Estimating the cardinality of the union of the conjugates of a proper subgroup.

Question

If $G$ is a finite group and $H$ is a nontrivial subgroup of $G$ such that $H^{a}\cap H=\{e\}$ for all $a\in G-H$, where $H^{a}=aHa^{-1}$ and $e$ is the neutral element of $G$, show that $$\left\lvert\bigcup_{a\in G}H^{a}\right\rvert\geq\frac{1}{2}|G|+1.$$ Conclude that if $H$ is a proper subgroup of $G$, then $$\bigcup_{a\in G}H^{a}\neq G.$$

Answer 1

The different conjugates $H^{a}$ all have $|H|$ elements exactly. Moreover, if $x\in H^{a}\cap H^b$ with $x\neq e$, then $x=bhb^{-1}=ah'a^{-1} $ for some $h,h'\in H\setminus\{e\}$. Then $$h=b^{-1}ah'a^{-1}b=(b^{-1}a)h'(b^{-1}a)^{-1}\in H\cap H^{b^{-1}a}.$$ Thus $b^{-1}a\in H$, and we have $a=b\hat{h}$ for some $\hat{h}\in H$. Then $$H^{a}=aHa^{-1}=b\hat{h}H\hat{h}^{-1}b^{-1}=bHb^{-1}=H^b.$$

So two distincts conjugate only have $\{e\}$ as a common element. This means their union contains $e$ and $|H|-1$ distincts elements for each $H^{a}$. Hence, if we denote $m$ the number of these distincts conjugate of $H$, $$\left\lvert\bigcup_{a\in G}H^{a}\right\rvert=1+m(|H|-1).$$

Now to find $m$ we will use the orbit-stabilizer property. Indeed, conjugation define an action of $G$ on the set of its subgroups; thus the set of conjugates $H^{a}$ is simply the orbit $H^G$, which is in bijection with the set of left cosets of the stabilizer. But the condition that $H^{a}\cap H=\{e\}$ for $a\notin H$ means that $H$ is its own stabilizer, and it follows that $m=\frac{|G|}{|H|}$. Thus $$\left\lvert\bigcup_{a\in G}H^{a}\right\rvert=1+\frac{|G|}{|H|}(|H|-1)=1+|G|-\frac{|G|}{|H|}.\label{1}\tag{1}$$ Because $H$ is non-trivial $|H|\geq 2$, and thus $\frac{|G|}{|H|}\leq \frac{|G|}{2}$. It follows that

$$\left\lvert\bigcup_{a\in G}H^{a}\right\rvert\geq 1+|G|-\frac{|G|}{2}=1+\frac{|G|}{2}.$$

Moreover $\bigcup_{a\in G}H^{a}=G$ implies that $\left|\bigcup_{a\in G}H^{a}\right|=|G|$. Comparing this to \eqref{1} shows that we must then have $\frac{|G|}{|H|}=1$, and thus $H=G$.