Product of conjugates is proper

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

Statement

Verbal statement

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

Symbolic statement

Let $H\leq G$ 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 of subgroup of finite index 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. $g\in HH^{g}$, so we can write $g=hk$ where $h\in H,k\in H^{g}$.
2. Thus, $H^{g}=H^{hk}=H^{k}$. This yields $H=(H^{g})^{k^{-1}}$.
3. But we know that $k\in H^{g}$, so we get $H=H^{g}$.
4. We thus get $G=HH^{g}=H$, contradicting the assumption that $H$ is a proper subgroup of $G$.