GhaSShee


Automorphism


Terminology - Aut(G) : automorphism : G -> G ; isomorphism - Int(G) : inner automorphism : G -> G where G -> Int(G) ; homomorphism - Out(G) : outer automorphism := Aut(G) \ Int(G) # $Aut(A_4)$ $\sim$ $S_4$ ![](/image/tetrahedron.png width=360) imagine the rotations of a tetrahedron to itself. Each rotation moves a point to the position which another point was. pick up any 2 faces and name a, b as 120˚ anti-clockwise rotations of each. (so, the above picture is clockwise i.e. it is mirrored and do not mind that.) $ A _ 4 $ is described only with a, b and their conpositions as id = aaa = bbb. $ A _ 4 $ = { ~~~ id, aab, aba, baa, a, aa, b, bb, abab, ab, baba, ba, ~~~ } if we define `c := abab`, and `d := baba`, then; $ A _4 $ = { ~~~ id, aab, aba, baa, <= 180˚ rotations a, aa, b, bb, c, cc, d, dd ~~~ } where ~~~ a, b, c, d is 120˚ anti-clockwise aa, bb, cc, dd is 240˚anti-clockwise or 120˚ clockwise ~~~ A member of Aut( $ A _ 4 $ ) should be isomorphism i.e. homomorphic, so it maps `a` to the 120˚ rotations. There is ; ~~~ a |-> a,b,c,d,aa,bb,cc,dd (8 patterns) ~~~ if we choose `bb` as the target, ~~~ a |-> bb ~~~ then because of homomorphism, `aa` is determined like, ~~~ aa |-> bbbb == b ~~~ Next, determine the target of `b` Because this function `a` $ \mapsto $ `bb` is `anti-clockwise` $ \mapsto $ `clockwise` `b` satisfies; ~~~ b |-> clockwise ~~~ so, ~~~ b |-> aa,cc,dd (3 patterns) ~~~ so Aut( $ A _ 4 $ ) has 8 x 3 = 24 elements. but why is this isomorphic to $ S _ 4 $? ![](/image/tetra_in_cube.png width=450) # Outer Automorphisms $ S _ 4 $ is said to be $ G( T _ 6 ) $, as $ A _ 4 $ to $ G( T _ 4) $.
This means we can regard $ S _ 4 $ as the rotaions of a cube to its self. A cube has at least one tetrahedron inside of itself as the above picture.
It means it exactly includes $ A _ 4 $ visibly. Then go back to Aut( $ A _ 4 $ ) .
Aut ( $ A _ 4 $ ) include $ A _ 4 $ itself as Int( $ A _ 4 $ ).
Aut ( $ A _ 4 $ ) consists of 24 elements, and $ A _ 4 $ has 12.
So there is the other 12 elements, which we call Out( $ A _ 4 $ ). Out( $ A _ 4 $ ) is consists of members which is not $ A _ 4 $, e.g. a mirror projection can map $ A _ 4 $ to itself.
In the above picture `z` $ \in $ $ S _ 4 $ and we can produce a mirror projection from `z`; ` ι(z)(-) = z * - * z` $ {} ^ {-1} $ z is 180 rotation and ι(z) is a mirror projection about xy-plain! (Note the above picture is quite misleading. ι(z) is not xz-plain reflection. ) If you rotate 180 degree with z axix fixed, then some operation and again rotate 180 degree with z fixed. It generates exactly its mirrored operation! As the same way, this `ι(-)` maps other elements of $ S _ 4 \setminus A _ 4 $ to Out( $ A _ 4 $ ). And this function is an isomorphism between $ S _ 4 $ and Aut( $ A _ 4 $ ). # Dihedral groups $D_n$ are also interesting. There is interesting facts; ~~~ Int(D3) == Aut(D3) Int(D5) != Aut(D5) ~~~ assume 'a' is a (360˚/n) clockwise rotation and 'b' is a 180˚flipping rotation then consider Automorphism in the same way; ~~~ b |-> b, ba, ... , ba^{n-1} (n patterns) ~~~ then ~~~ a |-> ? ~~~ since every member of Int(D5) is described as a D5 member, the codomain is clockwise or anti-clockwise (not a 180˚flipping rotation because of homomorphism). ~~~ Int(D5) 72˚ -> 72˚ (a |-> a) (anti-clockwise -> anti-clockwise), 72˚ -> -72˚ (a |-> aaaa) (anti-clockwise -> clockwise) Aut(D5) 72˚ -> 72˚, 72˚ -> 144˚, (a |-> aa) (anti-clockwise -> butterfly-shape) 72˚ -> -144˚, (a |-> aaa) (anti-clockwise -> (anti)butterfly-shape) 72˚ -> -72˚ ~~~ so, ~~~ |Int(D5)| = |Aut(D5)| / 2 ~~~ why not compare D3, D5, D7 ? ~~~ 1 * |Int(D3)| = |Aut(D3)| 2 * |Int(D5)| = |Aut(D5)| 3 * |Int(D7)| = |Aut(D7)| ~~~ more generally, ~~~ Aut(Dn) = n * φ(n) where φ(n) denotes Euler's number ~~~ # Galois The difference between $ D3 $ and $ D5 $ is
extended to $ A _ 5 $ $ \sim $ $ G( T _ {20} ) $ $ \sim $ $ G( T _ {12} ) $ As $ Aut( D _ 5 ) $ contains Butterfly-shape,
$ Aut( A _ 5 ) $ contains more complex-shape that is not included in A5 itself. So the structure of $ Aut( A _ 5 ) $ cause that we cannot solve algebraic quintic equation. # Haskell code ~~~ Haskell -- Type Definition data Mat a = Mat [[a]] -- Show Intance of Matrix showRow [] = "" showRow (x:xs) = show x ++ "\t\b" ++ showRow xs instance (Show a) => Show (Mat a) where show (Mat []) = "" show (Mat (x:xs)) = showRow x ++ "\n" ++ show (Mat xs) -- Define Lengths len_row x = maximum $ map length x len_col x = length x -- Define Operators instance (Num a) => Num (Mat a) where (Mat a) + (Mat b) = let m = max(len_row a)(len_row b) in let n = max(len_col a)(len_col b) in Mat [ [ (a!!i!!j) + (b!!i!!j) | j <- [0..(n-1)]] | i <- [0..(m-1)] ] (Mat a) - (Mat b) = let m = max(len_row a)(len_row b) in let n = max(len_col a)(len_col b) in Mat [ [ (a!!i!!j) - (b!!i!!j) | j <- [0..(n-1)]] | i <- [0..(m-1)] ] (Mat a) * (Mat b) = let l = len_row a - 1 in let m = len_col a - 1 in let n = len_col b - 1 in Mat [ [ sum [((a!!i!!k)*(b!!k!!j)) | k<-[0..m]] | j<-[0..n] ] | i<-[0..l] ] -------------------- an expample usage -------------------- -- Now we think of -- * Symmetry Group S_4 -- * Alternation Group A_4 -- A_4 is generated just by a & b -- S_4 is generated just by a, b & z -- the Matrix is like -- [0 0 1] -- [y z x] = [x y z][1 0 0] -- [0 1 0] -- here x,y or z axis implies each axis of 3 180˚rotations ----------------------------------------------------------- e = aaa a = Mat [ [ 0, 0, 1], [ 1, 0, 0], [ 0, 1, 0]] b = Mat [ [ 0, 0, 1], [-1, 0, 0], [ 0,-1, 0]] c = b*b*a*a d = a*a*b*b aa = a*a bb = b*b cc = c*c dd = d*d h = b*a*a i = a*b*a j = a*a*b z = Mat [ [ 0, 0,-1], [ 0,-1, 0], [-1, 0, 0]] za = z * a zb = z * b zc = z * c zd = z * d zaa = z * aa zbb = z * bb zcc = z * cc zdd = z * dd zh = z * h zi = z * i zj = z * j ~~~