First Sylow Theorem

This article was Featured Proof between 23 November 2008 and 1 December 2008.

Theorem

Let $p$ be a prime number.

Let $G$ be a group such that:

$\order G = k p^n$

where:

$\order G$ denotes the order of $G$

$p$ is not a divisor of $k$.

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

Corollary

Let $p$ be a prime number.

Let $G$ be a group.

Let:

$p^n \divides \order G$

where:

$\order G$ denotes the order of $G$

$n$ is a positive integer.

Then $G$ has at least one subgroup of order $p$.

Proof 1

Let $\order G = k p^n$ such that $p \nmid k$.

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

Let $N = \order {\mathbb S}$.

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:

$N = \dbinom {p^n k} {p^n} = \dfrac {\paren {p^n k} \paren {p^n k - 1} \cdots \paren {p^n k - i} \cdots \paren {p^n k - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$

From Binomial Coefficient involving Power of Prime:

$\dbinom {p^n k} {p^n} \equiv k \pmod p$

Thus:

$N \equiv k \pmod p$

Now let $G$ act on $\mathbb S$ by the rule:

$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$

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.

From Set of Orbits forms Partition, the orbits partition $\mathbb S$.

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

$\ds \mathbb S$

$=$

$\ds \Orb {S_1} \cup \Orb {S_2} \cup \ldots \cup \Orb {S_r}$

$\ds \size {\mathbb S}$

$=$

$\ds \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }$

If each orbit had length divisible by $p$, then $p \divides N$.

But this can not be the case, because, as we have seen:

$N \equiv k \pmod p$

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

Let $S \in \set {S_1, S_2, \ldots, S_r}$ be such that $\size {\Orb S)} = m: p \nmid m$.

Let $s \in S$.

It follows from Group Action on Prime Power Order Subset that:

$\Stab S s = S$

and so:

$\size {\Stab S} = \size S = p^n$

From Stabilizer is Subgroup:

$\Stab S \le G$

Thus $\Stab S$ is the subgroup of $G$ with $p^n$ elements of which we wanted to prove the existence.

$\blacksquare$

Proof 2

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

Now let $G$ act on $\Bbb S$ by the rule:

$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$

From Set of Orbits forms Partition, the orbits partition $\mathbb S$.

Let these orbits be represented by $\set {S_1, S_2, \ldots, S_n}$, so that:

$(1): \quad \size {\mathbb S} = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }$

Thus each $\Orb {S_i}$ is the orbit under $*$ of some $S_i$ whose order is $p^n$.

By the Orbit-Stabilizer Theorem:

$(2): \quad \order {\Orb {S_i} } = \dfrac {\order G} {\order {\Stab {S_i} } }$ for all $i \in \set {1, 2, \ldots, n}$

where $\Stab {S_i}$ is the stabilizer of $S_i$ under $*$.

Let $s \in S_i$ and $x \in \Stab {S_i}$.

Then $s x \in S_i$ because $S_i x = S_i$.

Let $s$ be fixed while $x$ runs over $\Stab {S_i}$.

Then:

$\forall s \in S_i: s \Stab {S_i} \subseteq S_i$

Therefore $S_i$ is the union of all the left cosets $s \Stab {S_i}$.

These left cosets may not all be distinct.

Let $r_i$ be the number of distinct left cosets of $s \Stab {S_i}$.

Then:

$\card {S_i} = r_i \card {\Stab {S_i} }$

because:

distinct left cosets are disjoint

all have the same cardinality $\card {\Stab {S_i} }$.

Thus:

$(3): \quad \card {\Stab {S_i} } \divides \card {S_i} = p^n$

Because the only divisors of $p^n$ are smaller powers of $p$:

$\card {\Stab {S_i} } = p^{n_i}$

for some $n_i \le n$, for $i = 1, 2, \ldots, r$.

We can write:

$g = \order G = k p^n$

where $p$ does not divide $k$.

From $(2)$ and $(3)$:

$\card {S_i} = k p^{d_i}$

where $d_i = n - n_i$ for $i = 1, 2, \ldots, r$.

From Cardinality of Set of Subsets:

$\card {\Bbb S} = \dbinom {k p^n} {p_n}$

We have from Binomial Coefficient involving Power of Prime that:

$\dbinom {k p^n} {p_n} \equiv k \pmod p$

But $p$ does not divide $k$.

Hence $p$ does not divide $\card {\Bbb S}$.

Hence by $(1)$ and $(4)$:

$p \nmid k \paren {p^{d_1} + p^{d_2} + \cdots p^{d_r} }$

If $d_i > 0$ for all $i = 1, 2, \ldots, r$, then $p$ would divide this integer.

So there must be some $i$ such that $d_i = 0$.

Then, for this $i$:

$n_i = n$

That is:

$\card {\Stab {S_i} } = p^n$

Thus $\Stab {S_i}$ is the subgroup of $G$ of order $p_n$ that was asserted.

$\blacksquare$

Examples

Alternating Group on $4$ Letters

The Alternating Group on 4 Letters $A_4$ is of order $12 = 2^2 \times 3$.

Thus the First Sylow Theorem tells us that $A_4$ has:

at least one subgroup of order $4$

at least one subgroup of order $2$

at least one subgroup of order $3$

These it has.

But it has no subgroup of order $6$, although $6 \divides 12$.

Also see

Sylow Theorems

Note

By Orbits of Group Action on Sets with Power of Prime Size, it was clear that $k \divides \size {\Orb S}$.

However here, since it is established that $\size {\Stab S} = p^n$, and by Orbit-Stabilizer Theorem, we also have $k = \size {\Orb S}$.

Source of Name

This entry was named for Peter Ludwig Mejdell Sylow.

Historical Note

When cracking open the structure of a group, it is a useful plan to start with investigating the prime subgroups.

The Sylow Theorems are a set of results which provide us with just the sort of information we need.

Ludwig Sylow was a Norwegian mathematician who established some important facts on this subject.

He published what are now referred to as the Sylow Theorems in $1872$.

The name is pronounced something like Soolof.

There is no standard numbering for the Sylow Theorems.

Different authors use different labellings.

Therefore, the nomenclature as defined on $\mathsf{Pr} \infty \mathsf{fWiki}$ is to a greater or lesser extent arbitrary.