Abstract:
Let $G$ be a finite group with subgroups $A$ and $B$. Denote by $M=M_G(A,B)$ (respectively, $m=m_G(A,B)$) the set of all minimal by inclusion (respectively, by order) intersections of the form $A\cap B^g$, where $g\in G$. Put $\min_G(A,B)=\langle m\rangle$ and ${\mathrm Min}_G(A,B)=\langle M\rangle$. In 1994 we proved that if $A$ and $B$ are abelian subgroups, then ${\mathrm Min}_G(A,B)\le F(G)$. In the present paper, we give other proof of this result. Futhermore, we construct a finite group $G$ such that it contan an abelian subgroup $A$, a minimal non-abelian subgroup $B$ and elements $g_1$ and $g_2$ with $A\cap B^{g_1}\le F(G)$, $A\cap B^{g_2}\not\le F(G)$, $|A\cap B^{g_1}|=|A\cap B^{g_2}|$ and $A\cap B^{g_1}$, $ A\cap B^{g_2}\in\min_G(A,B)$. We provide an example of a group $G$ such that $g_1, g_2\in G$$A\cap B^{g_1}$, $ A\cap B^{g_2}\in{\mathrm Min}_G(A,B)$, $A\cap B^{g_1}\le F(G)$, and $A\cap B^{g_2}\not\le F(G)$. Moreover, we show that there exists a group $G$ with nilpotent subgroups $A$ and $B$ such that $m\subset M$ and $\min_G(A,B) < {\mathrm Min}_G(A,B)$.
Key words:finite group, abelian subgroup, intersection of subgroups.
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