# Abstract Nonsense

## 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 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 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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$(|M|-1)\left[G:M\right]+\left(|N|-1\right)\left[G:N\right]+1$

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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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$2|G|+1-\left[G:M\right]-\left[G:N\right]$

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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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$|G|+1>\left[G:M\right]+\left[G:N\right]$

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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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References:

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