----- > [!proposition] Proposition. ([[A5 is the unique simple group of order 60]]) > The [[alternating group]] of degree $5$, $A_{5}$, is the unique [[simple group]] of order $60$ up to [[group isomorphism|isomorphism]]. > [!proposition] Lemma. > $A_{5}$ has $24$ elements of [[order of an element in a group|order]] $5$, $20$ elements of order $3$, and $15$ elements of order $2$. > [!proof] Proof of Lemma. > Let $\sigma \in A_{5}$ such that $|\sigma|=5$. Then $\sigma$ is an [[parity of a permutation|even permutation]] on $10$ letters for which $\sigma^{5}=e$. > [!proposition] Lemma 2. > Let $H$ be a [[normal subgroup]] of a finite group $G$ and let $x\in G$. If [[greatest common divisor]]$(|x|, |G / H|)=1$, then $x \in H$. > [!proof] > Consider the [[image]] $xH$ of $x$ under the [[kernel iff normal subgroup|natural projection homomorphism]] $\pi:G \to G / H$. By [[group homomorphisms preserve structure]] we know that $|xH|$ [[divides]] $|x|$ as well as $|G / H|$. Therefore the assumption that [[greatest common divisor]]$(|x|, |G / H|)=1$ implies $|xH|=1$. But this is only true if $xH=e_{G / H}=H$. Hence, $x \in H$. > [!proof]- Proof. ([[A5 is the unique simple group of order 60]]) > **Existence Proof 1. (From Gallian)** > > By [[Lagrange's Theorem]], if $A_{5}$ had a nontrivial proper [[normal subgroup]] $H$, then $|H| \in \{ 2,3,4,5,6,10,12,15,20,30 \}.$ > **Lemma 1** says that $A_{5}$ has $24$ elements of order $5$, $20$ elements of order $3$, and $15$ elements of order $2$. This matters because... > > If $|H| \in \{ 3,6,12,15 \}$, then $|A_{5} / H| \in \{ 20,10,5,4 \}$ is [[relatively prime integers|coprime]] to $3$ and therefore any element of order $3$ belongs to $H$ by **lemma 2** — i.e., $H$ contains $20$ elements of order $3$ (too many, not possible) > > **Existence Proof 2. (From Dummit & Foote)** > > **Claim 1.** Let $G$ be a [[group]] of order [[order of a group|order]] $15$, $20$, or $30$. Then $G$ must have a [[normal subgroup|normal]] [[p-Sylow subgroup|5-Sylow subgroup]]. > > **Proof of Claim 1.** > - Suppose $|G|=15$. The [[divides|factors]] of $15=5^{1} \cdot 3$ are $\{ 1,3,5,15 \}$; by the [[the Sylow theorems|third Sylow theorem]] $G$ contains just one $5$-Sylow [[subgroup]] and by the [[the Sylow theorems|second Sylow theorem]] it is [[normal subgroup|normal]]. > - Suppose $|G|=20$. The factors of $20=5^{1} \cdot 4$ are $\{ 1,2,4,5,10,20 \}$; by [[the Sylow theorems|third Sylow theorem]] $G$ contains just one $5$-Sylow [[subgroup]] and by the [[the Sylow theorems|second Sylow theorem]] it is [[normal subgroup|normal]]. > - Suppose $|G|=30$. Then by [[the Sylow theorems#^4691c0|the classification of groups of order 30]] $G$ has a [[normal subgroup]] of [[order of a group|order]] $15$. By the other-above bullet said [[normal subgroup]] itself has a [[normal subgroup|normal]] [[subgroup]] that is a $5$-Sylow subgroup. Because [[p-Sylow normality nests]], this means that said $5$-sylow subgroup is also a [[normal subgroup]] of $G$. > > **Claim 2.** If $|G|=60$, then $n_{5}$, the number of $5$-Sylow [[subgroup]]s in $G$, is either $1$ or $6$. > > **Proof of Claim 2.** This is just applying [[the Sylow theorems]]. The [[divides|factors]] of $60=5^{1} \cdot 12$ are $\{ 1,2,3,4,6,12,20,30,60 \}$; [[the Sylow theorems|the third Sylow theorem]] indicates that the only possible options for $n_{5}$ are $1$ and $6$. > > **Claim 3.** Suppose $|G|=60$ and $n_{5}=6$. Suppose $G$ contains a nontrivial proper [[subgroup]] $H$. Then $|H|$ must equal one of $2,3,4,6,12$. > > **Proof of Claim 3.** Via [[Lagrange's Theorem]], $H \in \{ 2,3,4,6,12,20,30\}$. If $|H|=20$ or $|H|=30$, then by *claim 1* $H$ contains a [[normal subgroup|normal]] $5$-sylow [[subgroup]]; i.e., a unique $5$-sylow subgroup. But then $n_{5}=1 \neq 6$, contradicting the premise. So $|H| \in \{ 2,3,4,6,12 \}$. > > **Claim 4.** Additionally, if there exists such an $H_{}$ of order $6$ or $12$, then there exists such an $H'$ with $H' \in \{ 2,3,4 \}$. > > **Proof of Claim 4.** Suppose $|H|=6=3^{1} \cdot 2=2^{1} \cdot 3$. Then $H$ contains a unique $3$-Sylow; hence a [[normal subgroup|normal]] $3$-sylow. By [[p-Sylow normality nests]] this implies $G$ contains that [[subgroup]] as a proper, nontrival [[normal subgroup|normal subgroup]] (of order $3$). Hence there can exist such an $H'$ with $H'=3$. For the same reasoning there can exist such an $H'$ with $H'=2$. Else suppose $|H|=12=2^{2} \cdot 3=3^{1} \cdot 4$. The factors of $12$ are $\{ 1,2,3,4,6,12 \}$, and so $H$ can contain a unique $2$-Sylow of order $4$. So there can exist such an $H'$ with $H'=2$. This is what we set out to show. > > **Claim 5.** Now the [[quotient group|quotient]] $G / H$ has order $15,20,$ or $30$. But this is not possible— why? Conclude the following: *If $G$ is any group of order $60$ with $n_{5}=6$, then $G$ is simple.* > > **Proof of Claim 5.** This is not possible by *Claim 1*, which implies that $G / H$ must have exactly one $5$-Sylow [[subgroup]] $K$. Using [[the correspondence theorem]] we can lift $K$ to some nontrivial $\overline{K} \triangleleft G$ [[group homomorphisms preserve structure|whose order is divisible by 5]]. Now, *claim 4* tells us this is impossible. We conclude that *if $G$ is any group of order $60$ with $n_{5}=6$, then $G$ is simple.* > > **Conclusion.** Now consider $A_{5}$. The factors of $60$ are $\{ 1,2,3,4,6,12,20,30,60 \}$. Since $|A_{5}|=60=5^1\cdot 12$, we see that $A_{5}$ has either $1$ $5$-Sylow [[subgroup]] or $6$ $5$-Sylow [[subgroup]]s. We see that any $5$-cycle in $A_{5}\textcolor{Apricot}{}$ generates a [[subgroup]] of order $5$, e.g. here are two (>1) explicit examples: $\langle 12345 \rangle=\{ (12345),(13524),(14253), (15432),e \} $ > or > > $⟨(12354)⟩=\{(12354),(12543),(15423),(14234),(13542),e\},$ > > > thus it can't have just $1$ $5$-Sylow. It must therefore have $5$, and therefore be simple. > > # Proof of Uniqueness > **Claim 1.** Suppose $G$ is a [[simple group]] of order $60$. Then $G$ cannot have a proper [[subgroup]] of [[index of a subgroup|index]] (strictly) less than $5$. > > **Proof of Claim 1.** We will show that any [[subgroup]] $H$ is [[normal subgroup|normal]] whenever $[G:H] \in \{ 1,2,3,4 \}$. In the case $[G:H]=1$ we have $G=H$ and so clearly $H$ is [[normal subgroup|normal]], so suppose $[G:H] \in \{ 2,3,4 \}$. We proceed by exhibiting that $H$ [[kernel iff normal subgroup|is the kernel]] of a [[group homomorphism]] $\psi:G \to \text{Perm}(S)$, where $S=G / H$ (left [[coset]]s of $H$). > > Suppose $2 \leq [G:H] \leq 4$. Define a [[group action]] $\begin{align} > G \times S \to & S \\ > (g,s) \mapsto & gs. > \end{align}$ > This [[group action]] induces a [[group homomorphism|homomorphism]] $\psi:G \to \text{Perm}(S)$ given mapping $g \in G$ to the [[permutation]] of $S$ given by multiplying the elements of $S$ by $G$. But $\text{Perm}(S) \cong S_{3}$, hence $|\text{Perm}(S)|=[G:H]! \in \{ 2, 12,24 \}$. In all cases, $|\text{Perm}(S)|<|G|$, and therefore $\psi$ must not be an [[injection]] by the [[pidgeonhole principle]]. It follows from [[group homomorphism is injective iff kernel is trivial iff is a monomorphism]] that $\psi$ has a nontrivial [[kernel]] that is a proper [[subgroup]] of $G$, and therefore $G$ has a nontrivial proper [[normal subgroup]]. > > > > **Claim 2.** *If* $G$ has a [[subgroup]] $H$ of [[index of a subgroup|index]] $5$, then $G$ is [[group isomorphism|isomorphic to]] $A_{5}$. > > **Proof of Claim Two.** The idea is similar to before: we look at the [[group action|action]] of $G$ on the set of [[coset]]s $G / H$, which induces a [[group homomorphism|homomorphism]] $\phi: G \to S_{5}$. $\psi$ is an [[injection]] into $S_{5}$ because if $\phi(g)=\id$, then $gH_{i}=H_{i}$ for all $H_{i} \in G / H$; this only is true when $g=e$. We want to show that we have an [[group isomorphism|isomorphism]] onto $A_{5}$. Since $|\phi(G)|=|G|=60=|A_{5}|$, it suffices to show $\phi(G)$ 'lands inside' $A_{5}$. The way to check that something like that happens is to take the composite map to $S_{5} / A_{5} \cong \{ -1,1 \}$, because we know the kernel of the [[kernel iff normal subgroup|quotient map]] to $S_{5} / A_{5}$ is $A_{5}$ ...$\overbrace{G \xhookrightarrow{\phi} S_{5} \twoheadrightarrow^{\text{sign}} \{ -1,+1 \}}^{\psi: G \to \{ \pm 1 \}}.$ > Now, $\psi$ needs to have a nontrivial [[kernel of a group homomorphism|kernel]] because the codomain is smaller than the domain. But it also cannot have a [[kernel]] which is a proper subset of $G$, because $G$ is [[simple group|simple]]! So we must have $\ker \psi = G$. Since $\phi$ is an [[injection]], this means $\ker \psi \subset \ker \text{sign}=A_{5}$. > > > > > **Claim 3.** $n_{2}=5$ or $n_{2}=15$. Assuming $n_{2}=5$, $G \cong A_{5}$. > > **Proof of Claim 3.** $60=2^{2} \cdot 15$. So $n_{2} | 15$ and $n_{2} \text{ mod }2=1$. The factors of $15$ are $\{ 1,3,5,15 \}$ and all of them are [[congruent]] to $1$ modulo $2$. Since $G$ is [[simple group|simple]], we can rule out the possibility of $1$. If there are $3$ 2-Sylows, then we can let $G$ [[group action|act on]] those $3$ [[subgroup]]s $\{ K_{1},K_{2},K_{3} \}$ via [[conjugate|conjugation]], giving rise to a [[group homomorphism]] $\psi:G \to \text{Perm}(S)$ with proper nontrivial (thus proper normal) [[kernel]] since $\psi$ cannot be an [[injection]] ($|G|=60>6=|\text{Perm}(S)|$). > > (can also argue using that [[index of a subgroup|index]] of [[normalizer of a subgroup|normalizer]] is number of [[p-Sylow subgroup|p-Sylow]] and apply claim 1) > > Now assume $n_{2}=5$. In this case we can define a [[group action]] on the $5$ $2$-Sylow [[subgroup]]s, inducing a [[group homomorphism|group homomorphism]] $G \to S_{5}$, to which the same argument as in **claim 2** may be applied to show $G \cong A_{5}$. > > **Claim 4.** $n_{2} \neq 15$, thus $G \cong A_{5}$. > > We use the following argument. Suppose $n_{2}=15$. Then there exist two distinct $2$-Sylow subgroups of order $4$ $P$ and $Q$ with $P\cap Q > (e)$. (To see this, just count.) Consider $N_{G}(P \cap Q)$. This must be strictly smaller than $G$, since $G$ is [[simple group|simple]]. (Else $P \cap Q$ would be normal in $G$.) We argue that its [[index of a subgroup|index]] in $G$ must be $5$. > > First, note that $N_G(P \cap Q)$ has to contain both $P$ and $Q$, since they are [[abelian group|abelian]] ([[classification of small groups|order 4]]). And certainly $|N_{G}(P \cap Q)| > |P|$ because it contains $Q$. So [[Lagrange's Theorem]] dictates that $|N_{G}(P \cap Q)| \in \{ 10,12, 20 ,30\}$. But it can't be $20$ or $30$, since then the [[index of a subgroup|index]] is too small and $G$ won't be simple (see previous arguments). It can't be $12$, either, since then we could argue that $G \cong A_{5}$ (but $A_{5}$ has exactly $5$ 2-Sylows so that can't be). Finally, it can't be $10$, because it contains a [[subgroup]] $P$ of [[order of a group|order 4]]. So we see that it is impossible to have $n_{2}=15$. > > ![[CleanShot 2023-10-04 at 14.31.54.jpg|500]] > > ----- #### ---- #### 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 > ```