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Let $G$ be a finite subgroup and $H$ a subgroup of index three in $G$, not necessarily normal. Put $n=|H|$. We choose representatives $a_1$ and $a_2$ such that $G$ is the disjoint union

$$ G=H \cup (a_1H) \cup (a_2H) $$

We can then define

$$ \begin{array}{lcl} H_1 &=& H \cap (a_1 H a_1^{-1}) \cap (a_1 H a_2^{-1}) \\ H_2 &=& H \cap (a_1 H a_1^{-1}) \cap (a_2 H a_2^{-1}) \\ H_3 &=& H \cap (a_2 H a_1^{-1}) \cap (a_1 H a_2^{-1}) \\ H_4 &=& H \cap (a_2 H a_1^{-1}) \cap (a_2 H a_2^{-1}) \\ \end{array}{} $$ Let us put $n_i=|H_i|$. Since $H$ is the disjoint union of the $H_i$, we have $n_1+n_2+n_3+n_4=|H|$. Are any further inequalities involving $n_1$, $n_2$, $n_3$ and $n_4$ known ?

Intuitively, the $n_i$ should be balanced, so that there should be a lower bound for $n_1+n_2,n_1+n_3$ etc that tends to $+\infty$ when $n\to +\infty$, but I don’t know how to show this.

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The key mathematical object here is the permutation action of $G$ on the left quotient set $G/H$. It allows us to define a homomorphism $\phi : G \to {\mathfrak S}_3$, whose kernel is $N=H_2$.

The image $\phi(G)$ is a transitive subgroup of ${\mathfrak S}_3$ ; it is therefore either $\lbrace id,\bar{\alpha},\bar{\alpha}^2 \rbrace$ where $\bar{\alpha}$ is a $3$-cycle, or the whole of ${\mathfrak S}_3$.

If the image is the whole of ${\mathfrak S}_3$, then there is a $\bar{\beta} \in {\mathfrak S}_3$ such that $\bar{\beta}^2=1$ and $\bar{\alpha}\bar{\beta}=\bar{\beta}\bar{\alpha}^2$. Let $\alpha,\beta\in G$ such that $\phi(\alpha)=\bar{\alpha}$ and $\phi(\beta)=\bar{\beta}$. Then $H$ is the disjoint union $N \cup \beta N$ and $\alpha\beta \in {\beta}\alpha^2 N$. In this case, we can take $a_1=\alpha$ and $a_2=\alpha^2$ ; then

$$ H_1=\emptyset, H_2=N, H_3=\beta N, H_4=\emptyset $$

Otherwise, the image is $\lbrace id,\alpha,\alpha^2 \rbrace$, and $\phi$ must be trivial on $H$, so $H$ is normal in $G$. Again, we can take $a_1=\alpha$ and $a_2=\alpha^2$ ; then

$$ H_1=\emptyset, H_2=H, H_3=\emptyset, H_4=\emptyset $$

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