Structure Structure Klein FOUR - GROUP of Klein four-group
Klein four yuan group $ K $ that satisfies the following multiplication table group
Klein four-group is the group satisfying the following multpilicative table
$$\begin{array}{|c|c|c|c|c|}\hline \cdot & 1 & a & b & c \\ \hline 1 & 1& a & b & c \\ \hline a & a & 1 & c & b \\ \hline b & b & c & 1 & a \\ \hline c & c & b & a & 1 \\ \hline \end{array}$$
Obviously, $ K \ cong \ mathbb {Z} / 2 \ times \ mathbb {Z} / 2 $, which is the smallest non-cyclic group.
Clearly, $K\cong \mathbb{Z}/2\times \mathbb{Z}/2$, which is the smallest non-cyclic group.
We know the following three facts
We know the following three facts
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Undoubtedly, $ K $ automorphism group is a third symmetry group $ \ mathfrak {S} _3 $.
There is no doubt that the automorphism group is the 3rd symmatric group $\mathfrak{S}_3$.
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TABLE FOUR - GROUP Klein characterized in that
The character table of Klein four group is
$$\begin{array}{|c|c|c|c|c|}\hline & 1 & a & b & c \\ \hline 1 & 1 & 1 & 1 & 1 \\ \hline \chi_a & 1 & 1 & -1 & -1 \\ \hline \chi_b & 1 & -1 & 1 & -1 \\ \hline \chi_c &1 & -1 & -1 & 1\\ \hline \end{array}$$
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Faithful representation is replaced with the minimum $ \ {1, (12) (34), (13) (24), (14) (23) \} \ subseteq \ mathfrak {S} _4 $.
The minimal faithful permutation representaion is $\{1, (12)(34), (13)(24), (14)(23)\}\subseteq \mathfrak{S}_4$.
Jieke explained by the above-described by the following. Consider a three-dimensional space in a $ (1,1,1), (1, -1, -1), (-1,1, -1), (- 1, - 1, 1) vertex $ tetrahedral , where $ a, b, c $ role above are around $ x, y, z $ axis $ 180 ^ \ circ $. this actually gives all of the regular tetrahedron reflection automorphism. Replacement is above the surface of the faithful representation.
All of above can be summarized as follow. Consider the tetrahedron in 3D space whose vertices are $(1,1,1), (1,-1,-1), (-1,1,-1),(-1,-,1,1)$, where the action of $a$, $b$ and $c$ is the rotation of $180^\circ$ along axis $x$, $y$ and $z$ respectively. These actually give rise to all of the automorphisms of reflection of the tetrahedron. The permutation of faces is exactly the faithful representation above.
The first application First Application
As shown in this blog post mentioned, there are red, yellow, blue sheep, 15, 19, 20, if two different colors of Mianyang collide, they fit into a sheep another color. Ask what the final color of the sheep will all become?
There are 15, 19 and 20 sheep of color red, yellow and blue respectively. If two sheep of distinct colors collide, then they will blend into one sheep in the other color. What kind of species will unify the farm at the end?
We can give a dyeing method, to three colors of sheep infected with Klein four yuan group in $ a, b, c $, then collided, sheep dyed product of all the elements is constant.
We can give a method through coloring---color three different kind of sheep with $a$, $b$ and $c$ in Klein four-group respectively. Then when collision happens, the product of elements colored is invariant.
And $ 15a + 19b + 20c = a + b = c $, so the final unified whole yellow sheep farm.
Note that 15a+19b+20c=a+b=c$, so the yellow sheep will unifty the whole farm.
The second application Second Application
Consider a regular octahedron, we installed a lamp on all vertices, a switch mounted on each side. Per switch, the lamp status of the three vertices of a plane will change - bright light is extinguished, the lamp is lit off. Now if only one light is on, we can do all the lights are switched off?
Consider a octahedron. We equip each vertice a bulb and each face a switch. Each usage of a switch, the three vertices of this face will change its state---the light one gets extincted, the light-off one turns off. Now there is only one bulb open, can we turn off all the bulbs?
Also, we present a method for dyeing. The three vertices are contracted diameter Klein group of four yuan $ a, b, c $, then every time the operation switch, the product of all the elements dyed bright light is constant.
Similarly, we shall present a coloring solution. Coloring the three pairs of antipodal vertices with $a$, $b$ and $c$ in Klein four-group respectively. Then no matter how the switches are operated, the product of elements of light bulbs is invariant.
Now, only a bright light, of course, not all become off state.
Now, there is only one bulb on, it cannot become the state with all distinct.
The third application Third Application
The third application is a classic, we show that the European version of the classic game of Peg Solitaire (also known as independent diamond) is no solution.
This application is classic. We will show that the game Peg Solitaire is unsolvable for European version.
Klein board strip to four yuan infected population of $ a, b, c $, each transition does not change the product of all the sub-elements.
Coloring the board with $a$, $b$ and $c$ in Klein four-group in striped pattern. Each skip does change the product of all elements.
| a | b | c | ||||
| a | b | c | a | b | ||
| a | b | c | a | b | c | a |
| b | c | a | b | c | a | b |
| c | a | b | c | a | b | c |
| b | c | a | b | c | ||
| a | b | c |
Now, the product of all the elements that $ 1 $, indicating the end, only one child of absence.
Now, the product is $1$. This shows that there does not exist the final state of only one chess.