# Abstract Algebra/Group Theory/The Sylow Theorems

In this section, we will have a look at the Sylow theorems and their applications. The Sylow theorems are three powerful theorems in group theory which allow us for example to show that groups of a certain order $d\in \mathbb {N}$ are not simple.
The proofs are a bit difficult but nonetheless interesting. Important remark: Wikipedia also has proofs of the Sylow theorems, see Wikipedia article on the Sylow theorems, which are shorter and more elegant. But here you can find other proofs. This is because the author wanted to avoid redundancy. So you can choose the proof you like, or read both :-)
Remark: The proofs below also teach a lot about how to apply group actions, so they might give you also an idea how to do this kind of stuff :-)

## The Sylow theorems

Definition 1: Let $G$ be a finite group of order $mp^{a}$, where $p$ is a prime, $a,m\in \mathbb {N}$ and $m$ is coprime to $p^{a}$. We say that a subgroup of $G$ is a Sylow $p$-subgroup iff it has order $p^{a}$.
Definition 2: Let H be a subgroup of a group G. We define the normalizer N[H] of H as follows:
    $N[H]:=\{g\in G:gHg^{-1}=H\}$ due to Fermat
Theorem 3 (Cauchy's theorem) Let G be a group and $p$ be a prime number such that $p$ divides $|G|$. Then there exists an element of G which has order p. In particular, there is a subgroup of order p of G, namely $<p>$.
Proof: Let X be the set of all tuples $(x_{1},\ldots ,x_{p}),\forall 1\leq i\leq p:x_{i}\in G$ for which $x_{1}\cdots x_{p}=1$. The cyclic group $\mathbb {Z} /p\mathbb {Z}$ acts on X with the action $d*(x_{1},\ldots ,x_{p})=(x_{d+1},\ldots ,x_{p},x_{1},\ldots ,x_{d})$. An example is $(1,\ldots ,1)$, for which the orbit contains only this same element ($(1,\ldots ,1)$).
We also have that $|X|=|G|^{p-1}$ since we can choose the first p-1 elements arbitrarily and $x_{p}=(x_{1}\cdots x_{p-1})^{-1}$, and therefore |X| is a multiple of p, because |G| is also divisible by p. Furthermore, we know by the orbit-stabilizer theorem (theorem 19 from the section about group actions), that $\forall x\in X:|\mathbb {Z} /p\mathbb {Z} *x||\mathbb {Z} /p\mathbb {Z} _{x}|=|\mathbb {Z} /p\mathbb {Z} |=p$. Since p is a prime number, we have for all $x\in X:$ that either $|\mathbb {Z} /p\mathbb {Z} *x|=1$ or $|\mathbb {Z} /p\mathbb {Z} *x|=p$. But since the orbits partition X (due to lemma 11 from the section about group actions), and $|X|$ is divisible by p, we need at least one (in the case p = 2, else even more elements) other element $x'\neq (1,\dots ,1)$ of X such that $|\mathbb {Z} /p\mathbb {Z} *x'|=1$. Let $x'=(x_{1}',\ldots ,x_{p}')$. We have $x_{1}'=\cdots =x_{p}'$, because otherwise x' would not be fixed by the action we defined. Then $x_{1}'^{p}=x_{1}'\cdots x_{p}'=1$. QED.
Theorem 4 (Sylow I): Let $G$ be a finite group of order $mp^{a}$, where $p$ is a prime, $a,m\in \mathbb {N}$ and $m$ is coprime to $p$. For every $0\leq i\leq a$, there is a subgroup of order $p^{i}$ of G. In particular, there exists a Sylow $p$-subgroup of G.
Proof: For this proof, we use induction. Let H be a p-subgroup of G, i. e. $|H|=p^{i}$ for some natural i. H acts on the sets of left cosets G/H by left multiplication. By corollary 23 from the section about group actions, we obtain that $|G/H|\equiv |Z|\mod p(*)$, where $Z=\{gH\in G/H:\forall h\in H:hgH=gH\}$. But also the following equivalences are true:
    $\forall h\in H:hgH=gH\Leftrightarrow \forall h\in H:hg\in gH\Leftrightarrow \forall h\in H:g^{-1}hg\in H$ $\Leftrightarrow gHg^{-1}\subseteq H\Leftrightarrow gHg^{-1}=H$ because $|H|=|gHg^{-1}|$ $\Leftrightarrow g\in N[H]$
But from this we can conclude that $Z=\{gH\in G/H:g\in N[H]\}=N[H]/H$. Therefore, (*) becomes $[G:H]\equiv [N[H]:H]\mod p$. From this we can conclude, that if $i<a$, and therefore p divides $[G:H]$ by the theorem of Lagrange, that then also p divides $[N[H]:H]$. And also: Since $H$ is a normal subgroup of $N[H]$, we know that $N[H]/H$ is a group. Therefore we can apply Cauchy's theorem: $N[H]/H$ has a subgroup $N$ of order p. But if we set $H'=\bigcup _{gH\in N}gH$, then $H'$ is a subgroup of $G$ of order $p^{i+1}$, because a) the intersection of two different cosets is the empty set, and b) $H'$ is a subgroup of G because for $gh,g'h'\in H'$ $gh(g'h')^{-1}=ghh'^{-1}g'^{-1}=gg'^{-1}h''h'''\in H'$ for some $h'',h'''\in H$, because $g\in N[H]$, the normaliser. QED.
Lemma 5 (order of the conjugate): Let G be a group with identity $Id_{G}$, and $g\in G$ an element of that group. Then $\forall x\in G:ord(g)=ord(xgx^{-1})$
Proof: First, we observe that $(xgx^{-1})^{n}=xg^{n}x^{-1}$ by induction: For n = 1, the claim is obviously true, and the calculation
    $(xgx^{-1})^{n+1}=xg^{n}x^{-1}xgx^{-1}=xg^{n+1}x^{-1}$
shows the induction step.
Therefore, $(xgx^{-1})^{ord(g)}=xg^{ord(g)}x^{-1}=Id_{G}$, which shows that $ord(xgx^{-1})\leq ord(g)$.
Let furthermore $k\in \mathbb {N}$. Then $(xgx^{-1})^{n}=xg^{k}x^{-1}=Id_{G}\Rightarrow g^{k}x^{-1}=x^{-1}\Rightarrow g^{k}=Id_{g}$, where the first implication is true because the inverse is uniquely determined, and the second implication is true because the identity is uniquely determined. Therefore $ord(xgx^{-1})\geq ord(g)$, implying with the former inequality that $ord(g)=ord(xgx^{-1})$ and finishing the proof. QED.
Lemma 6: Let $X=\{gPg^{-1}:g\in G\}$, and let G act on X by conjugation. Then $G_{p}$ is a subgroup of G, and any p-subgroup of $G_{P}$ is contained in P.
Proof: Conjugation of G on X is a transitive action: If $g,g'\in G$ are arbitrary, $h*gPg^{-1}=g'Pg'^{-1}$ by choosing $h=g'g^{-1}$. By the section about group actions, transitivity implies $G_{P}$ is really a group.
By the definition of X, we have that P is even a normal subgroup of $G_{P}$. Let now Q be an arbitrary p-subgroup of $G_{P}$. Then $QP$ is a subgroup due to the section about normal subgroups. Due to the second isomorphism theorem, we have that $QP/P\approx Q/(Q\cap P)$. Therefore, we also have by Lagrange's theorem, that $|QP/P|=|Q/(Q\cap P)|={\frac {|Q|}{|Q\cap P|}}=p^{s}$ for some $s\in \mathbb {N} _{0}$, because Q is a Sylow p-subgroup. Furthermore, Lagrange's theorem also assures that $|QP/P|={\frac {|QP|}{|P|}}$. Since P is a Sylow p-subgroup and QP is a subgroup of G and therefore $|QP|$ divides |G|, we know that p does not divide $|QP/P|$. Therefore, $|QP/P|$ must be the trivial subgroup, and therefore also $Q/(Q\cap P)$, which implies $Q\cap P=Q$ because $gH=H\Leftrightarrow g\in H$ due to the section about subgroups, QED.
Theorem 7 (Sylow II): If P is a Sylow p-subgroup of G, and Q is an arbitrary p-group of G, then $\exists g\in G:Q\subseteq gPg^{-1}$, so Q is contained in a Sylow p-group, since for arbitrary groups G if $H$ is an arbitrary subgroup of G, then also $gHg^{-1}$. In particular, all Sylow $p$-subgroup of $G$ are conjugate.
Proof: Let's choose $X=\{gPg^{-1}:g\in G\}$. P acts on X by conjugation. By the orbit-stabilizer theorem (corollary 19 of the section on group actions), we have that $\forall x\in X:|P*x|={\frac {|P|}{|P_{x}|}}$. But since P is a Sylow p-group, we know that $|P*x|\equiv 0\mod p$ or $|P*x|\equiv 1\mod p$. Since $\forall h\in P:hPh^{-1}=P$, we furthermore have $P*P=\{P\}$ and therefore $|P*P|=1$, because P is a single element in X.
But P is also the only element with trivial orbit: Let $gPg^{-1}\in X$. That $gPg^{-1}$ has trivial orbit means translated into the language of our group action, that $\forall x\in P:xgPg^{-1}x^{-1}=gPg^{-1}$. If we multiply this equation by $g$ on the right and $g^{-1}$ on the left, we obtain that $\forall x\in P:g^{-1}xg\in P_{P}$. Because of Lemma 5 we know that $|gxg^{-1}|=|x|$. Therefore $gPg^{-1}$ is a p-subgroup of $P_{P}$. Due to Lemma 6, we know that therefore $gPg^{-1}$ must be a subgroup of P, and since both sets contain the same number of elements, they must be equal.
We now recall the above formula $\forall x\in X:|P*x|={\frac {|P|}{|P_{x}|}}$ and note that since $|P*x|=p^{s},s\in \mathbb {N} _{0}$, all the other elements must have the property $|P*x|\equiv 0\mod p$, since their orbits are not trivial. Since the orbits partition X, we have that $|X|\equiv 1\mod p$.
Now we let Q act on X by conjugation, instead of P. Since $|X|\equiv 1\mod p$, we know that there is at least one orbit of length 1. So what does this mean?:
    $\exists g\in G:\forall x\in Q:x(gPg^{-1})x^{-1}=gPg^{-1}$
As before:
    $g^{-1}Qg\subseteq Q_{P}$
, and, by Lemma 6:
    $g^{-1}Qg\subseteq P$, and therefore $Q\subseteq gPg^{-1}$. QED.
Lemma 8: The normalizer of a subgroup is a subgroup.
Proof: Let H be a subgroup of G, and let G act on H by conjugation. Then the normalizer of H is the stabilizer of H in this action. Therefore, it is a subgroup due to Lemma 14 of the section about group actions, QED.
Theorem 9 (Sylow III*): Let again $n_{p}$ be the number of Sylow $p$-groups of $G$. Then $n_{p}=|G/N[P]|$, where $P$ is any Sylow $p$-group.
Proof: This follows from the proof of Sylow II and the thm. Sylow II itself: Choose $X$ as in the proof of Sylow II. Then by the theorem itself follows that $|X|=n_{p}$, and if we consider the group action of G on X of conjugation, then we have that $\forall x\in X:|G*P|=n_{p}$. But since due to the orbit-stabilizer theorem (thm. 19 in the section about group actions) $|G*P|={\frac {|G|}{|G_{P}|}}={\frac {|G|}{|N[P]|}}$ due to the definition of the normalizer, and ${\frac {|G|}{|N[P]|}}=|G/N[P]|$ due to the theorem of Lagrange (which is applicable since N[P] is a subgroup due to Lemma 8), the theorem follows. QED.
Theorem 10 (Sylow III): Let $G$ be a finite group of order $mp^{a}$, where $p$ is a prime, $a,m\in \mathbb {N}$ and $m$ is coprime to $p$. If $n_{p}$ is the number of Sylow $p$-subgroups of $G$, then $n_{p}\mid m$ and $n_{p}\equiv 1\mod p$.
Proof: Choose $X$ as in the proof of Sylow II. The proof for Sylow II shows that $|X|\equiv 1\mod p$, and the theorem Sylow II itself shows that $|X|=n_{p}$. This proves the second part. The first part follows from Sylow III* and the fact that $|G/N[P]|={\frac {|G|}{|N[P]|}}$ due to Lagrange (which we can apply here because N[P] is a subgroup due to Lemma 8): Since P is a subgroup of N[P], we know that $|P|$ divides $|N[P]|$. This means that $|G/N[P]|$ is not divisible by p. But since $|G/N[P]|$ divides $|G|=mp^{a}$, it must divide m. QED.

## How to show that groups of a certain order aren't simple

In this section, it will be shown how to show that groups of a certain order can not be simple using the Sylow theorems. This is a useful application of the Sylow theorems.
Example 11: Groups of order 340 are not simple.
Proof: Let |G| = 340 = 2$^2$ ⋅ 5 ⋅ 17. Due to Sylow III, we have that $n_{5}=1$, because $n_{5}\equiv 1\mod 5$ and $n_{5}|2^{2}\cdot 17$ has only this solution (this can be seen by computing all possible solutions, which are finitely many since the last condition implies that $n_{5}\leq 2^{2}\cdot 17$). But since, due to Sylow II, the conjugate of the only Sylow 5-group is again a Sylow 5-group, it is itself. This is the definition of normal subgroups. Therefore, by the definition of simple groups, groups of order 340 are not simple. QED.
Example 12: Groups of order 48 are not simple.
Proof: Let |G| = 48 = 2$^4$ ⋅ 3. Sylow III tells us that $n_{2}\equiv 1\mod 2$ and $n_{2}|3$. From this follows that either $n_{2}=1$ or $n_{2}=3$. If now $n_{2}=1$, then, in the same way as example 11, the (then only) Sylow 2-group is normal. In the other case, through conjugation on the set of the three Sylow 2-groups, we generate a homomorphism $\phi :G\to S_{3}$. This is due to theorem 2 from the section about group actions. The image can't be trivial, because all Sylow 2-groups are conjugate because of Sylow II. But since the kernel is a normal subgroup, and $|S_{3}|=6$, we have that due to the first isomorphism theorem and the theorem of Lagrange, that the kernel is a proper, non-trivial normal subgroup, which is why the group is not simple.
Example 13: Groups of order 108 are not simple.
Proof: Let |G| = 108 = 2 ⋅ 2 ⋅ 3$^3$ and S be a Sylow 3-group. We let G act on the cosets of S by left multiplication. Due to theorem 2 from the section about group actions, we know that this action generates a homomorphism $\varphi :G\to S_{4}$. Due to the first isomorphism theorem and Lagrange, we have that $|G|=|ker\varphi ||\varphi (G)|$ and therefore $|\varphi (G)|$ must divide 108. Since $\varphi (G)$ is a subgroup in $S_{4}$ and $|S_{4}|=24$, $|\varphi (G)|$ must divide 24. From this follows that $|\varphi (G)|\leq 12$, and therefore, again due to the formula $|G|=|ker\varphi ||\varphi (G)|$ from the first isomorphism theorem and Lagrange, we have that $|ker\varphi |\geq {\frac {108}{12}}=9$. If the kernel would be G, then the action would be trivial and therefore $S=G$, which is a contradiction, since G is no Sylow 3-group. Therefore, the kernel is a proper, non-trivial normal subgroup, which is why the group is not simple.