Intersections of Three Subgroups
$begingroup$
I encountered this elementary question which I didn't manage to prove or disprove successfully.
Let $G$ be some group and $K_1,K_2,K_3 leq G$ s.t.
$$G = <K_1,K_2,K_3>.$$
Furthermore, suppose that $forall sigma in S_3$ it holds that $$K_{sigma(1)} = <K_{sigma(1)} cap K_{sigma(2)},K_{sigma(1)} cap K_{sigma(3)}>,$$
i.e every subgroup is generated by its intersection with the two others.
Then for every three cosets $g_1K_1, g_2K_2, g_3K_3$, if every pair of cosets intersect then all three cosets intersect. That is, if $g_iK_i cap g_jK_j ne emptyset$ for all possible $i,j$, then $g_1K_1 cap g_2K_2 cap g_3 K_3 ne emptyset$.
I managed to reduce the question to showing the statement on $g_1=g_2=e$ - this is just because if $g_1K_1,g_2K_2$ intersect then the intersection is coset of $K_1 cap K_2$.
I also noticed that without the requirement that every subgroup is generated by the intersections, this statement is false - the counterexample is three (affine) lines in a plane that create a triangle. I don't see how to use the extra property though...
group-theory
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
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add a comment |
$begingroup$
I encountered this elementary question which I didn't manage to prove or disprove successfully.
Let $G$ be some group and $K_1,K_2,K_3 leq G$ s.t.
$$G = <K_1,K_2,K_3>.$$
Furthermore, suppose that $forall sigma in S_3$ it holds that $$K_{sigma(1)} = <K_{sigma(1)} cap K_{sigma(2)},K_{sigma(1)} cap K_{sigma(3)}>,$$
i.e every subgroup is generated by its intersection with the two others.
Then for every three cosets $g_1K_1, g_2K_2, g_3K_3$, if every pair of cosets intersect then all three cosets intersect. That is, if $g_iK_i cap g_jK_j ne emptyset$ for all possible $i,j$, then $g_1K_1 cap g_2K_2 cap g_3 K_3 ne emptyset$.
I managed to reduce the question to showing the statement on $g_1=g_2=e$ - this is just because if $g_1K_1,g_2K_2$ intersect then the intersection is coset of $K_1 cap K_2$.
I also noticed that without the requirement that every subgroup is generated by the intersections, this statement is false - the counterexample is three (affine) lines in a plane that create a triangle. I don't see how to use the extra property though...
group-theory
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
$endgroup$
add a comment |
$begingroup$
I encountered this elementary question which I didn't manage to prove or disprove successfully.
Let $G$ be some group and $K_1,K_2,K_3 leq G$ s.t.
$$G = <K_1,K_2,K_3>.$$
Furthermore, suppose that $forall sigma in S_3$ it holds that $$K_{sigma(1)} = <K_{sigma(1)} cap K_{sigma(2)},K_{sigma(1)} cap K_{sigma(3)}>,$$
i.e every subgroup is generated by its intersection with the two others.
Then for every three cosets $g_1K_1, g_2K_2, g_3K_3$, if every pair of cosets intersect then all three cosets intersect. That is, if $g_iK_i cap g_jK_j ne emptyset$ for all possible $i,j$, then $g_1K_1 cap g_2K_2 cap g_3 K_3 ne emptyset$.
I managed to reduce the question to showing the statement on $g_1=g_2=e$ - this is just because if $g_1K_1,g_2K_2$ intersect then the intersection is coset of $K_1 cap K_2$.
I also noticed that without the requirement that every subgroup is generated by the intersections, this statement is false - the counterexample is three (affine) lines in a plane that create a triangle. I don't see how to use the extra property though...
group-theory
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
$endgroup$
I encountered this elementary question which I didn't manage to prove or disprove successfully.
Let $G$ be some group and $K_1,K_2,K_3 leq G$ s.t.
$$G = <K_1,K_2,K_3>.$$
Furthermore, suppose that $forall sigma in S_3$ it holds that $$K_{sigma(1)} = <K_{sigma(1)} cap K_{sigma(2)},K_{sigma(1)} cap K_{sigma(3)}>,$$
i.e every subgroup is generated by its intersection with the two others.
Then for every three cosets $g_1K_1, g_2K_2, g_3K_3$, if every pair of cosets intersect then all three cosets intersect. That is, if $g_iK_i cap g_jK_j ne emptyset$ for all possible $i,j$, then $g_1K_1 cap g_2K_2 cap g_3 K_3 ne emptyset$.
I managed to reduce the question to showing the statement on $g_1=g_2=e$ - this is just because if $g_1K_1,g_2K_2$ intersect then the intersection is coset of $K_1 cap K_2$.
I also noticed that without the requirement that every subgroup is generated by the intersections, this statement is false - the counterexample is three (affine) lines in a plane that create a triangle. I don't see how to use the extra property though...
group-theory
group-theory
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
asked Dec 6 '18 at 8:48
Yotam DYotam D
18212
18212
add a comment |
add a comment |
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