Sylow Theory

A classical way of motivating Sylow theory is to find a weaker version of the converse statement to Lagrange’s theorem. Recall that Lagrange’s theorem implies that if $H \le G$, then the order of $H$ divides $G$. On the other hand, we might interested in asking if $k$ is some factor of $|G|$, then is there a subgroup $H$ of order $k$. In general, the answer to this question is no!

Consider the alternating group $A_4$. As it turns out, there are no subgroups of order $6$ which we can show with an ad hoc argument. For the sake of contradiction, suppose $A_4$ did have a subgroup of order $6$. Since $A_4$ has no elements of order 6, it follows that $H \cong S_3$. This implies that $H$ has 3 elements of order 2 and 2 elements of order 3. Since $A_4$ has exactly 3 elements of order $2$, these must be the elements of order 2 in $S_3$. But these elements form a subgroup of order $4$ so $H$ would have a subgroup of order 4, contradicting Lagrange’s theorem.

So then this brings up a natural question? Which groups satisfy a converse to Lagrange’s theorem? This is a very profound and deep question, which we will not answer here. We can, however, give an answer to a partial converse: if $k$ is a specific kind of factor of $|G|$, then we can assert the existence of subgroups of order $k$ in $G$.

Definition. We say that $G$ is a $p$-group provided it its order is a power of $p$. We say call subgroups that are $p$-groups $p$-subgroups.

Definition. Suppose that $G$ is a subgroup of order $p^am$ with $a \ge 1$ and $p$ and $m$ relatively prime. If $H$ is a subgroup of $G$ such that $|H| = p^a$, then we call $H$ a Sylow $p$-subgroup of $G$.

Sylow subgroups are well-studied. For $|G| = p^am$ as above, we can always assert the existence of a Sylow $p$-subgroup, describe the Sylow $p$-subgroups in relation to each other, and even make numerical claims about the number of Sylow $p$-subgroups present in $G$. As I do not have a deep enough understanding of Sylow theory to explain how these come about naturally, I will present a classical introduction of Sylow theory.

Theorem (Sylow I). Let $G$ is a group of order $p^a m$ where $a\ge 1$ and $p$ and $m$ are relatively prime. Then $G$ has a Sylow $p$-subgroup.

A classical way of proving Sylow I is by induction on the order of $G$ and the following lemma. This will amount to assuming the existence of a Sylow $p$-subgroup of order $p^{a-1}$ in a group of order $p^{a-1}m$ and then showing this implies the existence of a Sylow $p$-subgroup of order $p^{a}$ in a group of order $p^{a-1}m$. To do this, we mod out by a normal subgroup of order $p$ (the existence of which we take an appropriate subgroup of $Z(G)$). By Lagrange, the quotient has order $p^{a-1}m$ and induction implies the existence of a Sylow $p$-subgroup. We then take the preimage of this subgroup under the quotient map and then this group ends up being the Sylow $p$-subgroup of $G$ that we are looking for. This argument lies crucially on the fact that we can mod out $G$ by a normal subgroup of order $p$ and for that we need the following lemma:

Lemma (Cauchy’s theorem for abelian groups). Let $G$ be a finite abelian group whose order is a multiple of a prime $p$. Then $G$ has a subgroup of order $p$.

Proof. We proceed by induction on the order of $|G|$. If $|G| = p$, then $G$ is a subgroup of itself. Now suppose that $|G| > p$ and pick some nonidentity element $g$ of $G$. By Lagrange, the order of this element divides $|G|$. If the order of $g$ is divisible by $p$, then $g^{|g|/p}$ is an element of order $p$ which implies the desired result. If $|g|$ is not a multiple of $p$, then the order of $G/\langle g\rangle$ has at least one factor of $p$ since $|g|$ has none. Induction implies that $G/\langle g\rangle$ has a subgroup $Q$ of order $p$. TODO: finish