Proving 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.