there is a single nonabelian simple group of order less than 168

---- > [!theorem] Theorem. ([[there is a single nonabelian simple group of order less than 168]]) > ^335086 > [!proof]- Proof. ([[there is a single nonabelian simple group of order less than 168]]) > We've shown that [[there are no simple nonabelian groups of order less than 60]]. Also, every [[abelian group]] is not simple, so there are no nontrivial simple groups of order less than 60. We also showed that there is a [[A5 is the unique simple group of order 60|unique simple group of order 60]]: $A_{5}$. Let $|G|=n$. Here we will show that there are no nonabelian simple groups with [[order of a group|order]] satisfying $60 < n < 168$. To begin, we can remove all [[prime number]]s in the range, for [[cyclic group|groups of prime order are cyclic]] and [[cyclic group]]s are [[abelian group|abelian]]. Thus, our list of candidates is as follows: > >| | 62 | 63 | 64 | 65 | 66 | | 68 | 69 | 70 | >|----|----|----|----|----|----|----|----|----|----| | | 72 | | 74 | 75 | 76 | 77 | 78 | | 80 | | 81 | 82 | | 84 | 85 | 86 | 87 | 88 | | 90 | | 91 | 92 | 93 | 94 | 95 | 96 | | 98 | 99 |100 | | |102 | |104 |105 |106 | |108 | |110 | |111 |112 | |114 |115 |116 |117 |118 |119 |120 | |121 |122 |123 |124 |125 |126 | |128 |129 |130 | | |132 |133 |134 |135 |136 | |138 | |140 | | |142 |143 |144 |145 |146 |147 |148 | |150 | | |152 |153 |154 |155 |156 | |158 |159 |160 | |161 |162 | |164 |165 |166 | | | | | \ The proof that [[there are no simple nonabelian groups of order less than 60]] contains some useful facts: >- **Orange.** If $n=pq$ for distinct [[prime number|primes]] $p,q$, then $G$ is not simple. >- **Blue.** If $n=p^{r}q$ for some [[prime number|prime numbers]] $p,q$ with $p<q$, then $G$ is not simple. >- **Green.** If $n=pq^{r}$ for some [[prime number|prime numbers]] $p,q$ with $p<q$, then $G$ is not simple. >- **Red.** If $n=pqr$ for distinct primes $p,q,r$, then $G$ is not simple. >- **Yellow.** If $n=p^{r}$ for $p$ a [[prime number]], then $G$ is not simple. > After computing the [[Fundamental Theorem of Arithmetic|prime factorizations]] of the remaining $n$, we color the table accordingly: > | | 62 | 63 | 64 | 65 | 66 | | 68 | 69 | 70 | |----|----|----|----|----|----|----|----|----|----| | | 72 | | 74 | 75 | 76 | 77 | 78 | | 80 | | 81 | 82 | | 84| 85 | 86 | 87 | 88 | | 90 | | 91 | 92 | 93 | 94 | 95 | 96 | | 98 | 99 |100 | | | 102 | | 104 | 105 | 106 | | 108 | | 110 | |111 |112 | | 114 | 115 | 116 | 117 | 118 | 119 | 120 | | 121 | 122 | 123 | 124 | 125 | 126 | | 128 | 129 | 130 | | |132 | 133 | 134 | 135 | 136 | | 138 | | 140 | |141 |142 | 143 | 144 | 145 | 146 | 147 | 148 | | 150 | | |152 | 153 | 154 | 155 | 156 | | 158 | 159 | 160 | |161 |162 | | 164 | 165 | 166 | | | | | \ So the remaining candidates are: $\{ 72,84,90,10 0,108,120,126,132,140,144,150, 156 \}.$ Next, recall the following lemma (from Gallian's book): ![[the Sylow Test for Nonsimplicity#^0c4966]] \ ![[the Sylow Test for Nonsimplicity#^a79def]]Using this, we can weed out $84$, $10 0$, $126,$ $140$ and $156$. We are left considering $\{ 72,90,108,120, 132, 14 4, 150 \}.$\ Suppose $n=72$. $72=2^{3} \cdot 3^{3}$; by [[the Sylow theorems]] $G$ has $n_{3}=1$ or $n_{3}=4$ [[p-Sylow subgroup|3-Sylow subgroups]] of order $9$. If $n_{3}=1$ then [[the Sylow theorems|the second Sylow theorem]] implies $G$ as a [[normal subgroup|normal]] $3$-Sylow [[subgroup]]; so assume $n_{3}=4$. $G$ [[group action|acts on]] the set of [[p-Sylow subgroup|3-Sylow subgroups]] $S:=\{ H_{1},H_{2},H_{3},H_{4} \}$ by [[conjugate|conjugation]]: $\begin{align} G \times S \to S \\ (g,s) \mapsto gsg^{-1}, \end{align}$ and we know this induces a [[group homomorphism|homomorphism]] $\begin{align} G \to \text{Perm}(S) \cong S_{4} \\ g \mapsto c_{g} \end{align}$ where $c_{g}:S \to S$ denotes the [[inner automorphism|conjugation automorphism]] of $S$ by $g \in G$. Because $|G|=72 > 24 = |S_{4}|$, the [[pidgeonhole principle]] dictates that this [[group homomorphism]] must not be [[injection|injective]]. By [[group homomorphism is injective iff kernel is trivial]] it has nontrivial kernel. Because the [[group action]] is [[transitive group action|transitive]], the [[group homomorphism|homomorphism]] is nontrivial and so its [[kernel]] is not $G$ itself. This means that its [[kernel]] is a proper, nontrivial [[subgroup]] of $G$. By [[kernel iff normal subgroup]] we conclude that $G$ is not [[simple group|simple.]] \ Suppose $n=90=2 \cdot 3^{2} \cdot 5$. By [[the Sylow theorems]] $G$ has exactly $1$ $3$-Sylow, hence it has a proper nontrivial [[normal subgroup]] of order $9$ and is thus nonsimple. \ Suppose $n=108=2^{2} \cdot 3^{3}$. Then $G$ has $1$ or $4$ [[](group%20homomorphism%20is%20injective%20iff%20kernel%20is%20trivial%20iff%20is%20a%20monomorphism.md)=72$ to show that, because $|G|>|S_{4}|$, $G$ is not simple. \ Suppose $n=120=2^{3}\cdot 3 \cdot 5$. Then $n_{2} \in \{ 1,6 \}$. Assume $G$ is simple, so then $n_{5}=6$. As usual, let $G$ act on the $5$-Sylow [[subgroup]]s by [[conjugate|conjugation]] to obtain a [[group homomorphism|homomorphism]] $f:G \to S_{6}$. Because the [[group action]] is [[transitive group action|transitive]], $f$ clearly is nontrivial. Since $G$ is simple, $f$ has trivial [[kernel]]. Therefore $f$ is an [[group embedding|embedding]] of $G$ into $S_{6}$; $G$ is [[group isomorphism|isomorphic]] to a [[subgroup]] $H \leq A_{6}$ of order $120$ ($H$ is simple). Now we can let $A_{6}$ act on the cosets of $H$, inducing a [[group homomorphism|homomorphism]] $\ell:A_{6} \to S_{3}$. $\ell$ needs to have proper nontrivial kernel, but $A_{6}$ is simple— a contradiction. \ Suppose $n = 132=2^{2} \cdot 3 \cdot 11$. Then $n_{11} \in \{ 1,12 \}$. But it can't be 12 by counting (we'd need $n_{3}=4$, and $n_{2}>1$ but $1+10\cdot 12+2 \cdot 4=129$ and adding in the $n_{2}$s would put it over $132$; note that the $n_{2}$ do not nontrivially intersect the $n_{3}$ or $n_{4}$ (but can nontrivially intersect each other). Hence $n_{11}=1$ and thus we have a proper nontrival [[normal subgroup]] of $G$. \ Suppose $n=1 4 4=2^{4} \cdot 3^{2}$. Then $n_{3} \in \{ 1,4 ,16\}$. Suppose $G$ is simple. Then clearly $n_{3} \neq 1$ and $n_{3} \neq 4$ (else we could argue as we did when $n=72$). So there are $n_{3}=16$ $3$-Sylow [[subgroup]]s of [[order of a group|order]] $9$. If two of these [[subgroup]]s, $P$ and $Q$, intersect nontrivially, by [[Lagrange's Theorem]] their intersection $P \cap Q$ has $3$ elements. The [[normalizer of a subgroup|normalizer]] $N_{G}(P \cap Q)$ must (properly) contain $P$ and $Q$ since groups of order $9$ are [[abelian group|abelian]]. By [[Lagrange's Theorem]], its possible orders are $\{ 12,16,18,24,36,48,72 \}$. We can outlaw $36,48,$ and $72$ since they would yield [[subgroup]]s of [[index of a subgroup|index]] less than $5$, upon whose [[coset]] $G$ could [[group action|act]] (by left-multiplication) to produce a [[group homomorphism|homomorphism]] $G \to S_{m}$ for some $m<5$, necessarily with proper, nontrivial [[kernel]] (yielding a proper nontrivial [[normal subgroup]] of $G$). So we are left considering $|N_{G}(P \cap Q)| \in \{ 12,16,18,24 \}$. But observe that since $N_{G}(P \cap Q)$ contains both $P$ and $Q$, it needs to be a multiple of $9$. None of $12,16,18,24$ are multiples of 9, and hence we conlcude $n_{3} \neq 16$. So $G$ is not simple. \ Suppose $n=150=2 \cdot 3 \cdot 5^{2}$. $n_{5} \in \{ 1,6 \}$. Assume $G$ is [[simple group|simple]], so $n_{5}=6$. As usual, $G$ acts on its $5$-Sylow [[subgroup]]s via [[conjugate|conjugation]], inducing a [[group homomorphism|homomorphism]] $f:G \to S_{6}$. Because the [[group action]] is [[transitive group action|transitive]], $f$ is clearly not trivial. Since $G$ is simple, $f$ needs to have a trivial [[kernel]] and thus be [[injection|injective]]. But that means $|f(G)|=|G|=150$, which does not [[divides|divide]] $|S_{6}|=720$. So $G$ must not be simple. \ All cases have been shown. > Suppose $n=90=2 \cdot 3^{2} \cdot 5$. By [[the Sylow theorems]] $G$ has exactly $1$ $3$-Sylow or $9$ 3-Sylows, if $1$ it has a proper nontrivial [[normal subgroup]] of order $9$ and is thus nonsimple. So assume $n_{3}=9$. By counting, it must be the case that two $3$-Sylows $P$ and $Q$ nontrivially intersect; $|P \cap Q|=3$. Both $P$ and $Q$ are [[abelian group]]s ([[group of order p squared is abelian]]) so $N_{G}(P \cap Q)$ contains them both. By [[Lagrange's Theorem]], $|N_{G}(P \cap Q)| \in \{ 1,2,3,5,6,9,10,15,18,30,45,90 \}.$ > But actually since $PQ \in N_{G}(P \cap Q)$, it can't be less than $\frac{|PQ|}{|P \cap Q|}=27$. So the only options are $30,45,90$. But each of these would have index too small in $G$: $G$ could act via conjugation to give a homomorphism into $S_{m}$ for some $m$ which has nontrivial proper kernel. > ^b351fb ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as 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 > ```