证明4阶二面体群与4元群V同构及6阶二面体群与S₃同构

Alright, let's break down these two group isomorphism proofs step by step—they're classic foundational examples that help build intuition for how group isomorphisms work, so I'll make sure every step is explicit and easy to follow.

First, let's clarify the definitions of both groups to eliminate any ambiguity:

• 4-order dihedral group (( D_2 )): This is the symmetry group of a line segment (or a "2-sided polygon"). It has 4 elements: the identity operation ( e ), a 180° rotation ( r ), a reflection ( s ) over the segment's midpoint, and the composition ( rs ) (rotation followed by reflection). The group relations are: ( r^2 = e ), ( s^2 = e ), and ( rs = sr ) (since rotating 180° then reflecting is the same as reflecting then rotating 180°).
• Klein four-group (( V )): This is a 4-element abelian group, usually written as ( V = { e, a, b, c } ) with relations: ( a^2 = b^2 = c^2 = e ), and ( ab = c ), ( bc = a ), ( ca = b ) (every pair of non-identity elements multiplies to the third non-identity element).

Constructing the isomorphism map #

Define a function ( \phi: D_2 \to V ) as follows:

• ( \phi(e) = e )
• ( \phi(r) = a )
• ( \phi(s) = b )
• ( \phi(rs) = c )

Verifying ( \phi ) is an isomorphism #

We need to show two things: ( \phi ) is a bijection, and ( \phi ) preserves group operations.

1. Bijection: Since both groups have exactly 4 elements, we only need to confirm ( \phi ) is injective (no two distinct elements map to the same element). Looking at the definition, each element of ( D_2 ) maps to a unique element of ( V ), so ( \phi ) is injective. For finite groups, injective implies bijective.

2. Preserving operations: We check all key products (since the group is generated by ( r ) and ( s ), verifying their relations covers all possible products):

   • ( \phi(r \cdot r) = \phi(e) = e = a \cdot a = \phi(r) \cdot \phi(r) )
   • ( \phi(s \cdot s) = \phi(e) = e = b \cdot b = \phi(s) \cdot \phi(s) )
   • ( \phi(r \cdot s) = \phi(rs) = c = a \cdot b = \phi(r) \cdot \phi(s) )
   • ( \phi(s \cdot r) = \phi(rs) = c = b \cdot a = \phi(s) \cdot \phi(r) ) (since ( rs = sr ) in ( D_2 ) and ( ab = ba ) in ( V ))
   • ( \phi(r \cdot rs) = \phi(r^2 s) = \phi(s) = b = a \cdot c = \phi(r) \cdot \phi(rs) ) (since ( a \cdot c = a \cdot ab = b ))

All products are preserved, so ( \phi ) is a group isomorphism. Thus, ( D_2 \cong V ).

Again, start with clear definitions:

• 6-order dihedral group (( D_3 )): This is the symmetry group of an equilateral triangle. It has 6 elements: 3 rotations (identity ( e ), 120° rotation ( r ), 240° rotation ( r^2 )) and 3 reflections (( s ), ( rs ), ( r^2 s )). The key group relations are: ( r^3 = e ), ( s^2 = e ), and ( srs = r^{-1} = r^2 ) (reflecting, rotating, then reflecting is equivalent to rotating in the opposite direction).
• Symmetric group ( S_3 ): This is the group of all permutations of 3 elements (labeled 1, 2, 3). It has 6 elements: the identity permutation ( e = (1)(2)(3) ), three transpositions (2-cycles) ( (1\ 2) ), ( (1\ 3) ), ( (2\ 3) ), and two 3-cycles ( (1\ 2\ 3) ), ( (1\ 3\ 2) ).

Constructing the isomorphism map #

We map each symmetry operation in ( D_3 ) to the permutation it induces on the triangle's vertices (labeled 1, 2, 3 clockwise):

• ( \phi(e) = (1)(2)(3) ) (identity permutation)
• ( \phi(r) = (1\ 2\ 3) ) (120° rotation maps vertex 1→2, 2→3, 3→1)
• ( \phi(r^2) = (1\ 3\ 2) ) (240° rotation maps vertex 1→3, 3→2, 2→1)
• ( \phi(s) = (2\ 3) ) (reflection over the axis through vertex 1 swaps vertices 2 and 3)
• ( \phi(rs) = (1\ 2) ) (rotation ( r ) followed by reflection ( s ) swaps vertices 1 and 2)
• ( \phi(r^2 s) = (1\ 3) ) (rotation ( r^2 ) followed by reflection ( s ) swaps vertices 1 and 3)

Verifying ( \phi ) is an isomorphism #

1. Bijection: Both groups have exactly 6 elements. Every element of ( S_3 ) is covered by ( \phi ) (we mapped to all 3-cycles, transpositions, and the identity), so ( \phi ) is surjective. For finite groups, surjective implies bijective.

2. Preserving operations: Instead of checking all 36 possible products, we can verify that the generating relations of ( D_3 ) are preserved (since all group elements are combinations of ( r ) and ( s )):

   • ( \phi(r^3) = \phi(e) = (1)(2)(3) = (1\ 2\ 3)^3 = \phi(r)^3 ) (applying a 120° rotation three times gives the identity)
   • ( \phi(s^2) = \phi(e) = (1)(2)(3) = (2\ 3)^2 = \phi(s)^2 ) (reflecting twice gives the identity)
   • ( \phi(srs) = \phi(s)\phi(r)\phi(s) = (2\ 3)(1\ 2\ 3)(2\ 3) ). Calculating this permutation: first apply ( (1\ 2\ 3) ), then ( (2\ 3) ), which gives ( (1\ 3\ 2) = \phi(r^2) = \phi(r^{-1}) ), matching the relation ( srs = r^{-1} ).

We can also spot-check a product to be thorough: ( \phi(r \cdot s) = \phi(rs) = (1\ 2) ), and ( \phi(r)\phi(s) = (1\ 2\ 3)(2\ 3) = (1\ 2) ), which are equal. Another check: ( \phi(s \cdot r) = \phi(r^2 s) = (1\ 3) ), and ( \phi(s)\phi(r) = (2\ 3)(1\ 2\ 3) = (1\ 3) ), which matches the relation ( sr = r^2 s ).

All relations hold, so ( \phi ) is a group isomorphism. Thus, ( D_3 \cong S_3 ).

内容的提问来源于stack exchange，提问作者WoahImTired