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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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    $\begingroup$ Show your work? We are not here to give answers to your homework. $\endgroup$ – Chris Gerig Dec 29 '12 at 21:04
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    $\begingroup$ ok, I have just come back,I should say that this is not my homework. Actually this is to show the existence of the semidirect product of two cyclic group of order $p^{2} ,q$, since the Aut of the group of order q is of order q-1, and p divide q-1, so we can find the the homomorphism required for the semidirect product. $\endgroup$ – user53800 Dec 29 '12 at 22:18
  • $\begingroup$ 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. $\endgroup$ – user53800 Dec 29 '12 at 22:24
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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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  • $\begingroup$ I have a question, if $a^{p}\equiv 1 (mod q)$, then $y^{x^{p}}=1$, but I think it should be y $\endgroup$ – user53800 Dec 30 '12 at 2:14
  • $\begingroup$ @user53800 Oh yes you're right, I was going too fast. Let me fix that. $\endgroup$ – Alexander Gruber Dec 30 '12 at 3:15
  • $\begingroup$ Ok,I appreciate for that. Thanks $\endgroup$ – user53800 Dec 30 '12 at 17:46

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