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Let $p$ and $q$ be integer primes such that $p$ divides $q-1$.

(a) Show that there exists a group $G$ of order $p^{2}q$ with generators $x$ and $y$ such that $x^{p^{2}} =1$, $y^{q}=1$, and $xyx^{-1}=y^{a}$, with $1$ the identity element of $G$ and $a$ some integer such that $a\not\equiv 1 \pmod q$ but $a^{p}\equiv 1\pmod q$.

(b) Prove that Sylow $q$-subgroup $S_{q}$ is normal in $G$, $G/S_{q}$ is cyclic and deduce that $G$ has a unique subgroup $H$ of order $pq$.

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    I am really doing lots of problems recently preparing for the exams, I did try for a long time, but I am not familiar with some definition and theorem, so I have not idea of what is going on, so any hint and even recommendation is help for me. Thanks a lot.2012-12-29

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Let $Y$ be the cyclic group of order $q$ and $X$ be the cyclic group of order $p^2$, generated by $y$ and $x$ respectfully. $|\text{Aut}(Y)|=q-1$, so because $p\mid q-1$ we have by Cauchy's that there is an element $\sigma\in \text{Aut}(Y)$ of order $p$. (In particular, $x\mapsto x^a$, where $a$ is as described by the problem statement, is such an automorphism.) Since $X$ is cyclic, it contains a normal subgroup of order $p$ (namely $\langle x^p \rangle$), and thus we can define a homomorphism $\lambda:X\rightarrow \text{Aut}(Y)$ so that $X/\text{ker}(\lambda)\cong \langle \sigma \rangle$. Forming the semidirect product $G=Y\rtimes_\lambda X$ yields the desired group.

That answers part (a) and the first part of (b). For the second part of (b), we have that $\text{ker}(\lambda)=\langle x^p \rangle$ acts trivially on $Y$, so $\langle y \rangle \times \langle x^p\rangle$ is the unique subgroup of order $pq$ in $G$ by construction.

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    Ok,I appreciate for that. Thanks2012-12-30