# Abstract Nonsense

## Groups of Order pq (Pt. II)

Point of Post: This is a continuation of this post.

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Case 3 $q>p$ and $q\equiv 1\text{ mod }p$

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Lemma 1: Let $q$ be a prime, then $\text{Aut}(\mathbb{Z}_q)\cong U\left(\mathbb{Z}_q\right)\cong \mathbb{Z}_{q-1}$.\ (where $U\left(\mathbb{Z}_q\right)$ is the group of units of the field $\mathbb{Z}_q$)

Proof: Define $F:\text{Aut}\left(\mathbb{Z}_q\right)\to U\left(\mathbb{Z}_q\right)$ by $\phi\mapsto \phi(1)$. This is evidently a homomorphism since $F(\phi\theta)=(\phi\theta)(1)=\phi(\theta(1)\cdot 1)=\theta(1)\phi(1)$ it’s injective since $1$ is a generator of $\mathbb{Z}_q$ and it’s surjective because given any $k\in\mathbb{Z}^\times$ one has that $\phi(m)=mk$ is an automorphism of $\mathbb{Z}_q$.

Remark: Note there was no need for $q$ to be prime here, the above proof shows that for any $n$ one has that $\text{Aut}\left(\mathbb{Z}_n\right)\cong\mathbb{Z}_n^\times$.

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Thus, it remains to prove that $U(\mathbb{Z}_q)\cong\mathbb{Z}_{q-1}$. To do this (since $|U(\mathbb{Z}_q)|=q-1$) it suffices to prove that $U(\mathbb{Z}_q)$ is cyclic. To do this we note that if there does not exist an element of $x\in U(\mathbb{Z}_q)$ of order $q-1$ then there exists some $r such that $x^r=1$ for every $x\in U(\mathbb{Z}_q)$. But, this implies that the polynomial $x^r-1$ has more solutions than its degree which is impossible. The conclusion follows. $\blacksquare$

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Remark: The above proof that $U(\mathbb{Z}_q)$ is cyclic is a little wishy-washy. A more rigorous, and more general proof can be found in this post of mine.

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Lemma 2: Let $\varphi,\varphi':K\to\text{Aut}(H)$ be a homomorphism.  Suppose there are maps $\phi\in\text{Aut}(K)$ and $\theta\in\text{Aut}(H)$ such that $\varphi'=i_\theta\circ\varphi\circ\phi$ (where $i_\theta$ is the inner autormorphism induced by $\theta$ in $\text{Aut}(H)$). Then, $H\rtimes_\varphi K\cong H\rtimes_{\varphi' }K$.

Proof: This is mostly just grunt work. Namely, define $F:H\rtimes_\varphi K\xrightarrow{\;\;\approx\;\;} H\rtimes_{\varphi'} K$ by $(h,k)\mapsto (\theta(h),\phi^{-1}(k))$. As indicated, $F$ is an isomorphism. Indeed, to see it’s a homomorphism we note that

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\displaystyle \begin{aligned}F\left((h_1,k_1)(h_2,k_2)\right) &= F\left(\left(h_1\varphi_{k_1}(h_2),h_1h_2\right)\right)\\ &= \left(\theta\left(h_1\varphi_{k_1}(h_2)\right),\phi^{-1}\left(k_1k_2\right)\right)\\ &= \left(\theta(h_1)\theta\left(\varphi_{k_1}(h_2)\right),\phi^{-1}(k_1k_2)\right)\\ &=\left(\theta(h_1)\left(\theta\circ\varphi\circ\theta^{-1}\right)(\theta(h_2)),\phi^{-1}(k_1k_2)\right)\\ &= \left(\theta(h_1)(i_\theta\circ\varphi)(\theta(h_2)),\phi^{-1}(k_1,k_2)\right)\\ &= \left(\theta(h_1)\left(\varphi'\circ\phi^{-1}\right)(\theta(h_2)),\phi^{-1}(k_1)\phi^{-1}(k_2)\right)\\ &= \left(\theta(h_1),\phi^{-1}(k_1)\right)\left(\theta(h_2),\phi^{-1}(k_2)\right)\\ &= F\left((h_1,k_1)\right) F\left((h_2,k_2)\right)\end{aligned}

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But, evidently $F$ is a bijection since a quick check shows that $(h,k)\mapsto (\theta^{-1}(h),\phi(k))$ serves as an inverse. $\blacksquare$

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Lemma 3: If $K$ is cyclic and $\varphi,\varphi':K\to \text{Aut}(H)$ are such that $\varphi(K)$ is conjugate to $\varphi'(K)$ then $H\rtimes_\varphi K\cong H\rtimes_{\varphi'}K$.

Proof: By assumption $K=\langle x\rangle$ for some $x\in K$ and there exists some $\theta\in\text{Aut}(H)$ such that $i_\theta(\varphi(K))=\varphi'(K)$. Thus, $i_\theta(\varphi(k))=\varphi'(k)^m$ for some generator $\varphi'(k)^m$ of the cyclic $\varphi'(K)$. Since $\varphi'(k)^m$ is a generator we know that $\left(m,|\varphi'(K)|\right)=1$ and we may evidently then assume that $(m,|K|)=1$. We can then define $\phi:k^j\mapsto k^{m^{-1}j}$ and $\phi\in\text{Aut}\left(K\right)$. But, a quick check shows that $\varphi'=i_\theta\circ\varphi\circ\phi$ and so the lemma follows from lemma 2. $\blacksquare$

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We are now prepared to prove our main theorem. Namely:

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Theorem: Let $p$ and $q$ be primes such that $q\equiv 1\text{ mod }p$. Then, there exists a unique, up to isomorphism,  non-abelian group of order $pq$.

Proof: It follows by our previous argument that there is precisely one Sylow $q$-subgroup of $G$ and $q$ Sylow $p$-subgroups. If $Q$ represents the latter and $P$ one of the former then we have that $P\cap Q$ is trivial, $PQ=G$, and $Q\unlhd G$. Thus, $G\cong Q\rtimes_\varphi P$ for some homomorphism $\varphi:P\to\text{Aut}(Q)$. Thus, we desire to seek how many of these homomorphisms define non-isomorphic non-abelian structures $Q\rtimes_\varphi P$. But, since $Q\cong\mathbb{Z}_q$ and $P\cong\mathbb{Z}_p$ this evidently is equivalent to finding the number of non-isomorphic non-abelian $\mathbb{Z}_q\rtimes_\varphi\mathbb{Z}_p$ for some homomorphism $\varphi:\mathbb{Z}_p\to\text{Aut}\left(\mathbb{Z}_q\right)$. But, by our first lemma we know that $\text{Aut}(\mathbb{Z}_q)\cong\mathbb{Z}_{q-1}$ and so evidently every non-trivial $\varphi$ must have image with order $p$. But, since $\mathbb{Z}_{q-1}$ is cyclic there is precisely one subgroup of order $p$. Thus, for every non-trivial $\varphi,\varphi':\mathbb{Z}_p\to\text{Aut}\left(\mathbb{Z}_q\right)\cong\mathbb{Z}_{q-1}$ one has that $\varphi\left(\mathbb{Z}_p\right)=\varphi'\left(\mathbb{Z}_p\right)$ and so by lemma 3 we know that $\mathbb{Z}_q\rtimes_\varphi \mathbb{Z}_p\cong\mathbb{Z}_q\rtimes_{\varphi'}\mathbb{Z}_p$. The conclusion follows. $\blacksquare$

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Thus, every group $G$ with $|G|=pq$ where $pq$ are primes with $q\equiv1\text{ mod }p$ is either isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_q\cong\mathbb{Z}_{pq}$ or $\mathbb{Z}_q\rtimes_\varphi\mathbb{Z}_p$ where $\varphi:\mathbb{Z}_p\to\text{Aut}\left(\mathbb{Z}_q\right)$ is any non-trivial (and since $q\equiv 1\text{ mod }p$ there are evidently $p-1$ of them) homomorphism. $\blacksquare$

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References:

1. Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print

2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print

April 19, 2011 -