our main result and we show that there exist an efficient quantum algorithm for the HSP on the groups. In Section~\ref{conclusion}, we draw our conclusions. %\section{Preliminaries} \section{Semi-direct Product Groups}\label{semi-direct} The semi-direct product of two groups $A$ and $B$ is defined by a homomorphism $\phi : B \rightarrow \mbox{Aut}(A )$, where $\textup{Aut}(A)$ denotes the automorphism group of $A$. The semi-direct product $A\rtimes_{\phi} B$ is the set $\{(a,b):a\in A, b\in B\}$ with the group operation defined as $(a,b)(a',b')=(a+\phi(b)(a'),b+b')$. One easily checks that the group inversion operation satisfies $(a,b)^{-1}=(\phi(-b)(-a), -b)$. In this paper we consider the HSP on the semi-direct product groups $G= \Z_{N}\rtimes_{\phi }\ \Z_{q^s}$ for positive integers $N$ and $s$ and odd prime number $q$. We assume that the prime factorization of $N$ is $p_1^{r_1}\ldots p_n^{r_n}$ and there exist a $1\leq k \leq n$ such that $q^t$ divides $p_k-1$ and $q$ does not divide $p_i-1$ for all $i\neq k$. The parameter $t\in\{0,1,\ldots,s\}$ characterizes the group as shown in the following. The elements $x=(1,0)$ and $y=(0,1)$ generate the groups $\Z_{N}\rtimes_{\phi} \Z_{q^s}$. Since $\mbox{Aut}(\Z_{N})$ is isomorphic to $ \Z_{N}^{*}$, the homomorphism $\phi$ is completely determined by $\alpha:=\phi(1)(1)\in \Z_{N}^{*}$ and $\phi(b)(a)=a\alpha^b$ for all $a\in \Z_N$ and $b\in \Z_{q^s}$. Now, note that $\phi (0) =\phi (q^s):\Z_ {N}\rightarrow\Z_{N}$ is the identity element of the group Aut$(\Z_{N})$. Then $\alpha^{q^s} =\phi(q^s)(1)=1$. If the element $\alpha\in\Z_{N}^{*} $ satisfies the congruence relation $X^{q^s}= 1 \mod N$, then it defines the semi-direct product $\Z_ {N}\rtimes_{\alpha}\Z_ {q^s} $. In this case, we must have $\textup{ord}(\alpha)=q^{t}$ for some integer $0\le t \le s$. The case $t=0$ reduces to the direct product $\Z_{N}\times\Z_{q^s} $, which is an abelian group. An efficient solution for the HSP is known for this case~\cite{cris}. Since $\alpha \in \Z_{N}^{*}$, $q^t$ divides $\varphi(N)=p_1^{r_1-1}\ldots p_n^{r_n-1}(p_1-1)\ldots(p_n-1)$, where $\varphi$ is the Euler phi-function. Then, we can choose the option $q^t\mid p_n-1$ with no loss of generality. %Because $q^t$ must divide $p_1^{r_1}\ldots %p_n^{r_n}(p_1-1)\ldots(p_n-1)$ (the order of $\Z_N^{*}$), we %choose the option $q^t\mid p_n-1$ with no loss of generality. Now note that due to factorization of $N$, the group $\Z_{N}$ is isomorphic to product of cyclic groups $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}}$, which implies \begin{equation} \Z_{N}\rtimes_{\phi} \Z_{q^s}\cong(\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}})\rtimes_{\phi} \Z_{q^s}. \end{equation} %Since $\Z_{N}$ is isomorphic to $\Z_{p_1^{r_1}}\times \ldots %\times \Z_{p_n^{r_n}}$ the group $\Z_{N}\rtimes_{\phi} \Z_{q^s}$ %is isomorphic to $G=(\Z_{p_1^{r_1}}\times \ldots \times %\Z_{p_n^{r_n}})\rtimes_{\phi} \Z_{q^s}$. The elements of $(\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}})\rtimes_{\phi}\ \Z_{q^s}$ have the form $((a_1,\ldots,a_n),b)$, where $(a_1,\ldots,a_n)\in \Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}}$ and $b\in \Z_{q^s}$. For each $b$ in $\Z_{q^s}$ the element $\phi(b)$ is an automorphism on $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}}$ such that $\alpha=\phi(1)(1)$ is an element in $\Z_{p_1^{r_1}}^{*}\times \ldots \times \Z_{p_n^{r_n}}^{*}$ of order $q^t$. Note that $\Z_{p_i^{r_i}}$ is isomorphic to the subgroup $\mathcal{I}_1\times \Z_{p_i^{r_i}}\times \mathcal{I}_2$ of $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}}$, where $\mathcal{I}_1$ is the identity on $\Z_{p_1^{r_1}}\times\ldots\times\Z_{p_{i-1}^{r_{i-1}}}$ and $\mathcal{I}_2$ is the identity on $\Z_{p_{{i+1}}^{r_{i+1}}}\times\ldots\times\Z_{p_{n}^{r_{n}}}$, for all $i=1,\ldots,n$. Thus, we can identify an element $a_i$ in $\Z_{p_i^{r_i}}$ with the point $\overline{ a_i}$ in $\mathcal{I}_1\times \Z_{p_i^{r_i}}\times \mathcal{I}_2$ such that it has an integer value $a_i$ in the $i$-th coordinate and $0's$ elsewhere. %Using this notation we can state following two results Now we are ready to state the following results. \begin{lemmaTEMA}\label{lemma1} Let $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}}$ and $\Z_{q^s}$ be finite abelian groups with distinct odd prime numbers $p_1,\ldots,p_n, q$ and positive integers $r_1,\ldots,r_n$ and $s$. Define the semi-direct product group $(\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}})\rtimes_{\phi} \Z_{q^s}$. Then, for each $b\in \Z_{q^s}$ and $a_i \in\Z_{p_i^{r_i}}$ there exists a $ c_i\in\Z_{p_i^{r_i}}$ such that $\phi(b)(\overline{ a_i})= \overline{ c_i}$. \end{lemmaTEMA} \begin{proof} Let $e_i$ be elements in $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}}$ with all components equal zero except the $i$-th one which is 1. It is enough to show that $\phi(b)(e_i)=\overline{d_i}$, for some $d_i\in\Z_{p_i^{r_i}}$. In fact, let us suppose that $\phi(b)(e_i)=(d_1,\ldots,d_n)$. Note that \begin{eqnarray*} (0,\ldots,0)&=&\phi(b)(0,\ldots,0)=\phi(b)(0,\ldots,0,p_i^{r_i},0,\ldots,0)=p_i^{r_i}\phi(b) e_i\\ &=& (p_i^{r_i}d_1,\ldots,p_i^{r_i}d_n). \end{eqnarray*} Then, for all $j=1,\ldots,n$ we have $p_i^{r_i}a_j\,\equiv\,0\textup{ mod }p_j^{r_j}$ and this implies that $a_j\,\equiv\,0\textup{ mod }p_j^{r_j}$ for all $j\neq i$. Hence, $\phi(b)(e_i)=(0\ldots,d_i,0,\ldots,0)=\overline{d_i}$ as was to be shown. \end{proof} %Besides, since $\alpha$ is an element in $\Z_{N}^{*}$ of order %$q^t$ then $q^t$ must be a divisor of $\varphi(N)=p_1^{r_1}\ldots %p_n^{r_n}(p_1-1)\ldots(p_n-1)$, where $\varphi$ is the Euler %phi-function. Without loss of generality, we can suppose $q^t$ %divides $p_k-1$. \begin{lemmaTEMA} \label{lemma2}Let $N$ be a positive integer with prime factorization $p_1^{r_1}\ldots p_n^{r_n}$ and $q$ an odd prime such that $q\neq p_i$ and $s$ a positive integer. Define the semi-direct product group $G=\Z_{N}\rtimes_{\alpha} \Z_{q^s}$ for an $\alpha \in \Z_N^{*}$. Let $t\in\{1,\ldots,s\}$ be the smallest positive integer such that $\alpha^{q^t}=1$. Let us assume that there exist a $1\leq k \leq n$ such that $q^t \mid p_k-1$ and $q\nmid p_i-1$ for all $i\neq k$. By choosing $k=n$ (WLOG) we have %$$ %\Z_{N}\rtimes_{\alpha} \Z_{q^s}\cong $q^t\mid p_n-1$ %(\Z_{p_1}^{r_1}\times\ldots\times\Z_{p_{n-1}}^{r_{n-1}})\times ( %$$ \begin{equation} \Z_N\rtimes_{\phi}\Z_{q^s}\cong(\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_{n-1}^{r_{n-1}}})\times (\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}), \end{equation} for some homomorphism $\psi$ from $\Z_{q^s}$ into the group of automorphisms of $\Z_{p^{r}}$ and $p=p_n$ and $r=r_n$. \end{lemmaTEMA} \begin{proof} Note that $\phi(q^s)$ is the identity map $\mathcal{I}$ on $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_{n}^{r_{n}}}$. For all $i=1,\ldots,n-1$, follows from Lemma~\ref{lemma1} that $ e_i=\phi(q^s)e_i=(0\ldots,{d_i}^{q^s},\ldots,0)$. Then ${d_i}^{q^s}=1\textup{ mod } p_i^{r_i}$ and $d_i\in\Z_{p_i^{r_i}}^{*}$ has order $q^{t'}$, for some $t'\in \{1,\ldots,s\}$. Suppose $d_i\neq 1$. Since $q^{t'}$ divides the order of $\Z_{p_i^{r_i}}^{*}$ and $\textup{gcd}(p_i,q)=1$, we have that $q^{t'}$ divides $p_i-1$. But that leads to an absurd, hence $d_i$ must be 1 and $\phi$ acts trivially on $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_{n-1}^{r_{n-1}}}$. Thus, there exists a homomorphism $\psi$ from $\Z_{q^s}$ into the group of automorphisms of $\Z_{p^{r}}$ ($p=p_n$ and $r=r_n$), such that for all $b\in \Z_{q^s}$ and all $(a_1,\ldots,a_n)\in \Z_{p_1^{r_1}}\times\ldots \times\Z_{p_n^{r_n}}$ we have \begin{equation} \phi(b)(a_1,\ldots,a_n)=(a_1,\ldots,a_{n-1},\psi(b)(a_n)). \end{equation} Now for two elements $g=((a_1\ldots,a_n),b)$ and $g'=((a_1'\ldots,a_n'),b')$ in $(\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_n^{r_n}})\rtimes_{\phi}\ \Z_{q^s}$, the group operation is defined by \begin{eqnarray} gg'&=&((a_1,\ldots,a_n)+\phi(b)(a_1',\ldots,a_n'),b+b')\nonumber\\ &=&((a_1+a_1',\ldots,a_{n-1}+a_{n-1}',a_n+\psi(b)(a_n'),b+b'). \end{eqnarray} Define the map \begin{equation} \Gamma:\Z_{N}\rtimes_{\phi}\ \Z_{q^s}\rightarrow (\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_{n-1}^{r_{n-1}}})\times (\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}), \end{equation} such that $\Gamma(a_1,\ldots,a_{n},b))=((a_1,\ldots,a_{n-1}),(a_n,b))$. It is easy to show that this map is indeed an isomorphism. \end{proof} \begin{lemmaTEMA}\label{lemma3} Let $G_1$ and $G_2$ be finite groups with relatively prime orders. If $H$ is a subgroup of $G$ then $H=H_1\times H_2$ for some subgroups $H_1$ of $G_1$ and $H_2$ of $G_2$. \end{lemmaTEMA} \begin{proof} Let $\pi_i:G_1\times G_2 \rightarrow G_i$ such that $\pi_i(g_1,g_2)=g_i$, $i=1,2$. For any subgroup $H$ of $G_1\times G_2$ define $H_1=\pi_1(H)\leq G_1$ and $H_2=\pi_2(H)\leq G_2$. Then $H\leq H_1\times H_2$. We claim that $H=H_1\times H_2$. In fact, if $(h_1,h_2)\in H_1\times H_2$ it follows from definition of $H_1$ and $H_2$ that there exist $h_1'\in G_1$ and $h_2'\in G_2$ such that $(h_1,h_2'),(h_1',h_2)\in H$. From the fact that $\gcd(|G_1|,|G_2|)=1$ and by the Chinese remainder theorem, there exist integers $r_1$ and $r_2$ such that \begin{equation}\label{TCR} \left\{\begin{array}{c} r_1\equiv 1 \mod |G_1| \\ r_1\equiv 0 \mod |G_2| \end{array}\right.\;\mbox{and}\; \left\{\begin{array}{c} r_2\equiv 0 \mod |G_1|\\ r_2\equiv 1 \mod |G_2|. \end{array}\right. \end{equation} It follows from (\ref{TCR}) that there are integers $k_1,k_2,k_3,k_4$ such that \begin{equation*} \left\{\begin{array}{l} r_1=k_1|G_1|+1\\ r_1=k_2|G_2| \end{array}\right.\;\mbox{and}\; \left\{\begin{array}{l} r_2=k_3|G_1|\\ r_2=k_4|G_2|+1. \end{array}\right. \end{equation*} Thus, \begin{eqnarray*} (h_1,h_2')^{r_1}&=&(h_1^{r_1},h_2'^{r_1})=(h_1^{k_1|G_1|+1},h_2'^{k_2|G_2|})=(h_1,e_2) \in H\\ (h_1',h_2)^{r_2}&=&(h_1'^{r_2},h_2^{r_2})=(h_1'^{k_3|G_1|},h_2^{k_4|G_2|+1})=(e_1,h_2) \in H \end{eqnarray*} where $e_1$ and $e_2$ are the identities elements in the groups $G_1$ and $G_2$, respectively. Hence, $(h_1,h_2)=(h_1,e_2)(e_1,h_2)\in H$. \end{proof} \section{Quantum Algorithm for HSP in $\Z_{N}\rtimes_{\phi} \Z_{q^s} $} \label{sec3}In this section we present an efficient quantum algorithm that can solve the HSP in $\Z_{N}\rtimes_{\phi} \Z_{q^s} $, where $N$ is factorized as $N=p_1^{r_1}\ldots p_n^{r_n}$ and given a $1\leq t\leq s$, there exists a $1\leq k \leq n$ such that $q^t \mid p_k-1$ and $q\nmid p_i-1$ for all $i\neq k$. Before stating our main theorem, let us introduce the last two intermediate results. \begin{propTEMAi}\label{prop1} Let $G$ be a finite group, $H$ a subgroup of $G$ and $f:G\rightarrow X$ the oracle function that hides $H$ in $G$. For any subgroup $\widetilde{G}$ of $G$ we have $\widetilde{f}=f\Big|_{\widetilde{G}}:\widetilde{G}\rightarrow X$ hides $\widetilde{H}=H \cap\widetilde{G}$ in $\widetilde{G}$. \end{propTEMAi} \begin{proof} We must show that $\widetilde{f}(a)=\widetilde{f}(b)$ if and only if $a\widetilde{H}=b\widetilde{H}$, for all $a, b\in \widetilde{G}$. In fact, let $a, b\in \widetilde{G}$ such that $\widetilde{f}(a)=\widetilde{f}(b)$. Since $f$ hides $H$ in $G$, $aH=bH$ which implies $a=bh$, for some $h\in H$. Since $a, b\in \widetilde{G}$, we have $h=b^{-1}a\in \widetilde{G}$ which implies $h\in \widetilde{H}$, hence $a\widetilde{H}=b\widetilde{H}$. Conversely, if $a\widetilde{H}=b\widetilde{H}$, since $a\widetilde{H} \subset aH$ and $b\widetilde{H} \subset bH$ we have $aH\cap bH \neq \emptyset $ which implies $aH=bH$, or equivalently $\widetilde{f}(a)=\widetilde{f}(b)$. \end{proof} \begin{coroTEMAi}\label{coro1} Let $G_1$ and $G_2$ be finite groups with relatively prime orders. Then, an efficient solution to the HSP over $G_1$ and $G_2$ implies in an efficient solution to the HSP over the direct product $G_1\times G_2$. \end{coroTEMAi} \begin{proof} By Lemma~\ref{lemma3}, if $H$ is a subgroup of $G_1\times G_2$ then $H=H_1\times H_2$ for some subgroup $H_1$ of $G_1$ and subgroup $H_2$ of $G_2$. Let $f$ be the oracle function that hides $H$ in $G_1\times G_2$. By Proposition~\ref{prop1}, the restrictions of $f$ to $G_1$ and $G_2$ hide, respectively, $H_1$ and $H_2$. Since the HSP can solved efficiently over the groups $G_1$ and $G_2$ one can efficiently find generators to $H_1$ and $H_2$, or equivalently, generators to $H_1\times H_2$. \end{proof} Define $N'=N/p_n^{r_n}$ then $\Z_{p_1^{r_1}}\times \ldots \times \Z_{p_{n-1}^{r_{n-1}}}\cong \Z_{N'}$. By Lemma~\ref{lemma2}, $$ \Z_{N}\rtimes_{\phi} \Z_{q^s} \cong \Z_{N'}\times (\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}). $$ The orders of the groups in this direct product are relatively prime, then by Corollary~\ref{coro1}, the HSP over $ \Z_{N}\rtimes\Z_{q^s}$ can be solved efficiently on a quantum computer. In fact, %Hence, from Lemma 3 , if $H$ is a subgroup of $\Z_{N'}\times %(\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s})$ then $H=H_1\times H_2$, where %$H_1$ is a subgroup of $\Z_{N'}$ and $H_2$ is a subgroup of %$\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}$. The HSP on $ %\Z_{N}\rtimes\Z_{q^s}$ reduces to the HSP on each factor by the %following. % %Let $f$ be the oracle function that hides the subgroup $H$ in %$\Z_{N'}\times (\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s})$. For simplicity %of notation, let us call $A=\Z_{N'}$ and %$B=\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}$. Define oracle function $f_1$ %by the restriction of $f$ to $A$, which hides $H_1$ in $A$. %Analogously, define oracle function $f_2$ by the the restriction %of $f$ to $B$, which hides $H_2$ in $B$. %%$f_2=\mid_{\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}}$ (restriction of $f$ %%to $\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}$)which hides $H_2$ in %%$\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}$. %The solution of the HSP in the groups $A$ and $B$ with functions %$f_1$ and $f_2$ determines generators for the subgroups $H_1$ and %$H_2$, respectively. the group $\Z_{N'}$ is abelian and the group $\Z_{p^{r}}\rtimes_{\psi}\Z_{q^s}$ was addressed in~\cite{DRC1,DNGarXiv:1104.1361} and recently generalized by~\cite{wim14}. Therefore, we obtain the following result: % % \begin{thmTEMA} Let $N$ be a positive integer with prime factorization $p_1^{r_1}\ldots p_n^{r_n}$, $q$ an odd prime such that $q\neq p_i$ and $s$ a positive integer. Define the semi-direct product group $G=\Z_{N}\rtimes_{\alpha} \Z_{q^s}$ for an $\alpha \in \Z_N^{*}$. Let $t\in\{1,\ldots,s\}$ be the smallest positive integer such that $\alpha^{q^t}=1$. Let us assume that there exist a $1\leq k \leq n$ such that $q^t \mid p_k-1$ and $q\nmid p_i-1$ for all $i\neq k$. Then there exists an efficient quantum algorithm that solves the HSP in the semi-direct product groups $\Z_{N}\rtimes_{\alpha} \Z_{q^s} $. \end{thmTEMA} \section{Conclusion}\label{conclusion} We have addressed the HSP on the semi-direct product groups $G= \Z_{N}\rtimes\Z_{q^s}$ where $N$ is factorized as $N=p_1^{r_1}\ldots p_n^{r_n}$ and given a $1\leq t\leq s$, there exists a $1\leq k \leq n$ such that $q^t$ divides $p_k-1$ $q\nmid p_i-1$ for all $i\neq k$. 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