Home > Combinatorics, Group Theory, Higher Algebra, Linear Algebra > Existence of Sylow subgroups

Existence of Sylow subgroups


Let {G} be a finite group such that {|G|=mp^a} where {p} is a prime number, {a \geq 1} and {\gcd(m,p)=1}. Then there exists a subgroup {H \leq G} such that {|H|=p^a}. (such a subgroup is called a Sylow subgroup).

Proof:

1. Let {\Bbb{F}_p} be the finite field with {p} elements. For {n \geq 1} consider {G= GL(n,\Bbb{F}_p)} the space of invertible morphisms {u : \Bbb{F}_p^n \rightarrow \Bbb{F}_p^n} (or the space of {n \times n} matrices with elements in {\Bbb{F}_p} which are invertible). Then {G} is a finite groupe with cardinal equal to

\displaystyle |G| = (p^n-1)(p^n-p)...(p^n-p^{n-1}).

To see this note that every invertible matrix {A \in G} corresponds to a basis of {\Bbb{F}_p^n}. First we see that {G} is finite because there are at most {p^{n^2}} matrices with elements in {\Bbb{F}_p}. To find the cardinal of {G} we need to count the bases of {\Bbb{F}_p^n}. We can choose the first element in {p^n-1} ways because any element not equal to zero works.

Once we have chosen the first vector, the second cannot be a scalar multiple of the first (there are {p} such scalars), therefore we are left with {p^n-p} possibilities.

Once we have chosen {k} vectors, the space of linear combinations of these {k} vectors has {p^k} elements, so we have {p^n-p^k} choices at step {k}. Multiply all these for {k=0,1,..,n-1} and get the above formula.

We deduce that {|G| = p^{n(n-1)/2}m} with {\gcd(m,p)=1}. If we denote {A} the set of matrices which have {1} on the diagonal and zeros below the main diagonal, then {A} is a subgroup of {G} and {A} has {p^{n(n-1)/2}} elements. Therefore {A} is a {p}-Sylow subgroup of {G}.

2. Lemma Suppose {G} is a group with {|G|=mp^a,\ \gcd(m,p)=1} and {H} a subgroup of {G}. Suppose {G} has a {p}-Sylow subgroup {S}. Then there exists {a \in G} such that {aHa^{-1} \cap H} is a {p}-Sylow subgroup of {H}.

Proof of the lemma: {G} acts on {G/S} (left classes) by left translation, and the stabilizer of {aS} is {aSa^{-1}}. ({haS=aS \Leftrightarrow ha \in aS \Leftrightarrow h \in aSa^{-1}}) {H} acts also on {G/S} by left translation and the above argument proves that the stabilizer of {aS} under this action is {aSa^{-1}\cap H}.

These stabilizers {aSa^{-1}\cap H} are all {p}-groups since they are subgroups of {aSa^{-1}}. To finish it is enough to prove that for some {a} the cardinal {|H/(aSa^{-1}\cap H)|} is prime to {p}. But we know that {|H /(aSa^{-1}\cap H)|} is the cardinal of the orbit of {aS} under the action of {H} and the sum of cardinals of orbits equals the cardinal of the set {G/S}. Since {G/S} is not divisible by {p} ({S} is {p}-Sylow), it follows that there exists {a} such that {|H /(aSa^{-1}\cap H)|} is prime to {p}. Therefore {aSa^{-1} \cap H} is a {p}-Sylow subgroup of {H}.

3. In general, if {|G|=n=p^am,\ \gcd(m,p)=1} we know that {G} is isomorphic to a subgroup of {S_n} (Cayley’s theorem). Furthermore, {S_n} can be regarded as a subgroup of {GL(n,\Bbb{F}_p)} by associating to each permutation {\sigma \in S_n} the morphism defined on the canonical base {(e_i)} by

\displaystyle u_\sigma(e_i)=e_{\sigma(i)}.

In this way, we can see {G} is isomorphic to a subgroup of {GL(n,\Bbb{F}_p)}. Apply points 1. and 2. and we are done.

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: