First Sylow Theorem

Theorem

Let $p$ be a prime number, and let $G$ be a group such that $\left\vert{G}\right\vert = k p^n$ where $p \nmid k$.

Then $G$ has at least one Sylow $p$-subgroup.

Proof

Let $\left\vert{G}\right\vert = k p^n$ such that $p \nmid k$.

Let $\mathbb S = \left\{{S \subseteq G: \left\vert{S}\right\vert = p^n}\right\}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.

Let $N = \left\vert{\mathbb S}\right\vert$.

Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.

From Cardinality of Set of Subsets, this is given by:

$\displaystyle N = \binom {p^n k} {p^n} = \frac {\left({p^n k}\right) \left({p^n k - 1}\right) \cdots \left({p^n k - i}\right) \cdots \left({p^n k - p^n + 1}\right)} {\left({p^n}\right) \left({p^n - 1}\right) \cdots \left({p^n - i}\right) \cdots \left({1}\right)}$

From Binomial Coefficient involving Power of Prime, $\displaystyle \binom {p^n k} {p^n} \equiv k \left({\bmod\, p}\right)$.

Thus, $N \equiv k \left({\bmod\, p}\right)$.

• Now let $G$ act on $\mathbb S$ by the rule: $\forall S \in \mathbb S: g * S = g S = \left\{{x \in G: x = g s: s \in S}\right\}$.

That is, $g * S$ is the left coset of $S$ by $g$. From Group Action on Sets with k Elements, this is a group action.

• Now, let $\mathbb S$ have $r$ orbits under this action.

Each orbit is an equivalence class, and therefore the orbits partition $\mathbb S$.

Let these orbits be represented by $\left\{{S_1, S_2, \ldots, S_r}\right\}$, so that:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \mathbb S\)	\(=\)	\(\displaystyle \operatorname{Orb} \left({S_1}\right) \cup \operatorname{Orb} \left({S_2}\right) \cup \ldots \cup \operatorname{Orb} \left({S_r}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert{\mathbb S}\right\vert\)	\(=\)	\(\displaystyle \left\vert{\operatorname{Orb} \left({S_1}\right)}\right\vert \cup \left\vert{\operatorname{Orb} \left({S_2}\right)}\right\vert \cup \ldots \left\vert{\operatorname{Orb} \left({S_r}\right)}\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

If each orbit had length divisible by $p$, then $p \backslash N$. But this can not be the case, because, as we have seen, $N \equiv k \left({\bmod\, p}\right)$.

So at least one orbit has length which is not divisible by $p$.

Let $S \in \left\{{S_1, S_2, \ldots, S_r}\right\}$ be such that $\left\vert{\operatorname{Orb} \left({S}\right)}\right\vert = m: p \nmid m$.

It follows from Group Action on Prime Power Order Subset that $\operatorname{Stab} \left({S}\right) s = S$, and thus $\left\vert{\operatorname{Stab} \left({S}\right)}\right\vert = p^n$.

From Stabilizer is Subgroup, $\operatorname{Stab} \left({S}\right) \le G$.

Thus $\operatorname{Stab} \left({S}\right)$ is the subgroup of $G$ with $p^n$ elements of which we wanted to prove the existence.

$\blacksquare$

Source of Name

This entry was named for Peter Ludwig Mejdell Sylow.

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 6.5$: Example $121$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $25.18$
• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 56$
• John F. Humphreys: A Course in Group Theory (1996): $\S 11$: Theorem $11.4$