Isomorphisms. Sasha Patotski. Cornell University November 23, 2015

Transcription

1 Isomorphisms Sasha Patotski Cornell University November 23, 2015 Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

2 Last time Definition If G, H are groups, a map ϕ: G H is called a homomorphism if ϕ(a b) = ϕ(a) ϕ(b) For a homomorphism ϕ, necessarily ϕ(e G ) = e H and ϕ(a 1 ) = ϕ(a) 1. Definition The kernel of a homomorphism ϕ is ker ϕ = {a G ϕ(a) = e} The image of a homomorphism ϕ is im ϕ = {ϕ(a) H a G} ker ϕ and im ϕ are subgroups of G and H, respectively. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

3 Between any groups G, H there is a trivial homomorphism ϕ: G H, given by ϕ(g) = e H, for all g G. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

4 Between any groups G, H there is a trivial homomorphism ϕ: G H, given by ϕ(g) = e H, for all g G. The map n n( mod m) defines a homomorphism Z Z/m. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

5 Between any groups G, H there is a trivial homomorphism ϕ: G H, given by ϕ(g) = e H, for all g G. The map n n( mod m) defines a homomorphism Z Z/m. There are no nontrivial homomorphisms Z/m Z. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

6 Between any groups G, H there is a trivial homomorphism ϕ: G H, given by ϕ(g) = e H, for all g G. The map n n( mod m) defines a homomorphism Z Z/m. There are no nontrivial homomorphisms Z/m Z. For a fixed m Z, the map ϕ m : Z Z given by ϕ m (n) = nm defines a homomorphism. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

7 Between any groups G, H there is a trivial homomorphism ϕ: G H, given by ϕ(g) = e H, for all g G. The map n n( mod m) defines a homomorphism Z Z/m. There are no nontrivial homomorphisms Z/m Z. For a fixed m Z, the map ϕ m : Z Z given by ϕ m (n) = nm defines a homomorphism. For any abelian group G, the map ϕ m : G G given by g g m is a homomorphism. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

8 Between any groups G, H there is a trivial homomorphism ϕ: G H, given by ϕ(g) = e H, for all g G. The map n n( mod m) defines a homomorphism Z Z/m. There are no nontrivial homomorphisms Z/m Z. For a fixed m Z, the map ϕ m : Z Z given by ϕ m (n) = nm defines a homomorphism. For any abelian group G, the map ϕ m : G G given by g g m is a homomorphism. The same map for non-abelian group is not necessarily a homomorphism (can you give an example?). Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

9 The group of symmetries of an equilateral triangle is isomorphic to S 3. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

10 The group of symmetries of an equilateral triangle is isomorphic to S 3. Prove that the alternating group A 3 is isomorphic to Z/3. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

11 The group of symmetries of an equilateral triangle is isomorphic to S 3. Prove that the alternating group A 3 is isomorphic to Z/3. Prove that orientation preserving symmetries of the square form a subgroup of the group of all symmetries. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

12 The group of symmetries of an equilateral triangle is isomorphic to S 3. Prove that the alternating group A 3 is isomorphic to Z/3. Prove that orientation preserving symmetries of the square form a subgroup of the group of all symmetries. Prove that this subgroup is isomorphic to Z/4. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

13 The group of symmetries of an equilateral triangle is isomorphic to S 3. Prove that the alternating group A 3 is isomorphic to Z/3. Prove that orientation preserving symmetries of the square form a subgroup of the group of all symmetries. Prove that this subgroup is isomorphic to Z/4. Describe the group of orientation preserving symmetries of a regular n-gon. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

14 Tetrahedron Theorem The group G of symmetries of a regular tetrahedron is isomorphic to S 4. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

15 Tetrahedron Theorem The group G of symmetries of a regular tetrahedron is isomorphic to S 4. There is an obvious homomorphism ϕ: G S 4, sending a symmetry to corresponding permutation of vertices. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

16 Tetrahedron Theorem The group G of symmetries of a regular tetrahedron is isomorphic to S 4. There is an obvious homomorphism ϕ: G S 4, sending a symmetry to corresponding permutation of vertices. It is injective: if a symmetry fixes all vertices, it must be the identity symmetry. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

17 Tetrahedron Theorem The group G of symmetries of a regular tetrahedron is isomorphic to S 4. There is an obvious homomorphism ϕ: G S 4, sending a symmetry to corresponding permutation of vertices. It is injective: if a symmetry fixes all vertices, it must be the identity symmetry. Every transposition of neighbors is in the image im(ϕ). Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

18 Tetrahedron Theorem The group G of symmetries of a regular tetrahedron is isomorphic to S 4. There is an obvious homomorphism ϕ: G S 4, sending a symmetry to corresponding permutation of vertices. It is injective: if a symmetry fixes all vertices, it must be the identity symmetry. Every transposition of neighbors is in the image im(ϕ). But im(ϕ) is a subgroup. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5

19 Tetrahedron Theorem The group G of symmetries of a regular tetrahedron is isomorphic to S 4. There is an obvious homomorphism ϕ: G S 4, sending a symmetry to corresponding permutation of vertices. It is injective: if a symmetry fixes all vertices, it must be the identity symmetry. Every transposition of neighbors is in the image im(ϕ). But im(ϕ) is a subgroup. Since S 4 is generated by these transpositions, im(ϕ) = S 4. Done. Sasha Patotski (Cornell University) Isomorphisms November 23, / 5