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Let $G$ be a finite group, $H$ is proper subgroup and ${\cal H}(H)$ the set of all subgroups of conjugate H. Construct the graph $\Gamma$, with vertices ${\cal H}(H)$ and two subgroups adjacent if they have trival intersection.

Its true (if graph not trivial), that this graph regular and have diamert 2?

For example in case $G = A_5$, $H = S_3$ (generated by $(12)(34)$ and $(125)$), $\Gamma$ is complement of Petersen graph. For other $H$, $\Gamma$ is complete multipartite graph (complenet of $mK_n$ for some $n$ and $m$) (for $H \in \{C_2, C_3, C_2\times C_2, C_5\}$) and empty graph (for $H \in \{D_5, A_4\}$)

Next interest example gave $G = A_6$, and ${\cal H}(H) = Syl_2(A_6)$. Then a $\Gamma$ has order $45$, degree $32$, spectrum $\{32^1, 2^{19}, -1^{16}, -6^9\}$, $\Gamma$ is vertex transitive and has automorphism group isomorphic to $(A_{6} \rtimes C_{2}) \rtimes C_{2}$.

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    $\begingroup$ It's certainly always regular, because $G$ acts transitively by conjugation on its vertices, but why do you think that it should have diameter $2$? $\endgroup$ – Derek Holt Jun 9 at 16:30
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    $\begingroup$ Can't you just pick a group where all the conjugates intersect nontrvially? For example, they could be Sylow p-subgroups of a group $G$ with $O_p(G)\neq 1$. $\endgroup$ – verret Jun 9 at 21:07
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    $\begingroup$ Take for example $G=Q_8\rtimes C_3\cong SL(2,3)$, with the $C_3$ acting by cyclically permuting $\{i,j,k\}$. Then the group $\langle i\rangle$ has three conjugates in $G$, and they all intersect in $\langle -1\rangle$ so the graph is disconnected. $\endgroup$ – verret Jun 9 at 21:10

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