Fourth Sylow Theorem/Proof 1

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Theorem

The number of Sylow $p$-subgroups of a finite group is congruent to $1 \pmod p$.


Proof

Let $G$ be a finite group such that $\order G = k p^n$ where $p \nmid k$ and $n > 0$.

Let $r$ be the number of Sylow $p$-subgroups of $G$.


We want to show that $r \equiv 1 \pmod p$.


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

From the reasoning in the First Sylow Theorem, we have:

$\size {\mathbb S} = \dbinom {p^n k} {p^n}$


Let $G$ act on $\mathbb S$ by the group action defined in Group Action on Sets with k Elements::

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

From Orbits of Group Action on Sets with Power of Prime Size:

there are exactly as many Sylow $p$-subgroups as there are orbits whose length is not divisible by $p$.

Also by Orbits of Group Action on Sets with Power of Prime Size:

all the terms in the Partition Equation are divisible by $k$, perhaps also divisible by $p$.


We can write the Partition Equation as:

$\size {\mathbb S} = \size {\Orb {S_1} } + \size {\Orb {S_2} } + \cdots + \size {\Orb {S_r} } + \size {\Orb {S_{r + 1} } } + \cdots + \size {\Orb {S_s} }$

where the first $r$ terms are the orbits containing the Sylow $p$-subgroups:

$\Stab {S_i}$

For each of these:

$\order G = \size {\Orb {S_i} } \times \size {\Stab {S_i} } = p^n \size {\Orb {S_i} }$

Thus:

$\size {\Orb {S_i} } = k$

for $1 \le i \le r$.

Each of the rest of the orbits are divisible by both $p$ and $k$, as we have seen.

So:

$\size {\mathbb S} = k r + m p k$

where:

the first term corresponds to the $r$ orbits containing the Sylow $p$-subgroups
the second term corresponds to all the rest of the orbits
$m$ is some unspecified integer.

That is, there exists some integer $m$ such that:

$\size {\mathbb S} = \dbinom {p^n k} {p^n} = k r + m p k$

Now this of course applies to the special case of the cyclic group $C_{p^n k}$.

In this case, there is exactly one subgroup for each divisor of $p^n k$.

In particular, there is exactly one subgroup of order $p^n$.

Hence, in this case:

$r = 1$


So we have an integer $m'$ such that $\dbinom {p^n k} {p^n} = k + m' p k$.

We can now equate these expressions:

\(\ds k r + m p k\) \(=\) \(\ds k + m' p k\)
\(\ds \leadsto \ \ \) \(\ds r + m p\) \(=\) \(\ds 1 + m' p\) dividing by $k$
\(\ds \leadsto \ \ \) \(\ds r - 1\) \(=\) \(\ds p \paren {m' - m}\)
\(\ds \leadsto \ \ \) \(\ds r - 1\) \(\equiv\) \(\ds 0 \pmod p\)
\(\ds \leadsto \ \ \) \(\ds r\) \(\equiv\) \(\ds 1 \pmod p\)


and the proof is complete.

$\blacksquare$


Sources

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