----- > [!proposition] Proposition. ([[there are no simple nonabelian groups of order less than 60]]) > Let $G$ be a finite [[abelian group|nonabelian]] [[group]]. If $|G| < 60$, then $G$ is not [[simple group|simple]]. > [!proof]- Proof. ([[there are no simple nonabelian groups of order less than 60]]) > The [[center of a group]] is clearly a [[normal subgroup]], and [[p-groups have nontrivial centers]]. Therefore we can remove any [[p-group|p-groups]] from our list. > We are left to analyze $6, 10, 12, 14,15,18,20,21, 2 2, 24, 26, 28, 30, 33, 34,35,36,39,40,42$ > $4 4, 45, 46,48,51,52,54,56,57.$ > > > Next, we look at [[group|groups]] of order $p q$, where $p,q$ are both [[prime number|prime]] and $p<q$. By the [[the Sylow theorems|first and third Sylow theorems]], $G$ has a number $n_{q}$ of [[p-Sylow subgroup|q-Sylow subgroups]]s such that $n_{q}$ [[divides]] $p$ and $n_{q} \equiv 1 \text{ mod }q$. The only [[divides|divisors]] of $p$ are $1$ and $p$, so in particular we have either $n_{q}=1$ or $n_{q}=p$. Since $p<q$, $p \text{ mod }q=p \neq 1$, hence $n_{q}=1$. Now the [[the Sylow theorems|second Sylow theorem]] implies that this unique $q$-[[p-Sylow subgroup|Sylow subgroup]] $H$ is a [[normal subgroup]]. Thus we may remove all [[group|groups]] of order $pq$ from the list: > > $12,20, 24, 28, 30,36,40,42, 4 4, 45,48,52,54,56.$ > > Next we'll generalize this to orders of the form $pq^{2}$ or $q^{2}p$, again $p<q$. For $pq^{2}$ the result is straightforward: $n_{q}=1$ for same reasons as before. For $p^{2}q$ things are trickier. $n_{q}$ must be $1,p,\text{ or }p^{2}$. We can immediately rule out $p$, however, since $p<q$ so $p \text{ mod }q \neq 1$. For the $p^{2}$ case, we count: two $q$-Sylows intersect trivially (else [[Lagrange's Theorem]] is violated; recall that the [[intersection of subgroups is a subgroup]]), thus if there are $p^{2}$ $q$-Sylows, we get at least $p^{2}(q-1)$ elements in our group ($p^{2}$ spots left). We are guaranteed the existence of a $p$-Sylow [[subgroup]] of order $p^{2}$, and this [[subgroup]] does not intersect any of the $q$-Sylows ([[Lagrange's Theorem]]). We see that there is only space for one such $p$-Sylow, hence it is unique and [[normal subgroup|normal]] by the [[the Sylow theorems|second Sylow theorem]]. This argument easily generalizes to $pq^{n}$ and $p^{n}q$. > > We can also show that groups of order $pq r$ will not be [[simple group|simple]]. After this, > We are left with > > $24, 36, 48,56.$ > $56$ can be immediately counted out: $56=7^{1} \cdot 8$, such that $n_{7}=1$ or $n_{7}=8$. If $n_{7}=1$ we're done; so assume $n_{7}=8$. This [[subgroup]]s overlap trivially, thus they account for $8*(7-1)=48$ elements. There are $8$ elements left over; just enough to fit in a $2$-Sylow of order $8$ (which can't overlap any of the $7$-Sylows since $\text{gcd}(7,8)=1$). > > $24$, $36$, and $48$ are trickier to handle, as there is too much potential for substantial overlap that counting does not really work. However, we can recall another characterization of [[normal subgroup]]s: [[kernel iff normal subgroup]] and use [[group action]]s. > > Let's do $24=2^{3} \cdot 3$. We see that $n_{2} \in \{ 1,3 \}$ and $n_{3} \in \{ 1,4 \}$. Of these four cases, only the case with $n_{2}=3$ and $n_{3}=4$ is interesting, so assume that holds. Our goal is to form a [[group homomorphism|homomorphism]] $\begin{align} > \phi:G \to H \\ > G \times S \to S \\ > G \to \text{Perm}(S). > \end{align}$ > We can $S$ be? All we know about the structure of $G$ concerns its [[p-Sylow subgroup|Sylow subgroups]], so let's have $G$ act on those. There are $3$ $2$-Sylows, let them be $S:=\{ K_{1},K_{2},K_{3} \}$. Let's have $G$ [[group action|act on]] $S$ by [[conjugate|conjugation]], $G \xrightarrow{\phi} \text{ Perm}(S) \cong S_{3}.$ > $\phi$ is [[group homomorphism|nontrivial]], since it is [[transitive group action|transitive]] by [[the Sylow theorems|second Sylow theorem]]. So $\ker \phi \neq G$. And $\ker \phi \neq (e)$, because [[group homomorphism is injective iff kernel is trivial iff is a monomorphism|then it would be injective]], $|G|=24>12=|S_{3}|$ so the [[pidgeonhole principle]] forbids this. The conclusion is that $\ker \phi$ is a proper, nontrivial [[normal subgroup]] of $G$. > > The remaining possibilities are outlawed analogously. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```