The Group Properties Wiki (pre-alpha)

TIP: Read more about how the definition in Groupprops is structured

ABOUT US: Read our purpose statement and learn what makes us special

ALSO CHECK OUT: Diffgeom: The Differential Geometry Wiki

From Groupprops

This article describes a result of the form that argues that a subset constructed in a certain fashion is proper, viz it is not the whole group

Contents 1 Statement 1.1 Verbal statement 1.2 Symbolic statement 2 Related facts 2.1 Applications and similar facts 3 Proof

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
View other elementary non-basic facts

Statement

Verbal statement

Given any two proper subgroups of in a group that are conjugate to each other, their product is a proper subset of the group.

Symbolic statement

Let be a proper subgroup, and let H^g = g ^{− 1}Hg be a conjugate of H. Then HH^g is a proper subset of G.

Related facts

• Union of all conjugates is proper: This states that the union of all conjugates of a proper subgroup in a finite group is again proper.

Applications and similar facts

• Maximal implies normal or abnormal: The proof idea here is very similar.
• Maximal conjugate-permutable implies normal: This is an easy corollary.
• Maximal implies self-conjugate-permutable
• Conjugate-permutable and self-conjugate-permutable implies normal
• Conjugate-permutable implies subnormal in finite

Proof

Given: A finite group G, two proper conjugate subgroups H and H^g, where H^g = g ^{− 1}Hg.

To prove: HH^g is a proper subset of G.

Proof: Suppose not, i.e., suppose HH^g = G.

1. , so we can write g = hk where .
2. Thus, H^g = H^hk = H^k. This yields .
3. But we know that , so we get H = H^g.
4. We thus get G = HH^g = H, contradicting the assumption that H is a proper subgroup of G.