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A conjugacy class in an index-two normal subgroup remains one class or splits into two equal classes, with a centralizer criterion

Statement

Let $H ⊴ G$ have index $2$ , and let $x \in H$ . The $G$ -conjugacy class of $x$ is either one $H$ -conjugacy class or the disjoint union of two $H$ -conjugacy classes of equal cardinality. It remains one class if and only if $C_{G} (x)$ contains an element outside $H$ ; equivalently, it splits if and only if $C_{G} (x) \subseteq H$ .

Facts & Assumptions

Given: A normal subgroup $H ⊴ G$ of index $2$ and $x \in H$ .

[F1]

A normal subgroup is invariant under conjugation (Normal subgroup: invariance under conjugation).

[F2]

The conjugacy class and centralizer are ${g x g^{- 1} : g \in G}$ and ${g \in G : g x = x g}$ (The conjugacy class ${Cl}_{G} (x)$ and centralizer $C_{G} (x)$ of an element).

[F3]

Index $2$ means that the coset set has two elements (The coset set $G / H$ and the index $[G : H]$ of a subgroup).

Proof

technique · direct

1.1

Choose $t \in G ∖ H$ . By [F3], $G = H ⊔ t H$ .

F3choose

2.1

Let $C$ be the $H$ -class of $x$ . Using [F1] and step 1.1, the $G$ -class is $C \cup t C t^{- 1}$ ; these are $H$ -classes and therefore are equal or disjoint.

F1F2step 1.1algebra

3.1

Conjugation by $t$ is a bijection $C \to t C t^{- 1}$ , so in the disjoint case the two classes have equal cardinality.

step 2.1algebra

3.2

The classes agree exactly when $t x t^{- 1} = h x h^{- 1}$ for some $h \in H$ , which is equivalent to $h^{- 1} t \in C_{G} (x)$ . This element lies outside $H$ .

F2step 1.1step 2.1algebra

3.3

Conversely, if $c \in C_{G} (x) ∖ H$ , then $c = h^{- 1} t$ for some $h \in H$ , and $t x t^{- 1} = h x h^{- 1} \in C$ ; hence the two classes agree.

F2step 1.1step 2.1algebra

4.1

Steps 3.2--3.3 give the outside-centralizer criterion; negating it gives the equivalent containment criterion for splitting.

step 3.2step 3.3∎

Depends on

Used by

For n≥2, an Sₙ-class of an even permutation splits in Aₙ exactly when all cycle lengths, including 1-cycles, are odd and distinct Theorem

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Group Theory (standard reference, not scraped)