Sylow Theorems

theoremgroup-theory

Theorem 1

If $|G|=p^n\cdot m$, there exists $H\le G$ with $|H|=p^n$ for $p\ge 2$ prime, $n\ge 0$, $m\ge 1$. (Generalizes Cauchy’s Theorem)

Theorem 2

Every p-subgroup $H\le G$ is contained $H\le P\le G$, where $P$ is a Sylow P-subgroup. Every Sylow p-subgroup $P$ has the same size, $p^n$. If $P,P^{\prime }$ are Sylow p-subgroups, then there exists $g\in G$ such that $P^{\prime }=gP{g}^{-1}$.

Theorem 3

$n_p=\#$ Sylow p-subgroups, then

• $n_p\mid m$
• $n_p\equiv 1\mathrm{mod}p$
• $n_p=[G:N_G(P)]$ $P$ any Sylow p-subgroup $N_G(P)=$ normalizer of $P$ in $G$.

Theorem

Every group with $pq$ elements ($p,q$ prime) is cyclic isomorphic to $({\mathbb{Z}}_{pq},+)$.

Proof

Cauchy’s theorem implies there exists $x\in G$ such that ${\mathrm{ord}}_{}(x)=p$, and $y\in G$ such that ${\mathrm{ord}}_{}(y)=q$.

$H=⟨x⟩$ and $K=⟨y⟩$.

$H\cap K\le H$, $H\cap K\le K$ so by Lagrange’s theorem,$|H\cap K|\mid |H|=p$, and $|H\cap K|\mid |K|=q$, so $|H\cap K|=1$ and $H\cap K=\{e\}$.

Claim: $H,K$ are normal in $G$. $p\mid pq$, $p<q⟹p\ne q$, so $p^2\nmid pq$. Thus, $H$ is a Sylow $p$-subgroup, and $K$ is a Sylow $q$-subgroup.

Theorem 2 implies $n_p=\#$ conjugates of $P$ in $G$, $P⊴G⟺n_p=1$. We want $n_p=n_q=1$.

$n_p\mid q\stackrel{q\text{ prime}}{⟹}{n}_p=1$ or $n_p=q$, and $n_p\equiv 1\mathrm{mod}p$, so by assumption $q\not\equiv 1\mathrm{mod}p$, we have that $n_p=1⟹H$ (Sylow p-subgroup) is normal.

$n_q\mid p\stackrel{p\text{ prime}}{⟹}{n}_q=1$ or $n_q=p$, and $n_q\equiv 1\mathrm{mod}q$,

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