# Proving product of subgroups

This is a survey article related to:product of subgroups
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## Introduction

This article is about general strategies used to prove that a given group, is a product of two subgroups we have at hand. These techniques include situations where both subgroups are normal, where only one is normal, and where neither is normal.

## The group action method

Further information: Group equals product of transitive subgroup and isotropy of a point Suppose we need to prove that a given group $G$ can be expressed as the product of two subgroups $H$ and $K$. The group action method is as follows. We find an action of the group $G$ on some set $S$ such that:

• $K$ contains the isotropy subgroup of some point $s \in S$
• $H$ acts transitively on $S$ (the action being given by restriction from $G$)

Under these circumstances, we can show that $G = HK$, as follows:

• Pick any $g \in G$
• Find a $h \in H$ such that $gs = hs$
• Thus, $h^{-1}g$ fixes $s$, so $h^{-1}g \in K$. We'll thus get $g = hk$ for some $k \in K$

Note that since $G = HK$ is equivalent to $G = KH$, we can construct a group action with the roles of $H$ and $K$ reversed.

Described below are some applications of this method.

### Frattini's argument

Further information: Frattini's argument

The setup here is as follows: $G$ is a group, $H$ is a normal subgroup, and $P$ is an automorph-conjugate subgroup of $H$. We need to show that: $HN_G(P) = G$

Here, $N_G(P)$ is the normalizer of $P$ in $G$.

The set $S$ here is the set of all conjugates of $P$ in $G$. The $G$-action is by conjugation.

• The isotropy at $P$ is $N_G(P)$: This is by definition of normalizer
• Since $H$ is normal in $G$, any conjugate to $P$ in $G$ is an automorph of $P$ in $H$. Moreover, $P$ was chosen to be automorph-conjugate, so any conjugate of $P$ in $G$ is actually conjugate to it in $H$. Hence, $H$ acts transitively on $S$.

### Finding the Weyl group

Further information: Computing the Weyl group of the unitary group

This is the idea used to prove that in the unitary group $U(n)$, the normalizer of any torus $T$ is given by: $N(T) = TS_n$

where $S_n$ is the symmetric group on $n$ letters, embedded as matrices inside $U(n)$.

Pick an element in $T$ with distinct diagonal entries. Elementary linear algebra tells us that any conjugate of it in $T$, must have the same diagonal entries, possibly in a different order. Thus, its $N(T)$-orbit is precisely the set of $n!$ different diagonal matrices with exactly those diagonal entries.

Let $N(T)$ act on this finite set, by conjugation.

• The isotropy of any point in the set is exactly $T$. This is based on the linear algebra fact that any matrix that commutes with a diagonal matrix with distinct diagonal entries, must also be diagonal.
• The symmetric group $S_n$, viewed as permutation matrices, acts transitively by conjugation.

## Another variant of the group action method

In this variant, instead of computing the isotropy at one point, we do the following: Find an action of $G$ on a set $S$, such that:

• $K$ contains the kernel of the action (i.e. any element that acts trivially on $S$, lives in $K$)
• For any element of $G$, there exists an element of $H$ that has exactly the same action on $S$

Somewhere in between this and the previous approach is an approach where we look at the intersection of isotropies of some elements of $S$.

### Equivalence of definitions of central factor

Further information: Equivalence of definitions of central factor

There are two definitions of central factor:

1. $H$ is a central factor of $G$ if $HC_G(H) = G$ (where $C_G(H)$ is the centralizer of $H$ in $G$)
2. $H$ is a central factor of $G$ if every inner automorphism of $G$ restricts to an inner automorphism of $H$

We'll show here that (2) implies (1):

The set $S$ is the set of elements of $H$. The $G$ action is by conjugation. This action is well-defined because (2) in particular indicates that $H$ is normal in $G$.

• The kernel of the action is precisely $C_G(H)$
• By condition (2), we have that for any element of $G$, there exists an element in $H$ such that the action by conjugation on $H$ is the same for both.

This gives the result.

A similar technique is used to prove the equivalence of definitions of WC-subgroup.