# Abstract Nonsense

## 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 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 such that $M 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. 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.