First Sylow Theorem/Corollary
Corollary to First Sylow Theorem
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 $\order G = k p^r$ where $p \nmid k$.
From the First Sylow Theorem, $G$ has a subgroup $S$ of order $p^r$.
From (need to find it), $S$ itself has subgroups of order $p^n$ for all $n \in \set {1, 2, \ldots, r}$.
$\blacksquare$
Proof 2
This result can also be proved directly:
Let $\order G = m p^n$.
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:
- $N = \dbinom {p^n m} {p^n} = \dfrac {\paren {p^n m} \paren {p^n m - 1} \cdots \paren {p^n m - i} \cdots \paren {p^n m - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$
For $0 < i < p^n$, the highest power of $p$ which divides $m p^n - i$ is the same as that dividing $p^n - i$.
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Thus the power of $p$ in the numerator and denominator of $N$ cancel out, except those of $m$.
Thus $m$ and $N$ have the same number of divisors which are power of $p$.
Let $p^r$ be the largest power of $p$ which divides $m$ and $N$.
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.
If every orbit under $*$ were divisible by $p^{r - 1}$, then $p^{r + 1} \divides N$.
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But we have stated that $p^r$ is the largest power of $p$ which divides $N$.
Therefore there is at least one orbit under $*$ which can be represented as:
- $\set {S_1, S_2, \ldots, S_k}$
for which $p^{r + 1} \nmid k$.
Let $H = \Stab {S_1}$ denote the stabilizer of $S_1$.
From the Orbit-Stabilizer Theorem, we have that:
- $k = \index G H = \dfrac {\order g} {\order H}$
As $p^r \divides m$, we have:
- $p^{n + r} \divides p^n m$
But:
- $p^n m = \order G = k \order H$
and so:
- $p^{n + r} \divides k \order H$
But we also have that:
- $p^{r + 1} \nmid k$
which means:
- $p^n \divides \order H$
and that:
- $p^n \le \order H$
Now because $H = \Stab {S_1}$, we haveL
- $H g \subseteq S_1$
for all $g \in \S_1$
Thus:
- $\order H = \card {H g} \le \card {S_1} = p^n$
So we have $p^n \le \order H$ and $\order H \le p^n$ which means:
- $\order H = p^n$
and so $H$ is the subgroup of $G$ whose existence we were to demonstrate.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 44$. Some consequences of Lagrange's Theorem