Product of conjugates is proper

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 $$G$$.

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 $$G$$.