$p$-groups and Sylow theorems

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 14 October 2011

$p$-groups

Let $p$ be a prime $p\in {\mathbb{Z}}_{\ge 0}$.

• A $p$-group is a group of order $p^a$ with $a\in {\mathbb{Z}}_{>0}$.

If $G$ is a $p$-group then $G$ contains an element of order $p$.

If $G$ is a $p$-group then $Z\left(G\right)\ne \left\{1\right\}$.

Let $p$ be a prime and let $G$ be a group of order $p^2$. Then $G$ is abelian.

If $G$ is a $p$-group of order $p^a$, then there exists a chain, $$\left\{1\right\}\subseteq N_1\subseteq N_2\subseteq \cdots \subseteq N_{a-1}\subseteq G$$ of normal subgroups of $G$, such that $\left|N_i\right|=p^i$.

Sylow theorems

Let $G$ be a finite group of order $p^{a}b$ where $p$ is prime, $a\in {\mathbb{Z}}_{>0}$ and $p$ does not divide $b$.

• A $p$-Sylow subgroup of $G$ is a subgroup of order $p^a$.

(First Sylow theorem) $G$ has a subgroup of order $p^a$.

(Second Sylow theorem) All the $p$-Sylow subgroups of $G$ are conjugates of each other.

(Third Sylow theorem) The number of $p$-Sylow subgroups of $G$ is $1\text{mod}p$.

Examples

The second theorem implies that the number of $p$-Sylow subgroups of $G$ divides the order of $G$. This is because if we consider the action of $G$ on the $p$-Sylow subgroups by conjugation, the only orbit consists of a $p$-Sylow subgroup and all its conjugates, which by the second Sylow theorem is all the $p$-Sylow subgroups of $G$. Since the cardinality of the orbit must divide the order of $G$, the number of $p$-Sylow subgroups of $G$ divides the order of $G$.

Classifying the groups of order 21

By the third Sylow theorem, 1, 8, 15, 22, ... are the possibilities for the number of 7-Sylow subgroups, and 1, 4, 7, 10, 13, 16, ... are the possibilities for the number of 3-Sylow subgroups.
   The second Sylow theorem forces that there be exactly 1 7-Sylow subgroup and either 1 or 7 3-Sylow subgroups since the number of Sylow subgroups must divide 21, the order of the group.

Since there is only 7-Sylow subgroup of $G$, call it $K$, and all conjugates of $K$ equal $K$, $K$ is normal in $G$. Since $K$ has order 7, $K\simeq \mathbb{Z}/7\mathbb{Z}$.

Case 1. 1 3-Sylow subgroup.
If there is only 1 3-Sylow subgroup, call it $H$, then $H$ is also normal in $G$ and is isomorphic to $\mathbb{Z}/3\mathbb{Z}$. Now the intersection $K\cap H=\left\{1\right\}$ since any element in the intersection must have order dividing both 3 and 7, the only possibility being 1, the only element of order 1. Now, $HK$ is a subgroup of $G$ since $K$ is normal in $G$, and since $HK=\left\{1\right\}$, $\left|HK\right|=\left|H\right|\left|K\right|=3\cdot 7=21=\left|G\right|$. So $G=HK$. Then Theorem ??? gives that $G\simeq H\times K\simeq \mathbb{Z}/7\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$.

Case 2. 4 3-Sylow subgroups
Let $H$ be one of the 3-Sylow subgroups of $G$. Once again, $H\simeq \mathbb{Z}/3\mathbb{Z}$ and $H$ is normal in $G$. By the same reasoning as before, $H\cap K=\left\{1\right\}$ and $HK=G$. Theorem ??? states that this is enough to write $G$ as a semidirect product of $H$ and $K$. The number of ways to do this depends on how many different homomorphisms $\theta :H\to \mathrm{Aut}\left(K\right)$ there are. Suppose that $x$ is a generator of $H$ and $y$ is a generator of $K$. Then $\theta$ is completely determined by where $x$ goes i.e. what $x^{-1}yx$ is. We know that it is of the form $y^i$ since it is an element of $K$. Suppose that $x^{-1}yx=y^i$. Then $y=x^{-3}y{x}^3$ = yi3 forcing $i^3=1\text{mod}7$. The possiblities for $i$ are 2 and 4. The semidirect products obtained by these two possibilities are isomorphic since if $x^{-1}yx=y^2$ then $x^{-2}y{x}^2=y^4$, and since $x$ and $x^2$ are both generators of $H$ the map sending $x\mapsto x^2$, $y\mapsto y$ will be an isomorphism of the two semidirect products. So in this case $G\simeq \mathbb{Z}/3\mathbb{Z}{\times }_{\theta }\mathbb{Z}/7\mathbb{Z}$ and any two such semidirect products are isomorphic.

Groups of order $\le 10$

Order 1	$\left\{1\right\}$
Order 2	$\mathbb{Z}/2\mathbb{Z}$
Order 3	$\mathbb{Z}/3\mathbb{Z}$
Order 4	$\mathbb{Z}/4\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$
Order 5	$\mathbb{Z}/5\mathbb{Z}$
Order 6	$\mathbb{Z}/6\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$
	$S_3\simeq D_3$
Order 7	$\mathbb{Z}/7\mathbb{Z}$
Order 8	$\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$,
	the dihedral group $D_4$,
	the quaternion group $Q_8$,
Order 9	$\mathbb{Z}/9\mathbb{Z}$, $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}$
Order 10	$\mathbb{Z}/10\mathbb{Z}\simeq \mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$
	$\mathbb{Z}/5\mathbb{Z}{\times }_{\theta }\mathbb{Z}/2\mathbb{Z}$, where $\theta =\text{???}$

MAKE THIS TABLE A BIT PRETTIER

Proofs.

If $G$ is a $p$-group then $G$ contains an element of order $p$.

Proof.

• To show: There exists $g\in G$ such that $\mathrm{order}\left(g\right)=p$.
• Let $x\in G$, $x\ne 1$.
• Then $\mathrm{order}\left(x\right)\ne 1$. Since $\mathrm{order}\left(x\right)$ divides $\left|G\right|=p^a$ we know that $\mathrm{order}\left(x\right)=p^b$ for some $0<b\le a$.
• Then, since $\mathrm{order}\left(x\right)=p^b$, $$x^{p^{b-1}}\ne 1,\text{and}{\left(x^{p^{b-1}}\right)}^p=x^{p^{b-1}p}=x^{p^b}=1.$$
• So $\mathrm{order}\left(x^{p^{b-1}}\right)=p$.

$\square$

If $G$ is a $p$-group then $Z\left(G\right)\ne \left\{1\right\}$.

Proof.

• To show: $\mathrm{Card}\left(Z\right(G\left)\right)\ne 1$.
1. $p$ divides $\left|G\right|$ since $\left|G\right|=p^a$.
2. Let ${𝒞}_g$ be a conjugacy class in $G$. Then ${𝒞}_g$ is an orbit under the action of $G$ on itself by conjugation.
• So $\mathrm{Card}\left({𝒞}_g\right)$ divides $\left|G\right|=p^a$.
• If $\mathrm{Card}\left({𝒞}_g\right)>1$ then $p$ divides $\mathrm{Card}\left({𝒞}_g\right)$.
3. The class equation is $$\left|G\right|=\left|Z\right(G\left)\right|+\sum _{\left|{𝒞}_g\right|>1}\mathrm{Card}\left({𝒞}_g\right),$$ where the sum is over all distinct conjugacy classes such that $\mathrm{Card}\left({𝒞}_g\right)>1$.
• Since $p$ divides $\left|G\right|$ and $p$ divides every term in the sum we cannot have $\left|Z\right(G\left)\right|=1$.
• So $\left|Z\right(G\left)\right|\ne 1$.

$\square$

Let $p$ be a prime and let $G$ be a group of order $p^2$. Then $G$ is abelian.

Proof.

• To show: The order of the center of $G$ is $p^2$, $\left|Z\right(G\left)\right|=p^2$.
• By Proposition ???, we know that $\left|Z\right(G\left)\right|$ divides $\left|G\right|=p^2$.
• Case 1. $\left|Z\right(G\left)\right|=1$.
• By Proposition ???, $\left|Z\right(G\left)\right|\ne 1$.
• Case 2. $\left|Z\right(G\left)\right|=p$.
• Let $x\in G$, $x\notin Z\left(G\right)$.
• Since $Z=Z\left(G\right)$ is a normal subgroup of $G$, $G/Z\left(G\right)$ is a group and $$\left|G/Z\right|=p^2/p=p.$$
• So, by Proposition ???, $G/Z$ is cyclic.
• Since $x\notin Z$, $Z\ne xZ$ and so $xZ$ generates $G/Z$, $$G/Z=\{Z,xZ,x^{2}Z,\dots ,x^{p-1}Z\}.$$
• Let $g\in G$. Then there exists $k\in {\mathbb{Z}}_{\ge 0}$ such that $gZ=x^{k}Z$ and $k\le p-1$.
• So there exists $z\in Z$ such that $g=x^{k}z$.
• Then $$xg=x{x}^{k}z=x^{k}xz=x^{k}zx=gx.$$
• So $x\in Z\left(G\right)$.
• This is a contradiction to $x\notin Z\left(G\right)$.
• So $\left|Z\right(G\left)\right|\ne p$.
• So $\left|Z\right(G\left)\right|=p^2$.
• So $G$ is abelian.

$\square$

If $G$ is a $p$-group of order $p^a$, then there exists a chain, $$\left\{1\right\}\subseteq N_1\subseteq N_2\subseteq \cdots \subseteq N_{a-1}\subseteq G$$ of normal subgroups of $G$, such that $\left|N_i\right|=p^i$.

Proof.

• We know that $Z\left(G\right)$ of $G$ is a normal subgroup of $G$ of order at least $p$.
• $Z\left(G\right)$ contains a subgroup of order $p$ by Proposition ???.
• This subgroup $N_1$ is a normal subgroup of $G$ of order $p$.
• Doing the same argument on $G/N_1$ gives a normal subgroup $N_2/N_1$ of $G/N_1$ of order $p$.
• Then by the correspondence theorem this corresponds to a normal subgroup $N_2$ of $G$ of order $p^2$ that contains $N_1$.
• In general, since $G/N_i$ is a $p$-group of order $p^{a-1}$ it contains a normal subgroup of order $p$ in $G/N_i$ which corresponds to a normal subgroup $N_{i+1}$ of $G$ which contains $N_i$.

$\square$

(First Sylow theorem) $G$ has a subgroup of order $p^a$.

Proof.

• To show: There is a subgroup of order $p^a$.
• Let $𝒮$ be the set of subsets of $G$ with $p^a$ elements.
• Let $G$ act on $𝒮$ by left multiplication $$\begin{array}{ccc}G\times 𝒮& ⟶& 𝒮\\ (g,S)& ⟼& gS\end{array}\text{where}gS=\left\{gs\right|s\in S\}.$$

• To show:	1. $p$ does not divide $\mathrm{Card}\left(𝒮\right)$.
  	2. There exists $S\in 𝒮$ such that $p$ does not divide the order $\mathrm{Card}\left(GS\right)$ of the orbit $GS$.
  	3. If $S$ is as in (b) and $G_S$ is the stabilizer of $S$ then $\mathrm{Card}\left(G_S\right)\ge p^a$.
  	4. If $S$ is as in (b) then $\mathrm{Card}\left(G_S\right)\le p^a$.

• This will show that $G_S$ is a subgroup of order $p^a$.
1. $\mathrm{Card}\left(𝒮\right)$ is the number of subsets of $G$ with $p^a$ elements. $$\mathrm{Card}\left(𝒮\right)=\left(\begin{array}{c}\left|G\right|\\ p^a\end{array}\right)=\left(\begin{array}{c}{p}^{a}b\\ p^a\end{array}\right)=\frac{p^{a}b(p^{a}b-1)\cdots (p^{a}b-j)\cdots (p^{a}b-p^a+1)}{p^a(p^a-1)\cdots (p^a-j)\cdots 1}.$$
• Suppose $p^i$ divides $p^a-j$. Then $p^{i}k=p^{a}b-j$ for some $k$. So $j=p^{a}b-p^{j}k$ and $$p^a-j=p^a-p^{a}b+p^{i}k=p^i(p^{a-i}-p^{a-i}b+k).$$
• So $p^i$ divides $p^a-j$.
• This shows that any factors of $p$ in the numerator of $\mathrm{Card}\left(𝒮\right)=\left(\begin{array}{c}{p}^a\\ b\end{array}\right)$ of $p$ in the denominator.
• So $p$ does not divide $\mathrm{Card}\left(𝒮\right)$.
2. It follows from Proposition xxx that $$\mathrm{Card}\left(𝒮\right)=\sum _{\genfrac{}{}{0ex}{}{\mathrm{distinct}}{\mathrm{orbits}}}\mathrm{Card}\left(GS\right)$$ where the sum is over the distinct orbits $GS$ of $G$ acting on $𝒮$.
• Since $p$ does not divide $\mathrm{Card}\left(𝒮\right)$ we have that $p$ does not divide $\mathrm{Card}\left(GS\right)$ for some $S\in 𝒮$.
3. Fix $S\in 𝒮$ such that $p$ does not divide $\mathrm{Card}\left(GS\right)$.
• By Proposition xxx, $p^{ab}=\left|G\right|=\left|G_S\right|\mathrm{Card}\left(GS\right),$ where $GS$ is the stabilizer of $S\in 𝒮$.
• Since $p$ does not divide $\mathrm{Card}\left(GS\right)$ we must have $\left|G_S\right|=p^{a}k$ for some $k\ge 1$.
• So $\left|G_S\right|\ge p^a$.
4. Let $s\in S\subseteq G$. Then $G_{S}s\subseteq S$, since $G_{S}S=S.$
• Since all cosets of $G_S$ are the same size (Proposition xxx), $$\mathrm{Card}\left(G_{S}s\right)=\mathrm{Card}\left(G_S\right).$$
• Since $G_{S}s\subseteq S$, $\mathrm{Card}\left(G_{S}s\right)\le p^a.$
• So $\mathrm{Card}\left(G_S\right)\le p^a$.
• So $\mathrm{Card}\left(G_S\right)=p^a$.
• So $G$ contains a group of order $p^a$.

$\square$

(Second Sylow theorem) All the $p$-Sylow subgroups of $G$ are conjugates of each other.

Proof.

• Let $P$ be a $p$-Sylow subgroup of $G$.
• Let $H$ be another $p$-Sylow subgroup of $G$.
• To show: $H\subseteq gP{g}^{-1}$ for somme $g\in G$.
1. First we find the right $g\in G$.
• $H$ acts on $G/P$ by left multiplication, $$\begin{array}{ccc}H\times G/P& ⟶& G/P\\ (h,g_{1}P)& ⟼& h{g}_{1}P\end{array}$$
• The orbits are $H{g}_{1}P$, $g_1\in G$.
• By Proposition xxx, $$\mathrm{Card}\left(H{g}_{1}P\right)\text{divides}\mathrm{Card}\left(H\right)=p^a.$$
• So either $\mathrm{Card}\left(H{g}_{1}P\right)=1$ or $p$ divides $\mathrm{Card}\left(H{g}_{1}P\right)$.
• By Proposition xxx, $$b=\frac{p^{a}b}{p^a}=\frac{\left|G\right|}{\left|P\right|}=\left|\frac{G}{P}\right|=\sum _{\genfrac{}{}{0ex}{}{\mathrm{distinct}}{\mathrm{orbits}}}\mathrm{Card}\left(H{g}_{1}P\right).$$
• Since $p$ does not divide $b$, there is an orbit $HgP$, $g\in G$, such that $\mathrm{Card}\left(HgP\right)=1$.
2. Now show $H\subseteq gP{g}^{-1}$.
• Let $h\in H$. Since $\mathrm{Card}\left(HgP\right)=1$, $$HgP=gP.$$
• So $hg=gp$, for some $p\in P$.
• So $h=gp{g}^{-1}\in gP{g}^{-1}$.
3. Since $H\subseteq gP{g}^{-1}$ and $\mathrm{Card}\left(H\right)=\mathrm{Card}\left(gP{g}^{-1}\right)<\infty$, we have $H=gP{g}^{-1}$.
• So $H$ is a conjugate of $P$.

$\square$

(Third Sylow theorem) The number of $p$-Sylow subgroups of $G$ is $1\text{mod}p$.

Proof.

• Let $𝒮$ be the set of all $p$-Sylow subgroups of $G$.
• Let $P$ be a $p$-Sylow subgroup of $G$.
• $P$ acts on $𝒮$ by conjugation, $$\begin{array}{ccc}P\times S& ⟶& 𝒮\\ (p,Q)& ⟼& pQ{p}^{-1}\end{array}$$
• For each $Q\in 𝒮$ let $P*Q$ denote the orbit of $Q$ under this action.

• To show:	1. $\mathrm{Card}\left(𝒮\right)=\sum _{\genfrac{}{}{0ex}{}{\mathrm{distinct}}{\mathrm{orbits}}}\mathrm{Card}(P*Q)$
  	2. Either     $\mathrm{Card}(P*Q)=0\text{mod}p$     or     $$ \mathrm{Card}(P*Q)=1$.$
  	3. If $\mathrm{Card}(P*Q)=1$ then $Q=P$, so there is only one orbit with $\mathrm{Card}(P*Q)=1$.

1. This follows from Proposition xxx.
2. By Proposition xxx, $\mathrm{Card}(P*Q)$ divides $\left|P\right|=p^a$.
• So either $\mathrm{Card}(P*Q)=1$ or $p$ divides $\mathrm{Card}(P*Q)$.
3. Assume $\mathrm{Card}(P*Q)=1$.
• To show: $P=Q$.
• If $\mathrm{Card}(P*Q)=1$ then $pQ{p}^{-1}=Q$, for all $p\in P$.
• So, for every $p\in P$, $p\in N_Q$, the normalizer of $Q$.
• So $P\subseteq N_Q$.
• We know $Q\subseteq N_Q$ also. So $P$ and $Q$ are both $p$-Sylow subgroups of $N_Q$.
• So, by Theorem xxx, $P$ and $Q$ are conjugates in $N_Q$.
• So there exists some $n\in N_Q$ such that $nQ{n}^{-1}=P$.
• But, by Proposition xxx, $Q$ is normal in $N_Q$, so $nQ{n}^{-1}=Q$.
• So $P=Q$.
• So $\mathrm{Card}\left(𝒮\right)=1\text{mod}p$.

$\square$

Notes and References

These notes are a retyping of an old tex file of Arun Ram dated 13 January 1992.

References

[Ar] M. Artin, Algebra, ????, Prentice-Hall ???.

[BJN] P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic abstract algebra, Second Edition, Cambridge University Press 1994.

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1992-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre ?: ??????????? MR?????.

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