Proving that a subgroup is conjugacy-closed

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This article is about strategies to prove that a subgroup (either a chosen one, or one subject to certain constraints) is conjugacy-closed in the whole group.

The definition of conjuacy-closed subgroup

A subgroup $H$ of a group $G$ is termed self-centralizing in $G$ , if no two distinct conjugacy classes of $H$ get fused in $G$ . Equivalently, if $x,y \in H$ are such that $g x g - 1 = y$ , then there exists $h \in H$ such that $h x h - 1 = y$ .

Typical strategies used

When both groups are well-understood

When both $G$ and $H$ are well-understood groups, we use what we know about conjugacy classes in $G$ and in $H$ . In other words:

We determine a general criterion for elements of $G$ to be conjugate, and interpret this criterion when the two elements both happen to be from $H$ .
We determine a general criterion for elements of $H$ to be conjugate.
We then check that the two criteria are equal.

This usually involves a good understanding of both groups, which may or may not be available.

Finding a conjugate-dense subgroup, or subset

This is a strategy that works well, specially when the conjugacy classes in $H$ are well-understood. First, we find a subgroup $K$ of $H$ satisfying the following two conditions:

$K$ is conjugate-dense in $G$ : in other words, every element of $H$ is conjugate in $H$ , to some element in $K$ .
Any two elements of $K$ that are conjugate in $G$ are conjugate in $H$ . In other words, the fusion of $K$ in $G$ is completely controlled by $H$ .

This strategy works well when proving results about linear groups, because in these cases, the conjugacy classes are well-understood. In fact, we don't even need $K$ to be a subgroup above. All we need is a subset $K$ of $H$ satisfying:

Every element of $H$ is conjugate to some element of $K$
Any two elements of $K$ that are conjugate in $G$ are also conjugate in $H$

Here are some situations where we can apply this strategy:

General linear group of subfield is conjugacy-closed: In this case, we use the rational canonical form for the general linear group over the subfield, and argue that if two matrices in the rational canonical form over the subfield are conjugate in the big field, they're also conjugate in the subfield.
Unitary group is conjugacy-closed in general linear group: This states that the group $U(n,\mathbb{C})$ of unitary matrices over $\mathbb{C}$ is conjugacy-closed inside the group $GL(n,\mathbb{C})$ . For this, we use the spectral form theorem to first argue that any unitary matrix is conjugate, in $U(n,\mathbb{C})$ , to a diagonal unitary matrix. Thus, the subgroup of diagonal unitary matrices is conjugate-dense in the group $U(n,\mathbb{C})$ . Next, we show that any two diagonal unitary matrices that are conjugate in $GL(n,\mathbb{C})$ have the same entries upto permutation, hence they are conjugate by a permutation matrix, and hence they are conjugate in $GL(n,\mathbb{C})$ .