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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Motivation

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 map $f:G\to\text{Aut}(H):g\mapsto r_g$ 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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$ab=g^jhg^ih'=g^jg^ihh'=g^ig^jh'h=g^ig^jh'h=g^ih'g^jh=ba$

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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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References:

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

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September 13, 2011 - Posted by Alex Youcis | Uncategorized | Algebra, Automorphism Group, Classification, Cyclic Groups, Group, Group Theory, Groups, Maximal Subgroup, Necessary Condition, Simple Group, Sufficient Condition for Cyclic Group

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[...] 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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My name is Alex Youcis. I am currently a senior at the University of Maryland, College park getting my bachelor’s in mathematics. Before I came to UMD I was a math major at Drexel University in Philadelphia, Pennsylvania. I was born and raised in Lititz, Pennsylvania, a small town near Lancaster.

As is probably clear, I really enjoy doing math. It’s a passion for me, and I don’t means this in the weekend-hobby sense, I can’t imagine doing anything else other than mathematics. That said, I’m not one of those space cadets one often sees wandering around math departments, trying to solve others problems, or perpetrating some other equally creepy shenanigans. No, while math is a gargantuan aspect of my life, there are many, many other things that interest me. Perhaps the most consuming of these is my love of music. I find it difficult to go even one day without my iPhone, or grooveshark.com If you want to say something to me in the car, you have about eight seconds before my iPhone is plugged in and we’ve set sail on a musical adventure. I have hundreds, upon hundreds of bands that I intensely love, but at the time of writing I can emphatically say that Regina Spektor is my favorite artist–the woman is a musical Gauss. I also love dogs, friends, conversations about things, craft beer and the occasional online comic.

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Advanced Microeconomics

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