# Product of conjugates is proper

## Contents

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
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|

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 $H \le G$ be a proper subgroup, and let $H^g = g^{-1}Hg$ be a conjugate of $H$. Then $HH^g$ is a proper subset of $G$.

## 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 $G$.