# Frattini’s Argument

Being a useful exercise to review, let’s talk about Frattini’s argument.  We’ll approach the problem first from a standpoint that (hopefully) makes things intuitively clear, then we’ll look at the classical argument advanced by Frattini.

The topic that Frattini’s argument addresses begins with a relatively straightforward question:

Given a finite group $G$ with a nontrivial normal subgroup $K$, what is the relationship of $K$ to $G$?

Let’s begin at the top and observe that, as $K$ is nontrivial by assumption, it must have a Sylow-$p$ subgroup for some prime $p$.  Without fussing over what exactly $p$ is (in fact, this will hold for any choice of $p$ dividing the order of $K$), let’s just look at an arbitrary Sylow-$p$ subgroup $P$ of $K$.

Now, as $K$ is normal in $G$, let us observe that the second isomorphism theorem (frequently also called the ‘diamond’ isomorphism theorem) gives us a variety of nice things.  In general, for any subgroup $H$ of $G$, we have that:

• $HK = KH \leq G$;
• $H \bigcap K \unlhd H$;
• $(HK) / K \cong H / (H \bigcap K)$.

Looking at the first result, recall that $HK=KH$ is a necessary and sufficient condition for $HK$ to be a subgroup of $G$ (a fact which is occasionally referred to as the ‘product theorem’).  Taking some generous liberties with notation in order to make a point, note that we can ‘think of’ $HK = KH$ as $K = H^{-1}KH$ — at least, in the sense that, for some $h_1,h_2 \in H$ and $k \in K$,  assuming $HK = KH$ gives that $h_1k = kh_2$, or, equivalently, that $k=h_1^{-1}kh_2$.  So, the manner in which this fact follows from $K$‘s normality isn’t terribly surprising.

However, the second and third results are where the real “meat” of the concept lies.  As a general rule, if one is given a Sylow-$p$ group and asked to discern something about the group structure, it is often useful to recall that $\textrm{Syl}_p(K)$ constitutes a single conjugacy class.  Frequently, we may wish to consider the number of Sylow-$p$ subgroups, but for our purposes, let’s focus only on that basic fact.

With conjugacy classes in mind, we might wish to ask which elements of $K$ or $G$ fix $P$ under conjugation.  Of course, these are respectively given by $N_K(P)$ and $N_G(P)$.  However, note that $N_G(P) \bigcap K = N_K(P)$.  This is key — we have discerned potentially interesting subgroups of $G$, and their form makes them particularly well suited to the second isomorphism theorem.

Applying the second isomorphism theorem, we may see $N_G(P) \bigcap K=N_K(P) \unlhd N_G(P)$, and that $(N_G(P)K) / K \cong N_G(P) / N_K(P)$.  This seems useful: we have a subgroup in which $P$ is normal which is, itself, normal in a larger subgroup in which $P$ is also normal.  What may we do with this?

Well, for starters, let’s notice that $N_K(P)$ being normal in $N_G(P)$ isn’t terribly surprising.  Since $K$ is normal in $G$, and since $P \leq K$, we know that $P$ remains in $K$ after conjugation by any element of $G$.  But, this allows us to reach something very strong:  The number of $G$ conjugates of $K$ is the same as the number of $G$ conjugates of $P$, which is given by $[G:N_G(P)]$.  But, more than that, notice that the number of $K$ conjugates of $P$ is precisely $n_p(K)$, which is given by $[K:N_K(P)]$.  Thus, $[G:N_G(P)]=[K:N_K(P)]$.

Now, let’s apply Lagrange’s theorem to this, yielding

$[G:N_G(P)] = \frac{|G|}{|N_G(P)|} = \frac{|K|}{|N_K(P)|} = [K:N_K(P)]$

But, notice that this then gives $\frac{|N_G(P)|}{|N_K(P)|}=\frac{|G|}{|K|}$, and as we know $(N_G(P)K) / K \cong N_G(P) / N_K(P)$, it must follow that $N_G(P)K = G$.

So, we now have at least a reasonable answer to our initial question:  If $G$ is a finite group with nontrivial normal subgroup $K$, then $N_G(P)K=G$ for any $P \in \textrm{Syl}_p(K)$.

Diverging from the previous line of thinking briefly, let’s consider a few calculations which properly constitute Frattini’s argument.  These show the same fact that we discovered above, but from a more blunt and direct angle.

Applying the fact that $K$ is normal in $G$, let us note that, for any $g \in G$, we have that $P^g \leq K$ and, by fact that $\textrm{Syl}_p(K)$ forms a single conjugacy class, $P^g \in \textrm{Syl}_p(K)$.   Applying the conjugation argument yet again, we then yield that there must be $k \in K$ such that $(P^g)^k = P$.  But note that this is simply $k^{-1}g^{-1}Pgk$, so $gk \in N_G(P)$.  Moreover, $g \in N_G(P)K$, as $k \in K$ and, as $g$ was arbitrary, we have that $N_G(P)K = G$.

With this in mind, let’s conclude by reflecting briefly on what all of the above might give us.  For one, one might notice that we might apply this to any Sylow-$p$ subgroup of $H$, giving a whole host of different quotients of elements of $G$ isomorphic to the canonical quotient $G / H$.  Or, perhaps we might apply this to a subgroup of $G$, replacing $K$ with a subgroup now normal in the new, smaller group.  Among other things, we could apply this argument to show that $N_G(N_G(P))=N_G(P)$ for any $P \in \textrm{Syl}_p(G)$.

Most importantly though (in my mind, at least), it reveals a small part of just how much structure normality gives:  the way $G$ behaves everywhere except on a normal subgroup $K$ is precisely the way that the parts of $G$ not in $K$ which leave Sylow-$p$ subgroups of $K$ invariant behave.

(Note:  The image included was first posted by the stackexchange user p Groups in an excellent answer to this question regarding the second isomorphism theorem.)