$H$ be a proper subgroup of finite group $G$ such that $H \cap gHg^{-1}=\{e\} , \forall g \in G \setminus H$ , then $|\cup gHg^{-1}|>\dfrac 12 |G|+1$

Question

Let $H$ be a proper subgroup of finite group $G$ such that $H \cap gHg^{-1}=\{e\}$ for all $g \in G \setminus H$. Then is it true that $$|\cup_{g \in G \setminus H}gHg^{-1}|>\dfrac 12 |G|+1$$

If not, then is it at least true that

$$|\cup_{g \in G}gHg^{-1}|>\dfrac 12 |G|+1$$

(I know the upper bound $|\cup_{g \in G}gHg^{-1}|\le [G:H](|H|-1)+1$, but I am not aware of any lower bound. Please help. Thanks in advance.)

Answer 1

Claim Let $g_1H, \ldots, g_mH$ be all the cosets of $H$ in $G$. Then any two of the following $g_1Hg_1^{-1}, \ldots, g_kHg_m^{-1}$ intersect only in the identity.

Proof. This uses the hypothesis that $H\cap gHg^{-1}=\{e\}$ if $g\notin H$. Suppose $a\in g_1Hg_1^{-1}\cap g_2Hg_2^{-1}$. Then we have $$g_1^{-1}ag_1\in H\cap\ (g_1^{-1}g_2)H(g_1^{-1}g_2)^{-1}$$ Since $g_1H\neq g_2H$, we have $g_1^{-1}g_2\notin H$. Therefore, by hypothesis, we must have $g_1^{-1}ag_1=e$, giving $a=e$.

From the above claim, we conclude that $|\bigcup_{g\in G\setminus H}gHg^{-1}|$ has exactly $(|G:H|-1)(|H|-1)+1$ elements in it.

Answer 2

If $H$ is such a subgroup, then a Theorem of Frobenius tells us that there is $K \lhd G$ with $G = HK$ and $H \cap K = 1$. Furthermore, $K= \{1_{G} \} \cup (G \backslash \cup_{g \in G} g^{-1}Hg)$. Hence $|G| = |K| + [G:H](|H|-1).$ It follows that your seond inequality is violated precisely when $|K| \geq \frac{|G|}{2}.$ But $K$ is a subgroup of $G$, so the only possibility for the seond inequality to be violated (except the trivial case $G = K, H = \{1_{G} \}$) is that $|K| = \frac{|G|}{2}$ and $|H| = 2$.

Answer 3

As noted by Geoff Robinson groups which have such a subgroup are called Frobenius groups, and in his answer he gave you the conditions when your seond inequality is true and when not. His cited results, namely that $K$ (the Frobenius kernel) is a subgroup is a rather deep result. I add a more elementary proof.

Suppose $H \ne 1$ (otherwise the inequality is obviously false).

With group actions. Notice that $N_G(H) = H$, hence by considering the action of $G$ on $H$ by conjugation we have precisely $|G : H|$ different conjugates, hence $$ \left| \bigcup_{g\in G} H^g \setminus\{1\} \right| = |G : H|(|H| - 1) = \frac{|H|-1}{|H|} |G| $$ so that $|\bigcup_{g \in G} H^g| = \frac{|H|-1}{|H|} |G| + 1$ and $$ \frac{|H|-1}{|H|} |G| + 1 > \frac{|G|}{2} + 1 \Leftrightarrow |H| > 2 $$ so you see that if $|H| \in \{1,2\}$ the inequality fails, otherwise it holds.

By the way the counting argument could also be done without using group actions, this is similar to the way caffeinemachine did the counting. Note that the condition $N_G(H) = H$ gives $$ Hx = Hy \Leftrightarrow H^x = H^y $$ so we have exactly as many different conjugates as we have cosets. The implication $Hx = Hy \Rightarrow H^x = H^y$ is always true, as $Hx = Hy$ implies $xy^{-1} \in H$ and so $H^x = (H^{xy^{-1}})^y = H^y$ (or note that $xH \mapsto Hx^{-1}$ is a bijection, and so $H^x = x^{-1}(Hx) = x^{-1}(Hy) = (x^{-1}H)y = (y^{-1}H)y = H^y$). For the other implication if $H^x = H^y$, then $H^{xy^{-1}} = H$, hence we must have $xy^{-1} \in H$ as precisely the elements from $H$ normalise $H$. Now using the trivial intersection property we can count exactly like done above.