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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 HG have index 2, and let xH. 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 CG(x) contains an element outside H; equivalently, it splits if and only if CG(x)H.

Facts & Assumptions

Given: A normal subgroup HG of index 2 and xH.

[F1]

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

[F2]

The conjugacy class and centralizer are {gxg1:gG} and {gG:gx=xg} (The conjugacy class ClG(x) and centralizer CG(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 tGH. By [F3], G=HtH.

F3choose
2.1

Let C be the H-class of x. Using [F1] and step 1.1, the G-class is CtCt1; these are H-classes and therefore are equal or disjoint.

F1F2step 1.1algebra
3.1

Conjugation by t is a bijection CtCt1, so in the disjoint case the two classes have equal cardinality.

step 2.1algebra
3.2

The classes agree exactly when txt1=hxh1 for some hH, which is equivalent to h1tCG(x). This element lies outside H.

F2step 1.1step 2.1algebra
3.3

Conversely, if cCG(x)H, then c=h1t for some hH, and txt1=hxh1C; 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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 25 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources