example of a non-lattice homomorphism
ExampleOfCompleteLatticeHomomorphism
2007-04-24 11:13:43
2007-04-24 12:03:37
Example
lattice homomorphism
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Consider the Hasse diagram of the lattice of subgroups of the quaternion group
of order $8$, $Q_8$. [The use of $Q_8$ is only for a concrete realization of the lattice.]
\[
\begin{xy}<5mm,0mm>:<0mm,5mm>::
(0,3) +*{Q_8} = "Q8";
(-2,2) +*{\langle i\rangle} = "i";
(0,2) +*{\langle j\rangle} = "j";
(2,2) +*{\langle k\rangle} = "k";
(0,1) +*{\langle -1\rangle} = "-1";
(0,0) +*{\langle 1\rangle} = "1";
"1"; "-1" **@{-};
"-1"; "i" **@{-};
"-1"; "j" **@{-};
"-1"; "k" **@{-};
"i"; "Q8" **@{-};
"j"; "Q8" **@{-};
"k"; "Q8" **@{-};
\end{xy}
\]
To establish an order-preserving map which is not a lattice isomorphism
one can simply ``skip'' $\langle -1\rangle$, which we display graphically as:
\[
\begin{xy}<5mm,0mm>:<0mm,5mm>::
(-3,3) +*{Q_8} = "Q81";
(-5,2) +*{\langle i\rangle} = "i1";
(-3,2) +*{\langle j\rangle} = "j1";
(-1,2) +*{\langle k\rangle} = "k1";
(-3,1) +*{\langle -1\rangle} = "-11";
(-3,0) +*{\langle 1\rangle} = "11";
(3,2.5) +*{Q_8} = "Q82";
(1,1.5) +*{\langle i\rangle} = "i2";
(3,1.5) +*{\langle j\rangle} = "j2";
(5,1.5) +*{\langle k\rangle} = "k2";
(3,0.5) +*{\langle -1\rangle} = "-12";
(3,-0.5) +*{\langle 1\rangle} = "12";
"11"; "-11" **@{-};
"-11"; "i1" **@{-};
"-11"; "j1" **@{-};
"-11"; "k1" **@{-};
"i1"; "Q81" **@{-};
"j1"; "Q81" **@{-};
"k1"; "Q81" **@{-};
"12"; "-12" **@{-};
"-12"; "i2" **@{-};
"-12"; "j2" **@{-};
"-12"; "k2" **@{-};
"i2"; "Q82" **@{-};
"j2"; "Q82" **@{-};
"k2"; "Q82" **@{-};
"Q81"; "Q82" **@{..};
"i1"; "i2" **@{..};
"j1"; "j2" **@{..};
"k1"; "k2" **@{..};
"-11"; "12" **@{..};
"11"; "12" **@{..};
\end{xy}
\]
Since containment is still preserved the map is order-preserving. However, the intersection (meet) of
$\langle i\rangle$ and $\langle j\rangle$, which is $\langle -1\rangle$, is not perserved under this
map. Thus it is not a lattice homomorphism.