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.