Abstract Nonsense

Crushing one theorem at a time

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

Point of Post: In this post we conglomerate and extend a few exercises in Dummit and Foote’s Abstract Algebra which will prove that the only positive integers n for which the only group (up to isomorphism) of order n  is \mathbb{Z}_n are integers of the form n=p_1\cdots p_m are distinct primes with p_i\not\equiv 1\text{ mod }p_j for any i,j\in[m].

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This post will complete several lemmas/theorems which works towards proving not only that every group of order p_1\cdots p_m where p_i\not\equiv 1\text{ mod }p_j for any i,j\in[m] (greatly generalizing the statement that a group of pq for primes p<q with q\not\equiv 1\text{ mod }p is cyclic) but also that numbers of this form are the only numbers for which the converse is true (namely every group of order n is cyclic).

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The Classification

We will need several observations and lemmas to prove this result. The first is a lemma which, in and of itself, is a very powerful theorem:

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Theorem 1: Let G be a finite group and H\unlhd G be such that \left(|G|,|\text{Aut}(H)\right)=1, then H\leqslant \mathcal{Z}(G).

Proof: Note first that since H is normal each inner automorphism i_g:x\mapsto gxg^{-1} has a well-defined restriction r_g to H (i.e. for each g\in G the map r_g:H\to H:x\mapsto gxg^{-1} is a well-defined automorphism). We thus have a mapf:G\to\text{Aut}(H):g\mapsto r_g which is easily verified to be a homomorphism. Note then that by definition that \ker f=\bold{C}_G(H) where \bold{C}_G(H) is the centralizer of H in G (the largest subgroup of G for which H sits in the center) and so by the first isomorphism theorem we know that G/\bold{C}_G(H) embeds into \text{Aut}(H). Thus, we evidently have that |G/\bold{C}_G(H)| divides both |G| and \text{Aut}(H), but since (|G|,||\text{Aut}(H)|)=1 this implies that |G/\bold{C}_G(H)|=1 or G=\bold{C}_G(H). But, by definition this means that H\leqslant\mathcal{Z}(G). \blacksquare

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Another very common, and widely powerful theorem is the following:

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Theorem 2: Let G be a group and H\leqslant\mathcal{Z}(G), if G/H (note that containment in the center implies normality) is cyclic then G is abelian.

Proof: Let a,b\in G be arbitrary. Since G/H is cyclic we have that aH=g^jH and bH=g^iH where G/H=\langle gH\rangle and 1\leqslant i,j\leqslant [G:H]. We see then that a=g^jh and b=g^ih' for some h,h'\in H\subseteq\mathcal{Z}(G) and so

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since a,b\in G were arbitrary the conclusion follows. \blacksquare

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We now prove the last ‘common knowledge theorem’ (i.e. something that is widely relevant elsewhere):

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Theorem 3: Let G be a group of order p_1\cdots p_m where p_1,\cdots,p_m are distinct primes. If G is abelian then G is cyclic.

Proof: This follows trivially from the structure theorem. That said, there is a more simpleminded proof. Namely, by Cauchy’s Theorem there exists x_1,\cdots,x_m\in G with |x_i|=p_i. That said, it’s trivial fact of group theory that if one has a collection of pairwise commuting group elements with pairwise coprime orders then the order of their product is the product of their orders. In particular, since G is abelian and (p_i,p_j)=1 for all i,j we have that |x_1\cdots x_m|=p_1\cdots p_m=|G| and so G=\langle x_1\cdots x_m\rangle. \blacksquare

Recall now that a simple group G is one with no non-trivial proper normal subgroups and maximal subgroups of a group G are subgroups which are maximal in terms of containment among proper subgroups (i.e. subgroups M<G such that M<K implies K=G). Let us make the following observation

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Observation: If M is a maximal subgroup of a group G then N_G(M)=M or N_G(M)=G (where N_G(M) denotes the normalizer of M in G). In particular, if M is not normal in G then N_G(M)=M.

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With this simple observation we can conclude the following

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Theorem 4: Let G be a finite group and M a maximal non-normal subgroup of G. Then, the number of non-identity elements of G contained in the union over all conjugates of M is at most (|M|-1)[G:M] with equality precisely when all the unequal conjugates of M have trivial intersection.

Proof: One can quickly check that gMg^{-1}=hMh^{-1} if and only if h^{-1}gM(h^{-1}g)^{-1}=M or if h^{-1}g\in N_G(M), or gN_G(M)=hN_G(M). Thus, the number of distinct conjugates of M is equal to \left[G:N_G(M)\right], but we know that N_G(M)=M and so the number of distinct conjugates of M is [G:M]. Moreover, each of these contain precisely |M|-1 non-identity elements and so evidently the cardinality of their union is bounded above by (|M|-1)[G:M], and clearly equal if and only if all the unequal conjugates of M intersect trivially. \blacksquare

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With this we can prove the following fascinating theorem:

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Theorem 5: Let G be a finite group and H<G. Then,

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\displaystyle \bigcup_{g\in G}gHg^{-1}\subsetneq G

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Proof: Find a maximal subgroup M of G containing H (I haven’t proven that this is true, but it follows similarly to the proof of Krull’s theorem). We then have that

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\displaystyle \bigcup_{g\in G}gHg^{-1}\subseteq\bigcup_{g\in G}gMg^{-1}\quad\bold{(1)}

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and so it suffices to prove that the right hand side of \bold{(1)} is proper in G. To do this we have two choices, either M\unlhd G in which case the right hand side of \bold{(1)} is just M which is clearly proper in G. If M is not normal in G then by the previous theorem we have the right hand side is bounded above in cardinality by (|M|-1)[G:M]+1<|G| and so clearly can’t be all of G. Either way, we’re done. \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 | , , , , , , , , , ,

1 Comment »

  1. […] 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. […]

    Pingback by A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. II) « Abstract Nonsense | September 13, 2011 | Reply

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