## $p$-groups

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

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

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, ${1}⊆ N1⊆ N2⊆ ⋯⊆ Na-1⊆ G$ of normal subgroups of $G$, such that $|{N}_{i}|={p}^{i}$.

## Sylow theorems

Let $G$ be a finite group of order ${p}^{a}b$ where $p$ is prime, $a\in {ℤ}_{>0}$ 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 $1\phantom{\rule{0.2em}{0ex}}\text{mod}\phantom{\rule{0.2em}{0ex}}p$.

## 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, $K\simeq ℤ/7ℤ$.

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 $ℤ/3ℤ$. 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 $HK=\left\{1\right\}$, $|HK|=|H||K|=3\cdot 7=21=|G|$. So $G=HK$. Then Theorem ??? gives that $G\simeq H×K\simeq ℤ/7ℤ×ℤ/3ℤ$.

Case 2. 4 3-Sylow subgroups
Let $H$ be one of the 3-Sylow subgroups of $G$. Once again, $H\simeq ℤ/3ℤ$ 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 ${i}^{3}=1\phantom{\rule{0.2em}{0ex}}\text{mod}\phantom{\rule{0.2em}{0ex}}7$. 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 $x↦{x}^{2}$, $y↦y$ will be an isomorphism of the two semidirect products. So in this case $G\simeq ℤ/3ℤ{×}_{\theta }ℤ/7ℤ$ and any two such semidirect products are isomorphic.

#### Groups of order $\le 10$

 Order 1 $\left\{1\right\}$ Order 2 $ℤ/2ℤ$ Order 3 $ℤ/3ℤ$ Order 4 $ℤ/4ℤ$, $ℤ/2ℤ×ℤ/2ℤ$ Order 5 $ℤ/5ℤ$ Order 6 $ℤ/6ℤ$, $ℤ/2ℤ×ℤ/3ℤ$ ${S}_{3}\simeq {D}_{3}$ Order 7 $ℤ/7ℤ$ Order 8 $ℤ/8ℤ$, $ℤ/4ℤ×ℤ/2ℤ$, $ℤ/2ℤ×ℤ/2ℤ×ℤ/2ℤ$, the dihedral group ${D}_{4}$, the quaternion group ${Q}_{8}$, Order 9 $ℤ/9ℤ$, $ℤ/3ℤ×ℤ/3ℤ$ Order 10 $ℤ/10ℤ\simeq ℤ/5ℤ×ℤ/2ℤ$ $ℤ/5ℤ{×}_{\theta }ℤ/2ℤ$, 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 $|G|={p}^{a}$ we know that $\mathrm{order}\left(x\right)={p}^{b}$ for some $0.
• Then, since $\mathrm{order}\left(x\right)={p}^{b}$, $xpb-1 ≠1, and ( xpb-1 ) p = x pb-1p = xpb =1.$
• 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: $\mathrm{Card}\left(Z\left(G\right)\right)\ne 1$.
1. $p$ divides $|G|$ since $|G|={p}^{a}$.
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 $|G|={p}^{a}$.
• If $\mathrm{Card}\left({𝒞}_{g}\right)>1$ then $p$ divides $\mathrm{Card}\left({𝒞}_{g}\right)$.
3. The class equation is $|G| = |Z(G)| + ∑|𝒞g| >1 Card(𝒞g) ,$ where the sum is over all distinct conjugacy classes such that $\mathrm{Card}\left({𝒞}_{g}\right)>1$.
• Since $p$ divides $|G|$ and $p$ divides every term in the sum we cannot have $|Z\left(G\right)|=1$.
• So $|Z\left(G\right)|\ne 1$.
$\square$

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 ${p}^{2}$, $|Z\left(G\right)|={p}^{2}$.
• By Proposition ???, we know that $|Z\left(G\right)|$ divides $|G|={p}^{2}$.
• Case 1. $|Z\left(G\right)|=1$.
• By Proposition ???, $|Z\left(G\right)|\ne 1$.
• Case 2. $|Z\left(G\right)|=p$.
• 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 $|G/Z|= p2/p =p.$
• So, by Proposition ???, $G/Z$ is cyclic.
• Since $x\notin Z$, $Z\ne xZ$ and so $xZ$ generates $G/Z$, $G/Z={ Z,xZ,x2Z ,…, xp-1Z}.$
• Let $g\in G$. Then there exists $k\in {ℤ}_{\ge 0}$ such that $gZ={x}^{k}Z$ and $k\le p-1$.
• So there exists $z\in Z$ such that $g={x}^{k}z$.
• Then $xg=xxkz = xkxz= xkzx=gx.$
• So $x\in Z\left(G\right)$.
• This is a contradiction to $x\notin Z\left(G\right)$.
• So $|Z\left(G\right)|\ne p$.
• So $|Z\left(G\right)|={p}^{2}$.
• So $G$ is abelian.
$\square$

If $G$ is a $p$-group of order ${p}^{a}$, then there exists a chain, ${1}⊆ N1⊆ N2⊆ ⋯⊆ Na-1⊆ G$ of normal subgroups of $G$, such that $|{N}_{i}|={p}^{i}$.

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 $G×𝒮 ⟶ 𝒮 (g,S) ⟼ gS where gS={gs | s∈S}.$
•  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. $Card(𝒮) = ( |G| pa ) = ( pab pa ) = pab (pab-1) ⋯ (pab-j) ⋯ (pab- pa+1) pa (pa-1) ⋯ (pa-j) ⋯ 1 .$
• 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 $pa-j = pa-pab +pik = pi( pa-i - pa-i b+k).$
• 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 $Card(𝒮) = ∑ distinctorbits Card(GS)$ 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, ${p}^{ab}=|G|=|{G}_{S}|\mathrm{Card}\left(GS\right),$ where $GS$ is the stabilizer of $S\in 𝒮$.
• Since $p$ does not divide $\mathrm{Card}\left(GS\right)$ we must have $|{G}_{S}|={p}^{a}k$ for some $k\ge 1$.
• So $|{G}_{S}|\ge {p}^{a}$.
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), $Card(GSs) = Card(GS).$
• 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, $H×G/P ⟶ G/P (h,g1P) ⟼ hg1P$
• The orbits are $H{g}_{1}P$, ${g}_{1}\in G$.
• By Proposition xxx, $Card(Hg1P) divides Card(H) =pa.$
• 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, $b= pab pa = |G| |P| = |GP| = ∑ distinctorbits Card(Hg1P) .$
• 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 $1\phantom{\rule{0.2em}{0ex}}\text{mod}\phantom{\rule{0.2em}{0ex}}p$.

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, $P×S ⟶ 𝒮 (p,Q) ⟼ pQp-1$
• For each $Q\in 𝒮$ let $P*Q$ denote the orbit of $Q$ under this action.
•  To show: 1. $\mathrm{Card}\left(𝒮\right)=\sum _{\genfrac{}{}{0}{}{\mathrm{distinct}}{\mathrm{orbits}}}\mathrm{Card}\left(P*Q\right)$ 2. Either     $\mathrm{Card}\left(P*Q\right)=0\phantom{\rule{0.2em}{0ex}}\text{mod}\phantom{\rule{0.2em}{0ex}}p$     or     $\mathrm{Card}\left(P*Q\right)=1.$ 3. If $\mathrm{Card}\left(P*Q\right)=1$ then $Q=P$, so there is only one orbit with $\mathrm{Card}\left(P*Q\right)=1$.
1. This follows from Proposition xxx.
2. By Proposition xxx, $\mathrm{Card}\left(P*Q\right)$ divides $|P|={p}^{a}$.
• So either $\mathrm{Card}\left(P*Q\right)=1$ or $p$ divides $\mathrm{Card}\left(P*Q\right)$.
3. Assume $\mathrm{Card}\left(P*Q\right)=1$.
• To show: $P=Q$.
• If $\mathrm{Card}\left(P*Q\right)=1$ 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 $\mathrm{Card}\left(𝒮\right)=1\phantom{\rule{0.2em}{0ex}}\text{mod}\phantom{\rule{0.2em}{0ex}}p$.
$\square$

## 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?????.