Abstract Nonsense

Crushing one theorem at a time

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<q-1 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

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April 19, 2011 - Posted by | Algebra, Group Theory, Representation Theory | , , , ,

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