Abstract Nonsense

Crushing one theorem at a time

A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. II)

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

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With these theorems and observations we can begin the lemmas that will be used to prove the main theorem. Namely:

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Lemma 1: Let G be a finite simple group in which every proper subgroup is abelian. Prove then that if M and N are distinct maximal subgroups of G then M\cap N is trivial.

Proof: We begin by noticing that since M is abelian that gxg^{-1}=x for all g\in M and x\in M\cap N and so g normalizes M\cap N and similarly N normalizes M\cap N. It follows that M\subseteq N_G(M\cap N) and N\subseteq N_G(M\cap N) but since M and N are distinct this implies that M<N_G(M\cap N) and so N_G(M\cap N)=G and so by the simplicity of G we have that either M\cap N=G or M\cap N is trivial, but clearly the first is impossible (since they are both proper subgroups of G) and so we must conclude the second. \blacksquare

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From this we can prove

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Lemma 2: Let G be a non-abelian group in which every proper subgroup is abelian. Then, G is not simple.

Proof: Let x\in G-\{e\}, we then have that \{e\}<\langle x\rangle< G (otherwise G is cyclic, which is not possible since G is abelian). We may then find some maximal subgroup M of G with \langle x\rangle\leqslant M. By Theorem 5 we have that there exists some y\in G with y\notin gMg^{-1} for all g\in G. Since y\ne e we must have that \{e\}<\langle y\rangle<G and so there exists some maximal subgroup N of G with \langle y\rangle\leqslant N. If either N or M is normal in G we’re done, so assume not. Note first that since for each g\in G gMg^{-1} is maximal (this is clear since conjugation is order preserving both ways) we have by Lemma 1 that if gMg^{-1}\ne M then gMg^{-1}\cap M is trivial, and so by Theorem 4 we have that the number of non-identity elements of G contained in some conjugate to M is \left(|M|-1\right)\left[G:M\right]. A similar analysis shows that the number of non-identity elements of G contained in some conjugate of N is \left(|N|-1\right)\left[G:N\right].  Now, note that no element of G was double counted by counting the number of things in any conjugate of M and counting the number of things in any conjugate of N. Indeed, this would be to suppose that gMg^{-1}\cap hNh^{-1} is non-trivial for some g,h\in G, but by Theorem 4 this would imply that gMg^{-1}=hNh^{-1}, but this would imply that h^{-1}gM(h^{-1}g)^{-1}=N, and so y\in N=h^{-1}gM(h^{-1}g)^{-1} contradictory to assumption. Thus, we have not double counted as claimed. It follows then that we have counted

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distinct elements of G (the extra 1 comes from counting the identity element). But, expanding this gives

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We claim though that this number is greater than |G|. Indeed, this is equivalent to showing that

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or dividing through by |G| gives

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\displaystyle 1+\frac{1}{|G|}>\frac{1}{|M|}+\frac{1}{|N|}

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but since M,N are non-trivial we have that |M|,|N|\geqslant 2 from where this trivially follows.  Thus, we have produced a number of distinct elements of G greater than |G| which is clearly impossible. \blacksquare

With these this last lemma we can finally prove our main theorem:

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Theorem: Let S be the set of all integers n such that every group of order n is cyclic. Then, n\in S if and only if n=p_1\cdots p_m where p_1,\cdots,p_m are distinct primes with p_i\not\equiv 1\text{ mod }j for each j=1,\cdots,m.

Proof: We first show that each integer of the form described is actually in S. To do this we induct on the number of primes in the factorization of such an integer. If m=1 then the result follows trivially (since every group of prime order is cyclic). So, suppose that the result is true for each integer whose prime factorization is p_1\cdots p_m for distinct primes with p_i\not\equiv p_j\text{ mod }1 and let q_1,\cdots,q_{m+1} be distinct primes such that q_i\not\equiv 1\text{ mod }q_j for each i,j and G a group of order q_1\cdots q_{m+1}. By the induction hypothesis we have that every proper subgroup of G is cyclic, and so trivially abelian. So, either G is abelian and so trivially non-simple or it is non-ablelian from where it follows from Lemma 2 that G is non-simple. Either way, we have that there exists some non-trivial proper subgroup N\unlhd G. Say that N=q_1\cdots q_j (we may clearly reorder the primes so that this is true) we then have by assumption that N is cyclic that \left|\text{Aut}(N)\right|=(q_1-1)\cdots(q_j-1) (recalling theclassification of the automorphism groups of cyclic groups). But, by assumption this evidently implies that \left(|G|,\text{Aut}(N)\right)=1 and so by Theorem 1 we know that N\leqslant \mathcal{Z}(G). We see then that |G/N|=q_{j+1}\cdots q_{m+1} and so cyclic by the induction hypothesis. But, by Theorem 2 this implies that G is abelian. But, we may then conclude from Theorem 3 that G is cyclic, and so the induction is complete.

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Conversely, let n=p_1^{a_1}\cdots p_m^{a_m} not be of this form. We then have two choices, either a_j\geqslant 2 for some j  or there exists primes p_i,p_j such that p_i\mid p_j-1.  If we can produce non-cyclic groups G_{p} of order p^2 and H_{p,q} of order pq for these two cases then we’ll be done because we can form a group of order n by merely taking G_p or H_{p,q} cross \mathbb{Z}_{\frac{n}{p^2}} or \mathbb{Z}_{\frac{n}{pq}} respectively which is clearly not cyclic, because if it were then G_p and H_{p,q} would have to be cyclic. But, this is easy for G_p since evidently G_p=\mathbb{Z}_p\times\mathbb{Z}_p is not cyclic . For H_{p,q} the situation is a little harder, but not too bad. Namely, if p,q are distinct primes with p\mid q-1 we know there exists a non-trivial homomorphism \varphi:\mathbb{Z}_p\to\mathbb{Z}_{q-1} (namely, since p\mid q-1 we know that there exists some cyclic subgroup \langle x\rangle with order p from where the isomorphism is apparent) and since we know \text{Aut}(\mathbb{Z}_q)\cong\mathbb{Z}_{q-1} there is by extension a non-trivial homomorphism \varphi:\mathbb{Z}_p\to\text{Aut}(\mathbb{Z}_q). Consequently, the semidirect product \mathbb{Z}\rtimes_\varphi\mathbb{Z}_q is a group of order pq which is not isomorphic to \mathbb{Z}_p\times\mathbb{Z}_q\cong\mathbb{Z}_{pq}. The conclusion follows. \blacksquare

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1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.


September 13, 2011 - Posted by | Uncategorized | , , , , , , , ,

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