LEC ALG2 11

Sylow’s theorems

Note that the following lemma is a special case of Cauchy’s theorem, which we have already proved.

Lemma 100.1(Lemma).

If $G$ is a finite abelian group and $p$ is a prime dividing $|G|$, then $G$ has an element of order $p$.

Alternate proof using induction:

Proof.

Fix a prime $p$. Complete induction on $|G|$. Vacuously true for $|G|<p$. If $|G|=p$, $G$ is cyclic, and has an element of order $p$.

For $|G|>p$, pick $x\ne e$ in $G$. If $|x|=pn$, $x^n$ is an element of order $p$.

Thus, assume $p$ does not divide $|x|$. Since $G$ is abelian, $⟨x⟩◃G$. Now, $|G|=|G/⟨x⟩||⟨x⟩|$. Since $p$ does not divide $|⟨x⟩|$, it must divide $|G/⟨x⟩|$. Since $x\ne e$, $|G/⟨x⟩|<|G|$. By the induction hypothesis, $G/⟨x⟩$ has an element $y⟨x⟩$ of order $p$. Note that $y\notin ⟨x⟩$, since that would make $y⟨x⟩$ the identity in $G/⟨x⟩$. Now, since $y^p⟨x⟩=⟨x⟩$, $y^p\in ⟨x⟩$. Clearly, then, $⟨y⟩\ne ⟨y^p⟩$. This forces $|y^p|<|y|$, which in turn implies $p$ divides $|y|$. This takes us back to the previous case, so there exists a power of $y$ which has order $p$.□

Lemma 100.2(Lemma).

Let $P\in {\text{Syl}}_p(G)$. If $Q$ is any p-subgroup of $G$, then $Q\cap N_G(P)=Q\cap P$.

Proof.

Recall that$N_G(P)\le G$, and $P\le N_G(P)$. Also recall that the intersection of two subgroups is a subgroup. It follows then that $Q\cap P\le Q\cap N_G(P)$. It remains to show that $Q\cap N_G(P)\le Q\cap P$. This is equivalent to showing $Q\cap N_G(P)\le P$. Let $H\equiv Q\cap N_G(P)$. Note that $H\le N_G(P)$, so $PH\le G$. Also, $H\le PH$ and $P\le PH$. If we show that $PH\le P$, we will be done. We know that

$$\begin{array}{r}|PH|=\frac{|P||H|}{|P\cap H|}.\end{array}$$

Now, $H\le Q$, so $p$ divides $|H|$. Similarly, $p$ divides $|P\cap H|$. So, $PH$ is a p-subgroup. Moreover, $P\le PH$ and $P$ being a Sylow p-subgroup forces $P=PH$. Thus, $H\le P$.□

Definition 100.3(Definition).

Let $G$ be a group and $p$ be a prime.

1. A group of order $p^{\alpha }$ for some $\alpha ⩾1$ is called a $p$-group. Subgroups of $G$ which are $p$-groups are called $p$-subgroups.

2. If $G$ is a group of order $p^{\alpha }m$, where $p\nmid m$, then a subgroup of order $p^{\alpha }$ is called a Sylow $p$-subgroup of $G$.

3. The set of Sylow $p$-subgroups of $G$ is denoted by ${\text{Syl}}_p(G)$ and the number of Sylow $p$-subgroups of $G$ is denoted by $n_p(G)$ (or just $n_p$ when $G$ is clear from context).

Theorem 100.4(Theorem).

Let $G$ be a group of order $p^{\alpha }m$, where $p$ is a prime not dividing $m$. Then,

1. Sylow p-subgroups of $G$ exist.
2. If $P$ is a Sylow p-subgroup of $G$ and $Q$ is any p-subgroup of $G$, then there exists $g\in G$ such that $Q\le gP{g}^{-1}$, that is, $Q$ is contained in some conjugate of $P$. In particular, any two Sylow p-subgroups of $G$ are conjugate in $G$.
3. The number of Sylow p-subgroups of $G$ is of the form $1+kp$. Further, $n_p$ is the index of the normalizer $N_G(P)$ in $G$, for any Sylow p-subgroup $P$, hence, $n_p$ divides $m$.

Proof of 1
Induction on $|G|$. If $|G|=1$, there is nothing to prove. Assume inductively the existence of sylow p-subgroups for all groups of order less than $|G|$.

Let $G=p^{\alpha }m$, $p$ does not divide $m$. If $p$ divides $|Z(G)|$, since $Z(G)$ is abelian, it has an element $x$ of order $p$. Since $⟨x⟩\le Z(G)$, $⟨x⟩◃G$. It follows that $G/⟨x⟩$ is a group, which has order $p^{\alpha -1}m$. Form the inductive hypothesis, $G/⟨x⟩$ must have a sylow p-subgroup of order $p^{\alpha -1}$. From the correspondence theorem, this subgroup corresponds to a subgroup of $G$ of order $p^{\alpha }$, completing the proof for this case.

Consider the case when $p$ does not divide $|Z(G)|$. The class equation of $G$ is

$$|G|=|Z(G)|+\sum _{i=1}^n[G:C_G(g_i)],$$

where $g_i$, $1\le i\le n$ are representatives of the conjugacy classes of $G$. There must exist $g_i$ such that $p$ does not divide $[G:C_G(g_i)]$. We know form the orbit-stabilizer theorem that $|C_G(x)||{\theta }_x|=|G|$. In this context, the equation would read $|C_G(g_i)||G:C_G(g_i)|=|G|$. It follows that $p^{\alpha }$ must divide $|C_G(g_i)|$. $g_i\notin Z(G)$, so $|C_G(g_i)|\le |G|$. From the inductive hypothesis, $C_G(g_i)$ must have a sylow p-subgroup of order $p^{\alpha }$, which is of course a subgroup of $G$, completing the proof.

Proof of 2

Let $P$ be a Sylow-$p$ subgroup. Let $S=\{P_1,P_2,\dots ,P_r\}$ be the set of all conjugates of $P$. Let $Q$ be any p-subgroup of $G$.