MTH3003 Lecture 17

Lagrange’s theorem tells us only what subgroup orders are forbidden - the size of any subgroup must divide $|G|$. The converse fails: knowing $d\mid |G|$ doesn’t guarantee a subgroup of order $d$ ($A_4$ has order $12$ but no subgroup of order $6$). Sylow’s three theorems give a partial converse: they guarantee subgroups of any prime-power order dividing $|G|$, control how those subgroups are related, and count exactly how many there are. They are one of the most powerful structural results in finite group theory.

Definitions

$p$-Group, $p$-Subgroup

Fix a prime $p$. A group $H$ is a $p$-group if $|H|=p^r$ for some $r\ge 0$. If $H\le G$ is a $p$-group, $H$ is a $p$-subgroup of $G$.

Sylow $p$-Subgroup

Let $G$ be finite and let $p$ be a prime dividing $|G|$. Write

$$|G|=p^{r}t,p\nmid t.$$

A Sylow $p$-subgroup (or $p$-Sylow subgroup) of $G$ is a subgroup $H\le G$ with $|H|=p^r$ - i.e. the largest possible $p$-subgroup of $G$.

Examples

• $C_9$ has order $9=3^2$, so it is a $3$-group. As a subgroup of $S_9$, it is a $3$-subgroup.
• $S_9$ has order $9!$, which is not a prime power, so $S_9$ is not a $p$-group.
• The Klein four-group $K_4$ is a $2$-group ($|K_4|=4=2^2$), and a $2$-subgroup of $S_4$.
• $|S_5|=120=2^3\cdot 3\cdot 5$. A Sylow $2$-subgroup has order $8$, a Sylow $3$-subgroup has order $3$, a Sylow $5$-subgroup has order $5$. A subgroup of $S_5$ of order $4$ is a $2$-subgroup but not a Sylow $2$-subgroup (it’s not maximal).

Sylow’s Theorems

First Sylow Theorem

Let $G$ be a finite group. If a prime $p$ divides $|G|$, then $G$ has at least one Sylow $p$-subgroup.

Second Sylow Theorem

Let $G$ be finite with Sylow $p$-subgroup $H$. If $J$ is any $p$-subgroup of $G$ (not necessarily Sylow), then $J\le g^{-1}Hg$ for some $g\in G$. So all Sylow $p$-subgroups are conjugate, and every $p$-subgroup sits inside some Sylow $p$-subgroup.

Third Sylow Theorem

Let $a(p)$ be the number of Sylow $p$-subgroups of $G$. Then:

1. $a(p)\mid t$ (where $|G|=p^{r}t$ with $p\nmid t$);
2. $a(p)\equiv 1(\mathrm{mod}p)$.

Why these matter

• First Sylow guarantees existence of a maximum-size $p$-subgroup for each prime $p$ dividing $|G|$.
• Second Sylow says they’re all conjugate - there’s structurally only one “kind” of Sylow $p$-subgroup.
• Third Sylow counts them: $a(p)$ is constrained by two divisibility conditions, often pinning it down uniquely.

Combined, the three theorems often determine all groups of a given order. Hugely powerful.

Note

The First Sylow Theorem generalises Cauchy’s theorem, which says: if $p$ divides $|G|$, then $G$ has a subgroup of order $p$ (i.e. an element of order $p$). Sylow upgrades this to subgroups of order $p^r$.

Quick Examples

Example: Group of Order 54

Question

Let $|G|=54$. Show $G$ has a subgroup of order $27$ and one of order $2$.

$54=3^3\cdot 2$. By Sylow’s First Theorem applied with $p=3$, $G$ has a Sylow $3$-subgroup of order $3^3=27$. Applied with $p=2$, $G$ has a Sylow $2$-subgroup of order $2$. $■$

Example: Sylow Subgroups of $D_{10}$

$|D_{10}|=10=2\cdot 5$. So $D_{10}$ has Sylow $2$-subgroups of order $2$ and Sylow $5$-subgroups of order $5$.

• Sylow $5$-subgroup: $⟨(12345)⟩\cong C_5$.
• Sylow $2$-subgroup: any order-$2$ reflection generates one, $\cong C_2$.

Example: Sylow Subgroups of $S_6$

$|S_6|=720=2^4\cdot 3^2\cdot 5$. So Sylow $2$-subgroups have order $16$, Sylow $3$-subgroups order $9$, Sylow $5$-subgroups order $5$.

• $⟨(12345)⟩\cong C_5$ is a Sylow $5$-subgroup.
• $⟨(123)⟩\times ⟨(456)⟩\cong C_3\times C_3$ is a Sylow $3$-subgroup.
• $⟨(12)⟩\cong C_2$ is a $2$-subgroup but not Sylow (order $2$, not $16$).

Application: Every Group of Order 162 has a Normal Subgroup of Order 81

Example

Let $|G|=162$. Prove $G$ has a normal subgroup of order $81$.

$162=3^4\cdot 2$.

Claim 1. $G$ has a subgroup $H$ of order $81=3^4$. By the First Sylow Theorem, the Sylow $3$-subgroup exists.

Claim 2. $H$ is the only Sylow $3$-subgroup. By the Third Sylow Theorem, $a(3)\mid 2$ and $a(3)\equiv 1(\mathrm{mod}3)$. The first condition says $a(3)\in \{1,2\}$; the second forces $a(3)\in \{1,4,7,10,\dots \}$. The only common value is $a(3)=1$.

Claim 3. $H$ is normal. Fix $g\in G$. The conjugate $g^{-1}Hg$ is a subgroup (Quick Subgroup Test), with $|g^{-1}Hg|=|H|=81$, so it is a Sylow $3$-subgroup. Since there’s only one of those, $g^{-1}Hg=H$ for all $g\in G$. So $H⊴G$.

$■$

Recurring pattern

If a Sylow $p$-subgroup is unique (i.e. $a(p)=1$), then it is automatically normal - because conjugation permutes the Sylow $p$-subgroups, and there’s nothing else to permute it to.

Application: Every Group of Order 15 is Cyclic

Example

Let $|G|=15$. Prove $G\cong C_{15}$.

$15=3\cdot 5$.

Claim 1. Subgroups of $G$ have orders $1,3,5,15$ (Lagrange).

Claim 2. Only the trivial subgroup has order $1$, only $G$ has order $15$.

Claim 3. $G$ has only one subgroup of order $3$. By Sylow, every $3$-subgroup of $G$ is a Sylow $3$-subgroup (since $|G|=3^1\cdot 5$, no larger $3$-subgroup is possible). By the Third Sylow Theorem, $a(3)\mid 5$ and $a(3)\equiv 1(\mathrm{mod}3)$, giving $a(3)=1$.

Claim 4. $G$ has only one subgroup of order $5$. Same argument with $p=5$: $a(5)\mid 3$ and $a(5)\equiv 1(\mathrm{mod}5)$, so $a(5)=1$.

Claim 5. $G$ is cyclic. Let $J$ be the unique subgroup of order $3$ and $K$ the unique subgroup of order $5$. The four subgroups of $G$ are $\{e\},J,K,G$. By inclusion-exclusion,

$$|J\cup K|\le |J|+|K|-|J\cap K|=3+5-1=7.$$

Since $|G|=15>7$, there exists $g\in G$ with $g\notin J$ and $g\notin K$. Then $⟨g⟩\le G$, but $⟨g⟩\ne \{e\},J,K$, so $⟨g⟩=G$. Hence $G$ is cyclic. $■$

Proof of First Sylow (sketch - not examinable)

Note

Idea: act on subsets of size $p^r$ by left multiplication; use a divisibility argument and the Orbit-Stabiliser theorem to extract one whose stabiliser has order $p^r$.

Let $|G|=p^{r}t$, $p\nmid t$. Let $X$ be the set of all subsets of $G$ of size $p^r$. By a standard binomial-coefficient calculation,

$$|X|=\left(\genfrac{}{}{0ex}{}{p^{r}t}{p^r}\right)=\frac{p^{r}t}{p^r}\cdot \prod _{i=1}^{p^r-1}\frac{p^{r}t-i}{p^r-i}=t\cdot \prod _{i=1}^{p^r-1}\frac{p^{r}t-i}{p^r-i}.$$

A power of $p$ dividing $p^{r}t-i$ also divides $p^r-i$ (and vice versa), so all $p$-factors cancel between numerator and denominator. Hence $p\nmid |X|$.

$G$ acts on $X$ by $\lambda (g)W=gW$. Partition $X$ into orbits; since $p\nmid |X|$, at least one orbit $G{W}_i$ has $p\nmid |G{W}_i|$. Let $S={\mathrm{Stab}}_G(W_i)$.

By Orbit-Stabiliser theorem, $|G{W}_i|=|G|/|S|$. Write $|S|=p^{k}s$ with $s\mid t$; then $|G{W}_i|=p^{r-k}\cdot t/s$. For $p\nmid |G{W}_i|$ we need $r-k=0$, i.e. $|S|=p^{r}s$.

Now show $|S|\le |W_i|=p^r$. Fix $w\in W_i$ and consider the orbit $Sw$. If $g,h\in S$ with $gw=hw$, then $g=h$ (right-cancel by $w^{-1}$). So $|Sw|=|S|$. Also $Sw\subseteq W_i$ because $S$ stabilises $W_i$. Hence $|S|\le |W_i|=p^r$.

Combined: $|S|=p^{r}s\le p^r$, forcing $s=1$, so $|S|=p^r$. Thus $S$ is a Sylow $p$-subgroup. $■$

Proof of Second and Third Sylow (advanced - not examinable)

These rely on conjugation actions on the set of cosets $G/H$ (for the Second) and on the set of all Sylow $p$-subgroups (for the Third), combined with the Orbit-Stabiliser theorem and a fixed-point argument. Definitions used:

• Centraliser $C_G(h)=\{g\in G:gh=hg\}=\{g:g^{-1}hg=h\}$.
• Normaliser $N_G(S)=\{g\in G:g^{-1}Sg=S\}$ for $S\subseteq G$.
• Conjugacy classes: orbits of $G$ on itself under $\lambda (g)x=gx{g}^{-1}$.

Full proofs are in the lecture notes handout - examinable spirit is the application, not the derivation.

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Pre-Lecture Notes from mth3003 lecture notes 17.pdf

• $p$-group: order is a prime power $p^r$. Sylow $p$-subgroup of $G$: subgroup of size $p^r$ when $|G|=p^{r}t$, $p\nmid t$ - the maximum possible $p$-subgroup.
• First Sylow Theorem: existence - for every prime $p\mid |G|$, $G$ has at least one Sylow $p$-subgroup.
• Second Sylow Theorem: structure - all Sylow $p$-subgroups are conjugate; every $p$-subgroup sits inside some Sylow.
• Third Sylow Theorem: counting - $a(p)\mid t$ AND $a(p)\equiv 1(\mathrm{mod}p)$. Often pins $a(p)=1$ uniquely.
• Generalises Cauchy: prime $p\mid |G|\Rightarrow$ subgroup of order $p$ (and now: order $p^r$).
• Application 1: every group of order 162 has a normal subgroup of order 81. (Sylow uniqueness $\Rightarrow$ normality.)
• Application 2: every group of order $15$ is cyclic. (Two unique Sylow subgroups; counting gives an element of order $15$.)
• Standard pattern: count $a(p)$ via $a(p)\mid t,a(p)\equiv 1(\mathrm{mod}p)$, often forces $a(p)=1$, hence the Sylow is normal, often the group is a direct product or cyclic.
• Next lecture: more applications of Sylow - classifying small groups, finishing the course.